Complex numbers 1

Complex Numbers

N. B. Vyas

Department of Mathematics,Atmiya Institute of Tech. and Science,

Rajkot (Guj.)

N. B. Vyas Complex Numbers

Definition of Complex Number

Complex Numbers

A number of the form x+ iy, where x and y are

real numbers and i =√−1 is called a complex

number and is denoted by z. It is also denoted

by an ordered pair(x, y).

Thus z = x + iy or z = (x, y)



Complex Numbers

A number of the form x+ iy, where x and y are

real numbers and i =√−1 is called a complex

number and is denoted by z. It is also denoted

by an ordered pair(x, y).

Thus z = x + iy or z = (x, y)



The set of complex numbers is denoted by {.

If z = x+ iy is a complex number, thenx is called the real part of z and denoted byRe(z).y is called the imaginary part of z and isdenoted by Im(z).

If x = 0 and y 6= 0 then z = 0 + iy = iy

is called a purely imaginary number.

If x 6= 0 and y = 0 then z = x+ i0 = x

is called a real number.

If x = 0 and y = 0 then z = 0 + i.0 = 0is the zero complex number.




If z = x+ iy is a complex number, then

x is called the real part of z and denoted byRe(z).y is called the imaginary part of z and isdenoted by Im(z).









If z = x+ iy is a complex number, thenx is called the real part of z and denoted byRe(z).

y is called the imaginary part of z and isdenoted by Im(z).



















If x = 0 and y 6= 0 then z = 0 + iy = iyis called a purely imaginary number.









If x 6= 0 and y = 0 then z = x+ i0 = xis called a real number.







If x 6= 0 and y = 0 then z = x+ i0 = xis called a real number.



Definition of Conjugate Complex Number

Conjugate Complex Numbers

If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.

Thus x+ iy and x− iy are conjugate complexnumbers.

The conjugate of a complex number z is denotedby z̄.

The conjugate of real number is the real numberitself.




If two complex numbers differ only in the sign of theimaginary part then they are called conjugatecomplex numbers.Thus x+ iy and x− iy are conjugate complexnumbers.
















Properties of Complex Numbers

1 If x+ iy = 0 then x = 0, y = 0

2 If x1 + iy1 = x2 + iy2 then x1 = x2 and y1 = y2

3 If x1 + iy1 = x2 + iy2 then x1 − iy1 = x2 − iy2

4 Sum, difference and quotient(division) of any twocomplex numbers is a complex number.



1 If x+ iy = 0 then x = 0, y = 0






1 If x+ iy = 0 then x = 0, y = 0






1 If x+ iy = 0 then x = 0, y = 0






If z1 = x1 + iy1 and z2 = x2 + iy2 then their

Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)= (x1 + x2) + i(y1 + y2)

Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)= (x1 − x2) + i(y1 − y2)

Product: z1.z2 = (x1 + iy1).(x2 + iy2)= (x1x2 − y1y2) + i(x1y2 + x2y1)




Sum: z1 + z2 = (x1 + iy1) + (x2 + iy2)

= (x1 + x2) + i(y1 + y2)













Difference: z1 − z2 = (x1 + iy1)− (x2 + iy2)

= (x1 − x2) + i(y1 − y2)













Product: z1.z2 = (x1 + iy1).(x2 + iy2)

= (x1x2 − y1y2) + i(x1y2 + x2y1)








Polar form of a Complex Number

Let P (x, y) be the point which representsz = (x, y) = x+ iy

Let OP = r and ∠POM = θ.Then from ∆OPM .x = OM = rcosθ. y = PM = rsinθ.∴ z = x+ iy = r(cosθ + isinθ)



Let P (x, y) be the point which representsz = (x, y) = x+ iyLet OP = r and ∠POM = θ.

Then from ∆OPM .x = OM = rcosθ. y = PM = rsinθ.∴ z = x+ iy = r(cosθ + isinθ)



Let P (x, y) be the point which representsz = (x, y) = x+ iyLet OP = r and ∠POM = θ.Then from ∆OPM .

x = OM = rcosθ. y = PM = rsinθ.∴ z = x+ iy = r(cosθ + isinθ)



Let P (x, y) be the point which representsz = (x, y) = x+ iyLet OP = r and ∠POM = θ.Then from ∆OPM .x = OM = rcosθ. y = PM = rsinθ.

