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COMPLEX NUMBERS AND

QUADRATIC EQUATIONS

Introduction We know that the equation x2 + 1 = 0 has no real

solution as x2 + 1 = 0 gives x2 = ndash 1 and square of everyreal number is non-negative So we need to extend thereal number system to a larger system so that we can

find the solution of the equation x2 = ndash 1 In fact the mainobjective is to solve the equation ax2 + bx + c = 0 whereD = b2 ndash 4ac lt 0 which is not possible in the system of

real numbers

Complex NumbersLet us denote minus1 by the symbol i Then we have i2 = minus1 This means

that i is asolution of the equation x2 + 1 = 0

A number of the form a + ib where a and b are real numbers is defined to be a

complex number For example 2 + i3 (ndash 1) + i 3

Is a complex numbersFor the complex number z = a + ib a is called the real part denoted by

Re z andb is called the imaginary part denoted by Im z of the complex number z

For exampleif z = 2 + i5 then Re z = 2 and Im z = 5

Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d

Algebra of Complex NumbersIn this Section we shall develop the algebra of complex numbers Addition of two complex numbers Let z1 = a + ib and z2 = c + id

be any twocomplex numbers Then the sum z1 + z2 is defined as followsz1 + z2 = (a + c) + i (b + d) which is again a complex numberFor example (2 + i3) + (ndash 6 +i5) = (2 ndash 6) + i (3 + 5) = ndash 4 + i 8

The addition of complex numbers satisfy the following properties(i) The closure law The sum of two complex numbers is a complexnumber ie z1 + z2 is a complex number for all complex numbers

z1 and z2(ii) The commutative law For any two complex numbers z1 and z2

z1 + z2 = z2+ z1

Difference of two complex numbers Given any two complex numbers z1 and

z2 the difference z1 ndash z2 is defined as followsz1 ndash z2 = z1 + (ndash z2)

For example (6 + 3i) ndash (2 ndash i) = (6 + 3i) + (ndash 2 + i ) = 4 + 4iand (2 ndash i) ndash (6 + 3i) = (2 ndash i) + ( ndash 6 ndash 3i) = ndash 4 ndash 4i

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99 Multiplication of two complex numbers Let z1 = a + ib and z2 = c

+ id be anytwo complex numbers Then the product z1 z2 is defined as

followsz1 z2 = (ac ndash bd) + i(ad + bc)

For example (3 + i5) (2 + i6) = (3 times 2 ndash 5 times 6) + i(3 times 6 + 5 times 2) = ndash 24 + i28

The multiplication of complex numbers possesses the following properties which

we state without proofs

(i) The closure law The product of two complex numbers is a complex number

the product z1 z2 is a complex number for all complex numbers z1 and z2

(ii) The commutative law For any two complex numbers z1 and z2z1 z2 = z2 z1

(iii) The associative law For any three complex numbers z1 z2 z3

(z1 z2) z3 = z1 (z2 z3)(iv) The existence of multiplicative identity There exists the complex

number1 + i 0 (denoted as 1) called the multiplicative identity such that z1 = z

for every complex number z(v) The existence of multiplicative inverse For every non-zero complex

number z = a + ib or a + bi(a ne 0 b ne 0) we have the complex number

The square roots of a negative real numberNote that i2 = ndash1 and ( ndash i)2 = i2 = ndash 1

Therefore the square roots of ndash 1 are i ndash i However by the symbol minus we wouldmean i only

Now we can see that i and ndashi both are the solutions of the equation x2 + 1 = 0 or

x2 = ndash1Similarly ( ) ( ) 2 2

3i = 3 i2 = 3 (ndash 1) = ndash 3( )2

minus 3i = ( )2minus 3 i2 = ndash 3

Therefore the square roots of ndash3 are 3 i and minus 3i Again the symbol minus3 is meant to represent 3 i only ie minus3 = 3 i

