Solutions to Exercises 1 - University of...

Section 1.1 Complex Numbers 1

Solutions to Exercises 1.1

1. We have

−i = 0 + (−1)i.

So a = 0 and b = −1.

5. We have

(2 − i)2 = (2 + i)2 (because 2 − i = 2 − (−i) = 2 + i)

= 4 + 4i +

=−1︷︸︸︷(i)2 = 3 + 4i.

So a = 3 and b = 4.

9. We have

(12

+i

7

) (32− i

)=

12

32

+ i

(314

− 12

)= 1

7︷︸︸︷−i

i

7=

2528

− i27

So a = 2528 and b = −2

7 .

13. Multiplying and dividing by the conjugate of the denominator, i.e. by 2 − i = 2 + i weget

14 + 13i

2− i=

(14 + 13i)(2 + i)(2 − i)(2 + i)

=14 · 2 + 14 · i + 13i · 2 + 13i2

4 + 1

=28 + 14i + 26i− 13

5=

155

+405

i = 3 + 8i

So a = 3 and b = 8.

17. Applying the quadratic formula we get

x =0 ±

√−24

2= 0±

√−242

= 0 ± i√

6

21. Put u = x2. Then we get

u2 + 2u + 1 = 0.

By the quadratic formula

u =−2 ±

√4 − 4

2= −1.

Therefore we get −1 = x2 and the solution is x = ±i.

2 Chapter 1 A Preview of Applications and Techniques

25. Suppose w can be written in the form w = c + id. Then from the identity z + w = 0we have a + ib + c + id = 0, or

(a + c) + i(b + d) = 0.

It follows that a + c = 0 and b + d = 0. Hence c = −a and d = −b. And, this meansw = −a − ib.(b) Let es represent a complex number such that z + es = z for all complex z. Show thates = 0; that is, Re (es) = 0 and Im(es) = 0. Thus es = 0 is the unique additive identityfor complex numbers.

Solution. Let us put z = 0 into z + es = z. This gives 0 + es = 0, or if es = a + ib weget a + ib = 0 + i0. Since two numbers are equal if and only if their real and imaginaryparts are the same we have a = 0 and b = 0. So, es = 0.

29. (a) We have

anzn + an−1zn−1 + · · ·+ a1z + a0

= anzn + an−1zn−1 + · · ·+ a1z + a0

= an · zn + an−1 · zn−1 + · · ·+ a1 · z + a0

[Since a0, a1, · · · , an−1, an are all real]= anzn + an−1zn−1 + · · ·+ a1z + a0

[by using the property that zn = (z)n]= an(z)n + an−1(z)n−1 + · · ·+ a1(z) + a0

(b) Suppose that z0 is a root of the polynomial p(z) = anzn + an−1zn−1 + · · ·+ a1z + a0.

This means that

anzn0 + an−1z

n−10 + · · ·+ a1z0 + a0 = 0.

By using the part (a) we have

an(z0)n + an−1(z0)n−1 + · · ·+ a1z0 + a0 = anzn0 + an−1z

n−10 + · · ·+ a1z0 + a0 = 0.

Hence, the number z0 is also a root of the same polynomial.

33. Project Problem: The cubic equation. (a) Using the given change of variablesx = y − a

3 we get

x3 + ax2 + bx + c =(y − a

3

)3+ a

(y − a

3

)2+ b

(y − a

3

)+ c

= y3 − 3y2a

3+ 3y

(a

3

)2−

(a

3

)3+ ay2 − 2ay

a

3+ a

(a

3

)2

+by − ba

3+ c

= y3 + y2(−a + a) + y(a2

3− 2a2

3+ b) + (−a3

27+

a3

9− ab

3+ c)

= y3 + y(b− a2

3) + (

2a3

27− ab

3+ c)

= y3 + py + q = 0

Section 1.1 Complex Numbers 3

if we put p = b − a2

3 and q = 2a3

27 − ab3 + c.