∴ z = x+ iy = r(cosθ + isinθ)



Let P (x, y) be the point which representsz = (x, y) = x+ iyLet OP = r and ∠POM = θ.Then from ∆OPM .x = OM = rcosθ. y = PM = rsinθ.∴ z = x+ iy = r(cosθ + isinθ)



r is called the absolute value or the modulus ofz and is denoted by |z|.

∴ r = |z| =√x2 + y2 =

√z.z̄

Geometrically, |z| is the distance of point z fromthe origin.

θ is called the argument of z or amplitude of z.

It is denoted by argz or Ampz

∴ θ = argz = tan−1(yx

)θ is the directed angle from the positive X-axis toOP.

The value of θ lies in the interval −π < θ ≤ π iscalled principal value.



r is called the absolute value or the modulus ofz and is denoted by |z|.∴ r = |z| =

√x2 + y2 =

√z.z̄










√x2 + y2 =

√z.z̄










√x2 + y2 =

√z.z̄










√x2 + y2 =

√z.z̄





)

θ is the directed angle from the positive X-axis toOP.





√x2 + y2 =

√z.z̄










√x2 + y2 =

√z.z̄








Multiplication and Division of ComplexNumbers in Polar Form

Let z1 = r1(cosθ1 + isinθ1) and z2 = r2(cosθ2 + isinθ2)be two complex numbers in polar form. Then

Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.




Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)

= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.




Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]

= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.




Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]

which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.




Product: z1z2 = r1(cosθ1 + isinθ1)r2(cosθ2 + isinθ2)= r1r2[(cosθ1cosθ2 − sinθ1sinθ2) + i(cosθ1sinθ2 +sinθ1cosθ2)]= r1r2[cos(θ1 + θ2) + isin(θ1 + θ2)]which shows modulus and amplitude of the product oftwo complex numbers z1 and z2 is product of theirmoduli ( i.e. r1r2 ) and sum of their arguments ( i.eθ1 + θ2 ) respectively.



Division:z1

z2=r1(cosθ1 + isinθ1)

r2(cosθ2 + isinθ2)

=r1(cosθ1 + isinθ1)(cosθ2 − isinθ2)

r2(cosθ2 + isinθ2)(cosθ2 − isinθ2)

=r1

r2[(cosθ1cosθ2 + sinθ1sinθ2) + i(sinθ1cosθ2 −

cosθ1sinθ2)]

=r1

r2[cos(θ1 − θ2) + isin(θ1 − θ2)]

which shows that the modulus and amplitude of thequotient of two complex numbers z1 and z2 is the

quotient of their moduli ( i.e.r1

r2) and difference of

their arguments( i.e. θ1 − θ2 ) respectively.



Division:z1





=r1


cosθ1sinθ2)]

=r1

r2[cos(θ1 − θ2) + isin(θ1 − θ2)]







Division:z1





=r1


cosθ1sinθ2)]

=r1

r2[cos(θ1 − θ2) + isin(θ1 − θ2)]







Division:z1





=r1


cosθ1sinθ2)]

=r1

r2[cos(θ1 − θ2) + isin(θ1 − θ2)]







Division:z1





=r1


cosθ1sinθ2)]

=r1

r2[cos(θ1 − θ2) + isin(θ1 − θ2)]




their arguments( i.e. θ1 − θ2 ) respectively.N. B. Vyas Complex Numbers

Exponential form of Complex Numbers

We know that

sinθ = θ − θ3

3!+θ5

5!− θ7

7!. . .

cosθ = 1− θ2

2!+θ4

4!− θ6

6!. . . and

eθ = 1 + θ +θ2

2!+θ3

3!+θ4

4!. . .

Now θ = iθ

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!. . .

eiθ = 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!. . . {∵ i2 = −1, i3 = −i, i4 = 1

eiθ =

(1− θ2

2!+θ4

4!− . . .

