5 Complex Numbers Questions Sols

8/13/2019 5 Complex Numbers Questions Sols

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Transition to Mathematical ProofsChapter 5 - Complex Numbers Assignment Solutions

Question 1. Similar to how we obtained the double-angle formulae in thenotes, use the Euler equation to show the two angle-sum formulae hold:

sin(+) = sin cos+ sin cos;

cos(+) = cos cos sin sin .

Solution 1. We will use the following law of exponents:

ei(+) =ei ei .Using the Euler equation on ei(+), we have

ei(+) = cos(+) +i sin(+).

Using the Euler equation on the other side and FOILing, gives us

ei ei = (cos+i sin)(cos+i sin) =cos cos+i cos sin +i sin cos+i2 sin sin =

(cos cos sin sin) +i(cos sin+ sin cos).Since ei(+) =ei ei,we can equate the above results to get

cos(+) +i sin(+) =

(cos cos sin sin) +i(cos sin+ sin cos).Equating their real parts, we get

cos(+) = cos cos

sin sin .

Equating their imaginary parts, we get

sin(+) = cos sin+ sin cos.

Question 2.

(a) Show that|z|= Re(z) if and only ifz is a non-negative real number.

(b) Show that (z)2 = z2 if and only if z is purely real or purely imaginary(i.e., its real part is 0).

Solution 2a. To prove the above biconditional statement, we will prove thefollowing two conditional statements: If|z|= Re(z), then z is a non-negativereal number and Ifz is a non-negative real number, then|z|= Re(z)

For the first conditional statement, we assume that|z| = Re(z). Writingz= a+bi, we have that

a2 +b2 =a.

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Squaring both side, we get that a2 +b2 = a2 and thus b2 = 0. Thus, we canconclude that b = 0 and so z = a+ 0i is a real number. To prove that it isnon-negative, we note that since z = a= Re(z) =|z| and|z| 0, then z 0.Thus,z is a real, non-negative number.

For the second conditional statement, we assume that z is a non-negativereal number. Thus, z= a+ 0iwith a0. Thus,

|z|=|a+ 0i|=

a2 + 02 =a2 =|a|.

Since a0, then|a|= a. Thus,|z|=|a|= a = Re(z), as desired.Having proven both conditional statements, the original biconditional state-

ment |z|= Re(z) if and only ifz is a non-negative real number is also true.

Solution 2b. To prove the above biconditional statement, we will prove thefollowing two conditional statements: If (z)

2= z2, then z is purely real or

purely imaginary and Ifz is purely real or purely imaginary, then (z)2

=z2For the first conditional statement, we assume that z = a+ bi satisfies

(z)2

=z2. Thus,

(a

bi)2

= (a+bi)2.

Expanding both sides, we get

a2 2abi b2 =a2 + 2abi b2.Simplifying, we obtain that2abi = 2abi, which is equivalent to 4abi = 0.Dividing by 4i, we get ab = 0, and thus a = 0 or b = 0, giving two cases. Ifa = 0, then z = 0 +bi is purely imaginary. Ifb= 0, then z = a+ 0i is purelyreal.

For the second conditional statement, we will assume that z = a+biis purelyreal or purely imaginary, giving us two cases. For the first case, z is purely real.Since z is real, then z = z . Thus, (z)

2=z2,as desired. Ifz is purely imaginary,

then z= 0 +bi and thus z= bi =bi.Thus,(z)2 = (bi)2 = (1)2(bi)2 =z2,

as desired. In either case, (z)2

=z2.Having proven both conditional statements, the biconditional statement

(z)2

=z2 if and only ifz is purely real or purely imaginary is also true.

Question 3. The modulus of a complex number is, in many ways, a general-ization of the absolute value of a real number. Here, we give another propertyof the modulus that the absolute value of a real number already enjoys.Ifz, w C, show that

|z w|=|z| |w|in the following two ways:

(a) By using the cartesian form z = a+ bi and w = c+ di for the complexnumbers z and w.

(b) By using the polar formz = r1ei1 andw = r2e

i2 for the complex numbersz and w .

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Solution 3a. Let z= a+bi and w= c+di. Then,

|z w|=|(a+bi) (c+di)|=|(ac bd) +i(ad+bc)|

(ac bd)2 + (ad+bc)2 =

a2c2

2abcd+b2d2 +a2d2 + 2abcd+b2c2 =

a2c2 +b2d2 +a2d2 +b2c2.

If we instead compute|z| |w|, then we have

|z| |w|=|a+bi| |c+di|=

a2 +b2

c2 +d2 =

(a2 +b2)(c2 +d2) =

a2c2 +a2d2 +b2c2 +b2d2.

Notice that the above results for|z w| and|z| |w| are equal and thus

|z w|=|z| |w|.

Solution 3b. Let z= r1ei1 and w= r2e

i2 .Then,

|z w|= r1ei1 r2ei2=

(r1r2)ei(1+2) =r1r2.

If we instead compute|z| |w|, we then have

|z| |w|= r1ei1 r2ei2

= r1 r2.

Notice that the above results for|z w| and|z| |w| are equal and thus

|z w|=|z| |w|.

Question 4. Below, we will prove a remarkable fact about real polynomialsusing complex numbers. For the below, letz = a+biand w = c+dibe complexnumbers.

(a) Show that z+w = z +w.

(b) Show that z w= z w.

(c) Use (b) to show that zn = (z)n

for any natural number nN.(d) Consider the following polynomial p(z) with real coefficients:

p(z) = nzn +n1z

n1 + +1z+0,

where each i is a real number. Show that if a complex number w is aroot to the above polynomial with real coefficients, then its conjugate wis also a root to the same polynomial. That is, use (a) - (c) to show thatifp(w) = 0, then p(w) = 0.

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Solution 4a. Beginning with the left-hand side, we have

z+w = (a+bi) + (c+di) = (a+c) +i(b+d) =

(a+c) i(b+d) = (a bi) + (c di) = z +w,as desired.

Solution 4b. Expanding the z w, we have

z w= (a+bi)(c+di) = (ac bd) +i(ad+bc) = (ac bd) i(ad+bc).

Expanding z w, we have

z w= (a+bi) (c+di) = (a bi)(c di) =

ac adi bci bd= (ac bd) i(ad+bc).Since the above results for z wand z ware equal, thenz w= z w,as desired.

Solution 4c. In (b), we proved that z w= z w. Letting z = w, we havez2 =z z= z z= (z)2 .

Continuing in this fashion with ncopies ofz, we have

zn =z z z z= z z z= (z)n ,

as desired.

Solution 4d. Assume w is a root to p(z). Thus, p(w) = 0 and so

p(w) = nwn +n1w

n1 + +1w+0 = 0.

If we take the conjugate of the above equation, we get that

nwn +n1wn1 + +1w+0 = 0 = 0.

Using the above properties of the conjugate on the left-hand side of the equation,we have that

nwn +n1wn1 + +1w+0= 0nwn +n1wn1 + +1w+0= 0nwn +n1wn1 + +1w+0 = 0nw

n +n1wn1 + +1w+0 = 0.

Since the i are real, then i = i for all i. Thus, the above is equivalent to

nwn +n1w

n1 + +1w+0 = 0.

This is simply the polynomial p(z) with an input ofw. Thus,p(w) = 0 and wis a root.

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5 Complex Numbers Questions Sols