Generally if a is a positive real number minusa = a minus1 = a i

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

COMPLEX NUMBERS AND

QUADRATIC EQUATIONS

Introduction We know that the equation x2 + 1 = 0 has no real

solution as x2 + 1 = 0 gives x2 = ndash 1 and square of everyreal number is non-negative So we need to extend thereal number system to a larger system so that we can

find the solution of the equation x2 = ndash 1 In fact the mainobjective is to solve the equation ax2 + bx + c = 0 whereD = b2 ndash 4ac lt 0 which is not possible in the system of

real numbers

Complex NumbersLet us denote minus1 by the symbol i Then we have i2 = minus1 This means

that i is asolution of the equation x2 + 1 = 0

A number of the form a + ib where a and b are real numbers is defined to be a

complex number For example 2 + i3 (ndash 1) + i 3

Is a complex numbersFor the complex number z = a + ib a is called the real part denoted by

Re z andb is called the imaginary part denoted by Im z of the complex number z

For exampleif z = 2 + i5 then Re z = 2 and Im z = 5

Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d

Algebra of Complex NumbersIn this Section we shall develop the algebra of complex numbers Addition of two complex numbers Let z1 = a + ib and z2 = c + id

be any twocomplex numbers Then the sum z1 + z2 is defined as followsz1 + z2 = (a + c) + i (b + d) which is again a complex numberFor example (2 + i3) + (ndash 6 +i5) = (2 ndash 6) + i (3 + 5) = ndash 4 + i 8

The addition of complex numbers satisfy the following properties(i) The closure law The sum of two complex numbers is a complexnumber ie z1 + z2 is a complex number for all complex numbers

z1 and z2(ii) The commutative law For any two complex numbers z1 and z2

z1 + z2 = z2+ z1

Difference of two complex numbers Given any two complex numbers z1 and

z2 the difference z1 ndash z2 is defined as followsz1 ndash z2 = z1 + (ndash z2)

For example (6 + 3i) ndash (2 ndash i) = (6 + 3i) + (ndash 2 + i ) = 4 + 4iand (2 ndash i) ndash (6 + 3i) = (2 ndash i) + ( ndash 6 ndash 3i) = ndash 4 ndash 4i

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99 Multiplication of two complex numbers Let z1 = a + ib and z2 = c

+ id be anytwo complex numbers Then the product z1 z2 is defined as

followsz1 z2 = (ac ndash bd) + i(ad + bc)

For example (3 + i5) (2 + i6) = (3 times 2 ndash 5 times 6) + i(3 times 6 + 5 times 2) = ndash 24 + i28

The multiplication of complex numbers possesses the following properties which

we state without proofs

(i) The closure law The product of two complex numbers is a complex number

the product z1 z2 is a complex number for all complex numbers z1 and z2

(ii) The commutative law For any two complex numbers z1 and z2z1 z2 = z2 z1

(iii) The associative law For any three complex numbers z1 z2 z3

(z1 z2) z3 = z1 (z2 z3)(iv) The existence of multiplicative identity There exists the complex

number1 + i 0 (denoted as 1) called the multiplicative identity such that z1 = z

for every complex number z(v) The existence of multiplicative inverse For every non-zero complex

number z = a + ib or a + bi(a ne 0 b ne 0) we have the complex number

The square roots of a negative real numberNote that i2 = ndash1 and ( ndash i)2 = i2 = ndash 1

Therefore the square roots of ndash 1 are i ndash i However by the symbol minus we wouldmean i only

Now we can see that i and ndashi both are the solutions of the equation x2 + 1 = 0 or

x2 = ndash1Similarly ( ) ( ) 2 2

3i = 3 i2 = 3 (ndash 1) = ndash 3( )2

minus 3i = ( )2minus 3 i2 = ndash 3

Therefore the square roots of ndash3 are 3 i and minus 3i Again the symbol minus3 is meant to represent 3 i only ie minus3 = 3 i