)+ i

(θ − θ3

3!+θ5

5!− . . .

)eiθ = (cosθ + isinθ)



We know that

sinθ = θ − θ3

3!+θ5

5!− θ7

7!. . .

cosθ = 1− θ2

2!+θ4

4!− θ6

6!. . . and

eθ = 1 + θ +θ2

2!+θ3

3!+θ4

4!. . .

Now θ = iθ

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!. . .

eiθ = 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!. . . {∵ i2 = −1, i3 = −i, i4 = 1

eiθ =

(1− θ2

2!+θ4

4!− . . .

)+ i

(θ − θ3

3!+θ5

5!− . . .




We know that

sinθ = θ − θ3

3!+θ5

5!− θ7

7!. . .

cosθ = 1− θ2

2!+θ4

4!− θ6

6!. . . and

eθ = 1 + θ +θ2

2!+θ3

3!+θ4

4!. . .

Now θ = iθ

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!. . .

eiθ = 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!. . . {∵ i2 = −1, i3 = −i, i4 = 1

eiθ =

(1− θ2

2!+θ4

4!− . . .

)+ i

(θ − θ3

3!+θ5

5!− . . .




We know that

sinθ = θ − θ3

3!+θ5

5!− θ7

7!. . .

cosθ = 1− θ2

2!+θ4

4!− θ6

6!. . . and

eθ = 1 + θ +θ2

2!+θ3

3!+θ4

4!. . .

Now θ = iθ

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!. . .

eiθ = 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!. . . {∵ i2 = −1, i3 = −i, i4 = 1

eiθ =

(1− θ2

2!+θ4

4!− . . .

)+ i

(θ − θ3

3!+θ5

5!− . . .




We know that

sinθ = θ − θ3

3!+θ5

5!− θ7

7!. . .

cosθ = 1− θ2

2!+θ4

4!− θ6

6!. . . and

eθ = 1 + θ +θ2

2!+θ3

3!+θ4

4!. . .

Now θ = iθ

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!. . .

eiθ = 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!. . . {∵ i2 = −1, i3 = −i, i4 = 1

eiθ =

(1− θ2

2!+θ4

4!− . . .

)+ i

(θ − θ3

3!+θ5

5!− . . .




We know that

sinθ = θ − θ3

3!+θ5

5!− θ7

7!. . .

cosθ = 1− θ2

2!+θ4

4!− θ6

6!. . . and

eθ = 1 + θ +θ2

2!+θ3

3!+θ4

4!. . .

Now θ = iθ

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!. . .

eiθ = 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!. . . {∵ i2 = −1, i3 = −i, i4 = 1

eiθ =

(1− θ2

2!+θ4

4!− . . .

)+ i

(θ − θ3

3!+θ5

5!− . . .




We know that

sinθ = θ − θ3

3!+θ5

5!− θ7

7!. . .

cosθ = 1− θ2

2!+θ4

4!− θ6

6!. . . and

eθ = 1 + θ +θ2

2!+θ3

3!+θ4

4!. . .

Now θ = iθ

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!. . .

eiθ = 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!. . . {∵ i2 = −1, i3 = −i, i4 = 1

eiθ =

(1− θ2

2!+θ4

4!− . . .

)+ i

(θ − θ3

3!+θ5

5!− . . .

)

eiθ = (cosθ + isinθ)



We know that

sinθ = θ − θ3

3!+θ5

5!− θ7

7!. . .

cosθ = 1− θ2

2!+θ4

4!− θ6

6!. . . and

eθ = 1 + θ +θ2

2!+θ3

3!+θ4

4!. . .

Now θ = iθ

eiθ = 1 + iθ +(iθ)2

2!+

(iθ)3

3!+

(iθ)4

4!. . .

eiθ = 1 + iθ − θ2

2!− iθ

3

3!+θ4

4!. . . {∵ i2 = −1, i3 = −i, i4 = 1

eiθ =

(1− θ2

2!+θ4

4!− . . .

)+ i

(θ − θ3

3!+θ5

5!− . . .