(b) Now we substitute y = u + v into the equation we received in the previous step:

y3 + py + q = (u + v)3 + p(u + v) + q

= u3 + 3u2v + 3uv2 + v3 + p(u + v) + q

= u3 + v3 + 3uv(u + v) + p(u + v) + q

= u3 + v3 + (3uv + p)(u + v) + q = 0.

(c) We suppose that u and v satisfy the desired property 3uv + p = 0. Then we haveuv = −p/3 and if we cube both sides we get u3v3 = (−p/3)3. And, also using the propertythat 3uv + p = 0 and the equation from the part (b) we have

u3 + v3 + (3uv + p)(u + v) + q = u3 + v3 + q = 0.

So, it immediately follows that u3 + v3 = −q.(d) Let us show that U is a solutions of this quadratic equation. (Because of the symmetrythen V is also a solution!) Using the equation U+V = −β we get V = −β−U . We substituteit into the equation UV = γ. We get

0 = UV − γ = U(−β − U) − γ = −U2 − βU − γ,

or if we rewrite it and multiply both sides by −1 we have U2 + βU + γ = 0. And, thismeans that U is a solution of the given quadratic equation.

(e) Straightforward we can put β = −q and γ = p3

27 and use the part (d) to conclude thatu3 and v3 are the solutions of the quadratic equation X2 + qx − p3

27 = 0. By the quadraticformula we get that

u3 =−q +

√q2 + 4p3/272

and v3 =−q −

√q2 + 4p3/272

,

or u3 = −q

2+

√q2

2+

p

3and v3 = −q

2−

√q2

2+

p

3.

Now if we get the cubic roots of both sides of the equations we get that

u =3

√

−q

2+

√(q

2

)2+

(p

3

)3and v =

3

√

−q

2−

√(q

2

)2+

(p

3

)3.

(f) Substituting back the found values we have

x = y − a

3= u + v − a

3

=3

√

−q

2+

√(q

2

)2+

(p

3

)3+

3

√

−q

2−

√(q

2

)2+

(p

3

)3− a

3.

4 Chapter 1 Complex Numbers and Functions


1. z has coordinates (1,−1), (−z) has coordinates (−1, 1), and z has coordinates (−1,−1).5. Since z = 1 − i = 1 − i = 1 + i we have that z has coordinates (1, 1), (−z) has coordinates

(−1,−1), and z has coordinates (1,−1).

9. We use the property that |ab| = |a||b| twice below:

|(1 + i)(1 − i)(1 + 3i)| = |1 + i||(1− i)(1 + 3i)| =

√2︷︸︸︷

|1 + i|

√2︷︸︸︷

|1− i|

√10︷︸︸︷

|1 + 3i| = 2√

10.

13. Using the properties∣∣∣ z1z2

∣∣∣ = |z1||z2| and |z| = |z| we have

∣∣∣∣i

2 − i

∣∣∣∣ =|i|

|2− i|=

|i||2 − i|

=1√5

=√

55

.

17. The equation |z − i| = −1 does not have any solution because |z| is a distance from the point zto the origin. And clearly a distance is always non-negative.

21. Let z = x+iy satisfies the equation |2z|+|2z−1| = 4. Since 2z = 2x+i2y and 2z−1 = 2x−1+i2ywe have

√(2x)2 + (2y)2 +

√(2x − 1)2 + (2y)2 = 4, or

√(2x)2 + (2y)2 = 4 −

√(2x − 1)2 + (2y)2.

If we square both sides of the equation we get

(2x)2 + (2y)2 = 16− 8√

(2x − 1)2 + (2y)2 + (2x − 1)2 + (2y)2, or

(2x)2 + (2y)2 = 16 − 8√

(2x − 1)2 + (2y)2 + (2x)2 − 4x + 1 + (2y)2.

And if we subtract (2x)2 + (2y)2 from both sides of the equation and add 8√

(2x − 1)2 + (2y)2 weget

8√

(2x − 1)2 + (2y)2 = 17− 4x.