Generally if a is a positive real number minusa = a minus1 = a i

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Introduction We know that the equation x2 + 1 = 0 has no real

solution as x2 + 1 = 0 gives x2 = ndash 1 and square of everyreal number is non-negative So we need to extend thereal number system to a larger system so that we can

find the solution of the equation x2 = ndash 1 In fact the mainobjective is to solve the equation ax2 + bx + c = 0 whereD = b2 ndash 4ac lt 0 which is not possible in the system of

real numbers

Complex NumbersLet us denote minus1 by the symbol i Then we have i2 = minus1 This means

that i is asolution of the equation x2 + 1 = 0

A number of the form a + ib where a and b are real numbers is defined to be a

complex number For example 2 + i3 (ndash 1) + i 3

Is a complex numbersFor the complex number z = a + ib a is called the real part denoted by

Re z andb is called the imaginary part denoted by Im z of the complex number z

For exampleif z = 2 + i5 then Re z = 2 and Im z = 5

Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d

Algebra of Complex NumbersIn this Section we shall develop the algebra of complex numbers Addition of two complex numbers Let z1 = a + ib and z2 = c + id

be any twocomplex numbers Then the sum z1 + z2 is defined as followsz1 + z2 = (a + c) + i (b + d) which is again a complex numberFor example (2 + i3) + (ndash 6 +i5) = (2 ndash 6) + i (3 + 5) = ndash 4 + i 8

The addition of complex numbers satisfy the following properties(i) The closure law The sum of two complex numbers is a complexnumber ie z1 + z2 is a complex number for all complex numbers

z1 and z2(ii) The commutative law For any two complex numbers z1 and z2

z1 + z2 = z2+ z1

Difference of two complex numbers Given any two complex numbers z1 and

z2 the difference z1 ndash z2 is defined as followsz1 ndash z2 = z1 + (ndash z2)

For example (6 + 3i) ndash (2 ndash i) = (6 + 3i) + (ndash 2 + i ) = 4 + 4iand (2 ndash i) ndash (6 + 3i) = (2 ndash i) + ( ndash 6 ndash 3i) = ndash 4 ndash 4i

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99 Multiplication of two complex numbers Let z1 = a + ib and z2 = c

+ id be anytwo complex numbers Then the product z1 z2 is defined as

followsz1 z2 = (ac ndash bd) + i(ad + bc)

For example (3 + i5) (2 + i6) = (3 times 2 ndash 5 times 6) + i(3 times 6 + 5 times 2) = ndash 24 + i28

The multiplication of complex numbers possesses the following properties which

we state without proofs

(i) The closure law The product of two complex numbers is a complex number

the product z1 z2 is a complex number for all complex numbers z1 and z2

(ii) The commutative law For any two complex numbers z1 and z2z1 z2 = z2 z1

(iii) The associative law For any three complex numbers z1 z2 z3

(z1 z2) z3 = z1 (z2 z3)(iv) The existence of multiplicative identity There exists the complex

number1 + i 0 (denoted as 1) called the multiplicative identity such that z1 = z

for every complex number z(v) The existence of multiplicative inverse For every non-zero complex

number z = a + ib or a + bi(a ne 0 b ne 0) we have the complex number

The square roots of a negative real numberNote that i2 = ndash1 and ( ndash i)2 = i2 = ndash 1

Therefore the square roots of ndash 1 are i ndash i However by the symbol minus we wouldmean i only

Now we can see that i and ndashi both are the solutions of the equation x2 + 1 = 0 or

x2 = ndash1Similarly ( ) ( ) 2 2

3i = 3 i2 = 3 (ndash 1) = ndash 3( )2

minus 3i = ( )2minus 3 i2 = ndash 3

Therefore the square roots of ndash3 are 3 i and minus 3i Again the symbol minus3 is meant to represent 3 i only ie minus3 = 3 i