Laws of Complex Numbers

If z1 and z2 are two complex numbers, then

1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|2 |z1 − z2| ≥ ||z1| − |z2||3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)

4 |z1z2| = |z1||z2|

5

∣∣∣∣z1

z2

∣∣∣∣ =|z1||z2|




1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|

2 |z1 − z2| ≥ ||z1| − |z2||3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)

4 |z1z2| = |z1||z2|

5

∣∣∣∣z1

z2

∣∣∣∣ =|z1||z2|




1 Triangle Inequality: |z1 + z2| ≤ |z1|+ |z2|2 |z1 − z2| ≥ ||z1| − |z2||

3 Parellelogram equality:|z1 + z2|2 + |z1 − z2|2 = 2(|z1|2 + |z2|2)

4 |z1z2| = |z1||z2|

5

∣∣∣∣z1

z2

∣∣∣∣ =|z1||z2|





4 |z1z2| = |z1||z2|

5

∣∣∣∣z1

z2

∣∣∣∣ =|z1||z2|





4 |z1z2| = |z1||z2|

5

∣∣∣∣z1

z2

∣∣∣∣ =|z1||z2|





4 |z1z2| = |z1||z2|

5

∣∣∣∣z1

z2

∣∣∣∣ =|z1||z2|


Ex

1 Find complex conjugate of3 + 2i

1− i


Theorem

DeMoivre’s TheoremIf n is a rational number than the value or one of thevalues of (cosθ + isinθ)n is cosnθ + isinnθ

∴ z = x+ iy = r(cosθ + isinθ)zn = rn(cosθ + isinθ)n

= rn(cosnθ + isinnθ)E.g.:

(cosθ + isinθ)2 = cos2θ + isin2θ

(cosθ + isinθ)−1 = cos(−θ) + isin(−θ) = cosθ − isinθ


Theorem


∴ z = x+ iy = r(cosθ + isinθ)

zn = rn(cosθ + isinθ)n





Theorem







Theorem



= rn(cosnθ + isinnθ)

E.g.:




Theorem







Theorem







Theorem







Examples

Ex. Simplify

(cos2θ + isin2θ)23(cosθ − isinθ)2

(cos3θ − isin3θ)2(cos5θ − isin5θ)13


Examples

Ex. Evaluate (1 + i√

3)90 + (1− i√

3)90

Sol. Let x = 1 and y =√

3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√

3 = 2(cosπ3 + isinπ3 ) and 1− i√

3 = 2(cosπ3 − isinπ3 )

∴ (1 + i√

3)90 + (1− i√

3)90

=[2(cosπ3 + isinπ3 )

]90+[2(cosπ3 − isin

π3 )]90

= 290(cos30π + isin30π) + 290(cos30π − isin30π)

= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√

3 = 2(cosπ3 + isinπ3 )

and 1− i√


∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Examples


3)90 + (1− i√

3)90


3

r =√x2 + y2 =

√1 + 3 =

√4 = 2

θ = tan−1(yx

)= tan−1

(√3

1

)=π

3

∴ 1 + i√



∴ (1 + i√

3)90 + (1− i√

3)90



π3 )]90


= 290(2cos30π)

= 291cos30π

= 291(1) = 291


Example

Ex. Evaluate

(1 + sinθ + icosθ

1 + sinθ − icosθ

)n


Sol. We have sin2θ + cos2θ = 1

∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ= (sinθ − icosθ)(sinθ + icosθ)

Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)

∴1 + sinθ + icosθ

1 + sinθ − icosθ= sinθ + icosθ

= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]


Sol. We have sin2θ + cos2θ = 1∴ 1 = sin2θ + cos2θ = sin2θ − i2cos2θ

= (sinθ − icosθ)(sinθ + icosθ)




= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]







= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]




Now1 + sinθ + icosθ

= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)= (sinθ + icosθ)(sinθ − icosθ + 1)



= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]




Now1 + sinθ + icosθ= (sinθ − icosθ)(sinθ + icosθ) + (sinθ + icosθ)

= (sinθ + icosθ)(sinθ − icosθ + 1)



= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]







= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]







= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]







= cos(π2 − θ

)+ isin

(π2 − θ

)

∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]







= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n

= cos[n(π2 − θ

)]+ isin

[n(π2 − θ

)]







= cos(π2 − θ

)+ isin

(π2 − θ

)∴

(1 + sinθ + icosθ


)n=[cos(π2 − θ

)+ isin

(π2 − θ

)]n= cos

[n(π2 − θ

)]+ isin

[n(π2 − θ

)]


Example

Ex. If z1 = eiθ1 , z2 = eiθ2 and z1 − z2 = 0 prove that1

z1− 1

z2= 0


Example

Sol. Herez1 = eiθ1 = cosθ1 + isinθ1

z2 = eiθ2 = cosθ2 + isinθ2Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0

Now1

z1− 1

z2=

1

cosθ1 + isinθ1− 1

cosθ2 + isinθ2

=cosθ1 − isinθ1cos2θ1 + sin2θ1

− cosθ2 − isinθ2cos2θ2 + sin2θ2


Example

Sol. Herez1 = eiθ1 = cosθ1 + isinθ1z2 = eiθ2 = cosθ2 + isinθ2

Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0

Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Example


Given that z1 − z2 = 0

(cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0

Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Example


Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0

(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0

Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Example


Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0

Comparing real and imaginary parts on both the sides, we getcosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0

Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Example


Given that z1 − z2 = 0 (cosθ1 + isinθ1)− (cosθ2 + isinθ2) = 0(cosθ1 − cosθ2) + i(sinθ1 − sinθ2) = 0Comparing real and imaginary parts on both the sides, we get

cosθ1 − cosθ2 = 0 and sinθ1 − sinθ2 = 0

Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Example



Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Example



Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Example



Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Example



Now1

z1− 1

z2=

1


cosθ2 + isinθ2




Ex. (1 + i)n + (1− i)n = 2n2+1cosnπ4


Root of a Complex Number

De Moivre’s theorem is useful for finding the roots of a complexnumber.

If n is any positive integer, then by De Moivre’s theorem(cos

θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ

nis one of the nth root of cosθ + isinθ, i.e.,

(cosθ + isinθ)1n = cos

θ

n+ isin

θ

nThe remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,

cos(2kπ + θ) = cosθ and sin(2kπ + θ) = sinθ

Hence the general form of complex number cosθ+ isinθ form restof all the roots.




If n is any positive integer, then by De Moivre’s theorem

(cos

θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ



θ

n+ isin

θ








θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ



θ

n+ isin

θ








θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ



θ

n+ isin

θ








θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ



θ

n+ isin

θ

n

The remaining roots may be obtained by periodic nature of thetrigonometric functions, i.e.,







θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ



θ

n+ isin

θ








θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ



θ

n+ isin

θ








θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ



θ

n+ isin

θ








θ

n+ isin

θ

n

)n= cos

(θ

nn

)+ isin

(θ

nn

)= cosθ + isinθ

Thus cosθ

n+ isin

θ



θ

n+ isin

θ






(cosθ + isinθ)1n = [cos(2kπ + θ) + isin(2kπ + θ)]

1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n

)which gives all the roots of (cosθ + isinθ)

1n for

k = 0, 1, 2, . . . (n− 1).

This roots are as follows:

For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n

]The further values of k will give the same roots as above in order.




1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n

)

which gives all the roots of (cosθ + isinθ)1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n





1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n





1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n





1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n





1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)

k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n





1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

)

. . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n





1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n





1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n





1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n

]

The further values of k will give the same roots as above in order.




1n

= cos

(2kπ + θ

n

)+ isin

(2kπ + θ

n


1n for

k = 0, 1, 2, . . . (n− 1).


For k = 0, cosθ

n+ isin

θ

n

k = 1, cos

(2π + θ

n

)+ isin

(2π + θ

n

)k = 2, cos

(4π + θ

n

)+ isin

(4π + θ

n

). . . . . .

. . . . . .

k = n− 1, cos

[2(n− 1)π + θ

n

]+ isin

[2(n− 1)π + θ

n



Education

Complex numbers 1