If we square both sides of the equation one more time we get the equation of the ellipse we arelooking for:

64((2x − 1)2 + (2y)2) = (17− 4x)2, or

64(2x − 1)2 + 176y2 = (17− 4x)2.

25. (a) Let us recall that |a − b| is the distance between a and b. So, |z − i − 1| is the distancebetween z and 1 + i. Also Re(z) is the distance between z and 1 + i. Also Re(z) + 1 is the signeddistance from z to the line Re(z) = −1. Hence, if z satisfies the equation |z − 1 − i| = Re(z) + 1 itfollows that the distances from the line Re(z) = −1 to z and from (1 + i) to z are equal. So, thisequation is a parabola equation which directrix is the line Re(z) = −1 and the focus is (1 + i).

Section 1.2 The Complex Plane 5

(b) Similar |z − z0| is the distance from z to z0, and Re(z) − a is the distance from z to theline whose equation is Re(z) = a. So, it is also a parabola with the directrix whose equation isRe(z) = a and the focus is z0.

29. Let us concentrate on the circle which center is the origin and the radius is r = ||z1| − |z2||.Obviously the absolute value of any complex number lying outside of the disc bounded by this circlewill be greater than r. Therefore, since from the picture it follows that the number z1 − z2 whichis the tip of the vector −z2 on the picture lies outside (or on the circle of) the disc described abovethen the absolute value which is |z1 − z2| is greater than (or equal to ) r. And, if we write thisalgebraically we get the needed inequality:

|z1 − z2| ≥ ||z1| − |z2||.

33. We get the estimate| cos θ + i sin θ| ≤ | cos θ| + |i sin θ|

simply by the triangle inequality. Now we notice that

|i sinθ| = |i| | sin θ| = 1| sin θ| = | sin θ|.

So,| cos θ| + |i sin θ| = | cos θ| + | sin θ|.

We know from algebra that | sin θ| ≤ 1, and | cos θ| ≤ 1 Therefore we have

| cos θ| + | sin θ| ≤ 1 + 1 = 2,

and this justifies the last step of the estimation above.(b) By definition of the absolute value we know that if z = x + iy then |z| =

√x2 + y2.

Therefore if z = cos θ + i sin θ then |z| =√

(cos θ)2 + (sin θ)2. But from trigonometry we know theformula

(cos θ)2 + (sin θ)2 = 1

is true for any number θ. Therefore we have

| cos θ + i sin θ| =√

(cos θ)2 + (sin θ)2 = 1.

37. We want to know something about | 1z−4 | given some information about |z − 1|. Notice that an

upper bound of the quantity | 1z−4 | is given by the reciprocal of any lower bound for |z − 4|. So, first

we can find some information about |z − 4|. For this we use some ideas from the example 8 in thissection. We notice that z − 4 = (z − 1) − 3. Therefore by the inequality |z1 − z2| ≥ ||z1| − |z2|| wehave

|z − 4| = |(z − 1) − 3| ≥ ||z − 1| − 3|

Now since we know |z − 1| ≤ 1 we get that ||z − 1| − 3| ≥ |1 − 3| = 2. Hence |z − 4| ≥ 2. Finallytaking reciprocals of the sides of the inequality we will reverse the sign of the inequality and get theneeded upper estimate:

1|z − 4|

≤ 12.

41. (a) By triangle inequality we have∣∣∣∣∣∣

n∑

j=1

vjwj

∣∣∣∣∣∣≤

n∑

j=1

|vjwj|.


And now we use the properties |z1z2| = |z1||z2| and |z| = |z| and get

|vjwj| = |vj||wj| = |vj||wj|.

Using this identity in the triangle inequality above and recalling the assumption that we alreadyproved (??) we have

∣∣∣∣∣∣

n∑

j=1

vjwj

∣∣∣∣∣∣≤

n∑

j=1

|vjwj|

=n∑

j=1

|vj||wj| ≤

√√√√n∑

j=1

|vj|2√√√√

n∑

j=1

|wj|2.

(b) Using the hint we start from the obvious inequality

0 ≤n∑

j=1

(|vj| − |wj|

)2

.