Generally if a is a positive real number minusa = a minus1 = a i

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Complex NumbersLet us denote minus1 by the symbol i Then we have i2 = minus1 This means

that i is asolution of the equation x2 + 1 = 0

A number of the form a + ib where a and b are real numbers is defined to be a

complex number For example 2 + i3 (ndash 1) + i 3

Is a complex numbersFor the complex number z = a + ib a is called the real part denoted by

Re z andb is called the imaginary part denoted by Im z of the complex number z

For exampleif z = 2 + i5 then Re z = 2 and Im z = 5

Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d

Algebra of Complex NumbersIn this Section we shall develop the algebra of complex numbers Addition of two complex numbers Let z1 = a + ib and z2 = c + id

be any twocomplex numbers Then the sum z1 + z2 is defined as followsz1 + z2 = (a + c) + i (b + d) which is again a complex numberFor example (2 + i3) + (ndash 6 +i5) = (2 ndash 6) + i (3 + 5) = ndash 4 + i 8

The addition of complex numbers satisfy the following properties(i) The closure law The sum of two complex numbers is a complexnumber ie z1 + z2 is a complex number for all complex numbers

z1 and z2(ii) The commutative law For any two complex numbers z1 and z2

z1 + z2 = z2+ z1

Difference of two complex numbers Given any two complex numbers z1 and

z2 the difference z1 ndash z2 is defined as followsz1 ndash z2 = z1 + (ndash z2)

For example (6 + 3i) ndash (2 ndash i) = (6 + 3i) + (ndash 2 + i ) = 4 + 4iand (2 ndash i) ndash (6 + 3i) = (2 ndash i) + ( ndash 6 ndash 3i) = ndash 4 ndash 4i

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99 Multiplication of two complex numbers Let z1 = a + ib and z2 = c

+ id be anytwo complex numbers Then the product z1 z2 is defined as

followsz1 z2 = (ac ndash bd) + i(ad + bc)

For example (3 + i5) (2 + i6) = (3 times 2 ndash 5 times 6) + i(3 times 6 + 5 times 2) = ndash 24 + i28

The multiplication of complex numbers possesses the following properties which

we state without proofs

(i) The closure law The product of two complex numbers is a complex number

the product z1 z2 is a complex number for all complex numbers z1 and z2

(ii) The commutative law For any two complex numbers z1 and z2z1 z2 = z2 z1

(iii) The associative law For any three complex numbers z1 z2 z3

(z1 z2) z3 = z1 (z2 z3)(iv) The existence of multiplicative identity There exists the complex

number1 + i 0 (denoted as 1) called the multiplicative identity such that z1 = z

for every complex number z(v) The existence of multiplicative inverse For every non-zero complex

number z = a + ib or a + bi(a ne 0 b ne 0) we have the complex number

The square roots of a negative real numberNote that i2 = ndash1 and ( ndash i)2 = i2 = ndash 1

Therefore the square roots of ndash 1 are i ndash i However by the symbol minus we wouldmean i only

Now we can see that i and ndashi both are the solutions of the equation x2 + 1 = 0 or

x2 = ndash1Similarly ( ) ( ) 2 2

3i = 3 i2 = 3 (ndash 1) = ndash 3( )2

minus 3i = ( )2minus 3 i2 = ndash 3

Therefore the square roots of ndash3 are 3 i and minus 3i Again the symbol minus3 is meant to represent 3 i only ie minus3 = 3 i

Generally if a is a positive real number minusa = a minus1 = a i

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Algebra of Complex NumbersIn this Section we shall develop the algebra of complex numbers Addition of two complex numbers Let z1 = a + ib and z2 = c + id

be any twocomplex numbers Then the sum z1 + z2 is defined as followsz1 + z2 = (a + c) + i (b + d) which is again a complex numberFor example (2 + i3) + (ndash 6 +i5) = (2 ndash 6) + i (3 + 5) = ndash 4 + i 8

The addition of complex numbers satisfy the following properties(i) The closure law The sum of two complex numbers is a complexnumber ie z1 + z2 is a complex number for all complex numbers

z1 and z2(ii) The commutative law For any two complex numbers z1 and z2

z1 + z2 = z2+ z1

Difference of two complex numbers Given any two complex numbers z1 and

z2 the difference z1 ndash z2 is defined as followsz1 ndash z2 = z1 + (ndash z2)

For example (6 + 3i) ndash (2 ndash i) = (6 + 3i) + (ndash 2 + i ) = 4 + 4iand (2 ndash i) ndash (6 + 3i) = (2 ndash i) + ( ndash 6 ndash 3i) = ndash 4 ndash 4i

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99 Multiplication of two complex numbers Let z1 = a + ib and z2 = c

+ id be anytwo complex numbers Then the product z1 z2 is defined as

followsz1 z2 = (ac ndash bd) + i(ad + bc)

For example (3 + i5) (2 + i6) = (3 times 2 ndash 5 times 6) + i(3 times 6 + 5 times 2) = ndash 24 + i28

The multiplication of complex numbers possesses the following properties which

we state without proofs

(i) The closure law The product of two complex numbers is a complex number

the product z1 z2 is a complex number for all complex numbers z1 and z2

(ii) The commutative law For any two complex numbers z1 and z2z1 z2 = z2 z1

(iii) The associative law For any three complex numbers z1 z2 z3

(z1 z2) z3 = z1 (z2 z3)(iv) The existence of multiplicative identity There exists the complex

number1 + i 0 (denoted as 1) called the multiplicative identity such that z1 = z

for every complex number z(v) The existence of multiplicative inverse For every non-zero complex

number z = a + ib or a + bi(a ne 0 b ne 0) we have the complex number

The square roots of a negative real numberNote that i2 = ndash1 and ( ndash i)2 = i2 = ndash 1

Therefore the square roots of ndash 1 are i ndash i However by the symbol minus we wouldmean i only

Now we can see that i and ndashi both are the solutions of the equation x2 + 1 = 0 or

x2 = ndash1Similarly ( ) ( ) 2 2

3i = 3 i2 = 3 (ndash 1) = ndash 3( )2

minus 3i = ( )2minus 3 i2 = ndash 3

Therefore the square roots of ndash3 are 3 i and minus 3i Again the symbol minus3 is meant to represent 3 i only ie minus3 = 3 i

Generally if a is a positive real number minusa = a minus1 = a i

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Difference of two complex numbers Given any two complex numbers z1 and

z2 the difference z1 ndash z2 is defined as followsz1 ndash z2 = z1 + (ndash z2)

For example (6 + 3i) ndash (2 ndash i) = (6 + 3i) + (ndash 2 + i ) = 4 + 4iand (2 ndash i) ndash (6 + 3i) = (2 ndash i) + ( ndash 6 ndash 3i) = ndash 4 ndash 4i

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 99 Multiplication of two complex numbers Let z1 = a + ib and z2 = c

+ id be anytwo complex numbers Then the product z1 z2 is defined as

followsz1 z2 = (ac ndash bd) + i(ad + bc)

For example (3 + i5) (2 + i6) = (3 times 2 ndash 5 times 6) + i(3 times 6 + 5 times 2) = ndash 24 + i28

The multiplication of complex numbers possesses the following properties which

we state without proofs

(i) The closure law The product of two complex numbers is a complex number

the product z1 z2 is a complex number for all complex numbers z1 and z2

(ii) The commutative law For any two complex numbers z1 and z2z1 z2 = z2 z1

(iii) The associative law For any three complex numbers z1 z2 z3

(z1 z2) z3 = z1 (z2 z3)(iv) The existence of multiplicative identity There exists the complex

number1 + i 0 (denoted as 1) called the multiplicative identity such that z1 = z

for every complex number z(v) The existence of multiplicative inverse For every non-zero complex

number z = a + ib or a + bi(a ne 0 b ne 0) we have the complex number

The square roots of a negative real numberNote that i2 = ndash1 and ( ndash i)2 = i2 = ndash 1

Therefore the square roots of ndash 1 are i ndash i However by the symbol minus we wouldmean i only

Now we can see that i and ndashi both are the solutions of the equation x2 + 1 = 0 or

x2 = ndash1Similarly ( ) ( ) 2 2

3i = 3 i2 = 3 (ndash 1) = ndash 3( )2

minus 3i = ( )2minus 3 i2 = ndash 3

Therefore the square roots of ndash3 are 3 i and minus 3i Again the symbol minus3 is meant to represent 3 i only ie minus3 = 3 i

Generally if a is a positive real number minusa = a minus1 = a i

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

(i) The closure law The product of two complex numbers is a complex number

the product z1 z2 is a complex number for all complex numbers z1 and z2

(ii) The commutative law For any two complex numbers z1 and z2z1 z2 = z2 z1

(iii) The associative law For any three complex numbers z1 z2 z3

(z1 z2) z3 = z1 (z2 z3)(iv) The existence of multiplicative identity There exists the complex

number1 + i 0 (denoted as 1) called the multiplicative identity such that z1 = z

for every complex number z(v) The existence of multiplicative inverse For every non-zero complex

number z = a + ib or a + bi(a ne 0 b ne 0) we have the complex number

The square roots of a negative real numberNote that i2 = ndash1 and ( ndash i)2 = i2 = ndash 1

Therefore the square roots of ndash 1 are i ndash i However by the symbol minus we wouldmean i only

Now we can see that i and ndashi both are the solutions of the equation x2 + 1 = 0 or

x2 = ndash1Similarly ( ) ( ) 2 2

3i = 3 i2 = 3 (ndash 1) = ndash 3( )2

minus 3i = ( )2minus 3 i2 = ndash 3

Therefore the square roots of ndash3 are 3 i and minus 3i Again the symbol minus3 is meant to represent 3 i only ie minus3 = 3 i

Generally if a is a positive real number minusa = a minus1 = a i

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

The square roots of a negative real numberNote that i2 = ndash1 and ( ndash i)2 = i2 = ndash 1

Therefore the square roots of ndash 1 are i ndash i However by the symbol minus we wouldmean i only

Now we can see that i and ndashi both are the solutions of the equation x2 + 1 = 0 or

x2 = ndash1Similarly ( ) ( ) 2 2

3i = 3 i2 = 3 (ndash 1) = ndash 3( )2

minus 3i = ( )2minus 3 i2 = ndash 3

Therefore the square roots of ndash3 are 3 i and minus 3i Again the symbol minus3 is meant to represent 3 i only ie minus3 = 3 i

Generally if a is a positive real number minusa = a minus1 = a i

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

We already know that atimes b = ab for all positive real number a and b This

result also holds true when either a gt 0 b lt 0 or a lt 0 b gt 0 What if a lt 0 b lt 0

Let us examineNote that

COMPLEX NUMBERS AND QUADRATIC EQUATIONS 101i2 = minus1 minus1= (minus1) (minus1) (by assuming atimes b = ab for all real numbers)

= 1 = 1 which is a contradiction to the fact that i = minus Therefore atimes bne ab if both a and b are negative real numbersFurther if any of a and b is zero then clearly atimes b= ab= 0

537 Identities We prove the following identity( )2 2 2

z1+z2 =z1+z2+2z1z2 for all complex numbers z1 and z2

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

The Modulus and the Conjugate of a Complex NumberLet z = a + ib be a complex number Then the modulus of z

denoted by | z | is definedto be the non-negative real number a2+b2 ie | z | = a2+b2

and the conjugateof z denoted as z is the complex number a ndash ib ie z = a ndash

ibFor example 3+i =32+12=10 2minus5i = 22+(minus5)2= 29

and 3+i=3minusi 2minus5i=2+5i minus3i minus5 = 3i ndash 5

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Summary1048720A number of the form a + ib where a and b are real numbers is

called acomplex number a is called the real part and b is called the

imaginary partof the complex number

1048720Let z1 = a + ib and z2 = c + id Then(i) z1 + z2 = (a + c) + i (b + d)

(ii) z1 z2 = (ac ndash bd) + i (ad + bc)1048720For any non-zero complex number z = a + ib (a ne 0 b ne 0) there

exists thecomplex number 2 2 2 2

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

1048720For any integer k i4k = 1 i4k + 1 = i i4k + 2 = ndash 1 i4k + 3 = ndash i1048720The conjugate of the complex number z = a + ib denoted by z is given by

z = a ndash ib1048720The polar form of the complex number z = x + iy is r (cosegrave + i sinegrave) where

r = x2+ y2 (the modulus of z) and cosegrave =x

r sinegrave =y

r (egrave is known as theargument of z The value of egrave such that ndash eth lt egrave le eth is called the principal

argument of z

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

SUMSbull Convert the given complex

number in polar form ndash3bull Answer bull Discussion bull ndash3bull Let r cos θ = ndash3 and r sin θ = 0bull On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Convert the given complex number in polar form ndash3Answer Discussion

ndash3Convert the given complex number in polar form ndash3Answer

Discussion ndash3

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Let r cos θ = ndash3 and r sin θ = 0On squaring and adding we obtain

Historical NoteThe fact that square root of a negative

number does not exist in the real numbersystem was recognised by the Greeks But

the credit goes to the Indianmathematician Mahavira (850 AD) who first

stated this difficulty clearly ldquoHementions in his work lsquoGanitasara Sangraharsquo

as in the nature of things a negative(quantity) is not a square (quantity)rsquo it has

therefore no square rootrdquo Bhaskaraanother I

Indian mathematician also writes in his work Bijaganita written in

1150 AD ldquoThere is no square root of a negative quantity for it is not a

squarerdquo Cardan (1545 AD) considered the problem of solving

x + y = 10 xy = 40He obtained x = 5 + minus15 and y = 5 ndash minus15 as the

solution of it whichwas discarded by him by saying that these numbers

are lsquouselessrsquo Albert Girard(about 1625 AD) accepted square root of negative

numbers and said that this

will enable us to get as many roots as the degree of the polynomial equation

Euler was the first to introduce the symbol i for minus1 and WR Hamilton

(about 1830 AD) regarded the complex number a + ib as an ordered

pair ofreal numbers (a b) thus giving it a purely mathematical definition and

avoidinguse of the so called lsquoimaginary

numbersrsquo

Indian mathematician also writes in his work Bijaganita written in

1150 AD ldquoThere is no square root of a negative quantity for it is not a

squarerdquo Cardan (1545 AD) considered the problem of solving

x + y = 10 xy = 40He obtained x = 5 + minus15 and y = 5 ndash minus15 as the

solution of it whichwas discarded by him by saying that these numbers

are lsquouselessrsquo Albert Girard(about 1625 AD) accepted square root of negative

numbers and said that this

will enable us to get as many roots as the degree of the polynomial equation

Euler was the first to introduce the symbol i for minus1 and WR Hamilton

(about 1830 AD) regarded the complex number a + ib as an ordered

pair ofreal numbers (a b) thus giving it a purely mathematical definition and

avoidinguse of the so called lsquoimaginary

numbersrsquo

will enable us to get as many roots as the degree of the polynomial equation

Euler was the first to introduce the symbol i for minus1 and WR Hamilton

(about 1830 AD) regarded the complex number a + ib as an ordered

pair ofreal numbers (a b) thus giving it a purely mathematical definition and

avoidinguse of the so called lsquoimaginary

numbersrsquo