Expanding the right hand side of this inequality we have

0 ≤n∑

j=1

(|vj| − |wj|

)2

=n∑

j=1

(|vj|2 − 2|vj||wj| + |wj|2

)

=n∑

j=1

(|vj|2 + |wj|2

)− 2

n∑

j=1

|vj||wj|.

Thereforen∑

j=1

|vj ||wj| ≤ 12

n∑

j=1

(|vj|2 + |wj|2

)=

12

( n∑

j=1

|vj|2 +n∑

j=1

|wj|2)

=12(1 + 1) = 1 =

√1 ·

√1 =

√√√√n∑

j=1

|vj|2√√√√

n∑

j=1

|wj|2.

(c) Using the hint we can consider v = (v1, v2, . . . , vn) and w = (w1, w2, . . . , wn) and look atthem as at vectors. If we define ‖v‖ =

√∑nj=1 |vj|2 and ‖w‖ =

√∑nj=1 |wj|2. Then it turns out we

need to prove the following inequality:n∑

j=1

|vj| |wj| ≤ ‖v‖‖w‖.

In order to do so we define new vectors U = 1‖u‖u and W = 1

‖w‖w. So, the coordinates of thesevectors are correspondingly Uj = uj

‖u‖ and Wj = wj

‖w‖ . Then we have

n∑

j=1

|Uj|2 =n∑

j=1

∣∣∣∣uj

‖u‖

∣∣∣∣2

=1

‖u‖2

n∑

j=1

|uj|2 =1

‖u‖2‖u‖2 = 1,

and similar we get∑n

j=1 |Wj|2 = 1. Therefore we can apply part (b) to the vectors V and W . Wehave

n∑

j=1

|Vj| |Wj| ≤

√√√√n∑

j=1

|Vj|2√√√√

n∑

j=1

|Wj|2.

Section 1.2 The Complex Plane 7

The left side is equal to

n∑

j=1

|vj |‖v‖

|wj|‖w‖ =

1‖v‖‖w‖

n∑

j=1

|vj||wj|.

The right side is equal to√√√√

n∑

j=1

|vj|2‖v‖2

√√√√n∑

j=1

|wj|2‖w‖2

=

√√√√ 1‖v‖2

n∑

j=1

|vj|2√√√√ 1

‖w‖2

n∑

j=1

|wj|2

=1

‖v‖‖w‖

√√√√n∑

j=1

|vj|2√√√√

n∑

j=1

|wj|2

=1

‖v‖‖w‖‖u‖‖v‖ = 1.

Hence we can rewrite the inequality we received in the form:

1‖v‖‖w‖

n∑

j=1

|vj||wj| ≤ 1.

So, if we multiply both sides by ‖v‖‖w‖ we get the needed inequality.



1. We need to present the number given in its polar form in the form with the real and imaginaryparts z = x + iy. We have

z = 3(

cos7π

12+ i sin

7π

12

)= 3 cos

7π

12+ i 3 sin

7π

12

So, the number can be presented in the Cartesian coordinates as the point (3 cos 7π12 , 3 sin 7π

12 ).

5. Let z = −3 − 3 i. Then we have r =√

(−3)2 + (−3)2 =√

9 · 2 = 3√

2. Also, we can find theargument by evaluating

cos θ =x

r=

−33√

2= −

√2

2and sin θ =

y

r=

−33√

2= −

√2

2.

From the Table 1 in the Section 1.3 we see that θ = 5π4 . Thus, arg z = 5π

4 + 2kπ. Since 5π4 is not

from the interval (−π, π] we can subtract 2π and get that Arg z = 5π4 − 2π = −3π

4 . So, the polarrepresentation is

−3 − 3 i = 3√

2(

cos(−3π

4

)+ i sin

(−3π

4

)).

9. Again we can denote z = − i2 . Then we have r =

√(−1

2 )2 = 12 . We can evaluate

cos θ =x

r=

01/2

= 0 and sin θ =y

r=

−1/21/2

= −1.

And, from the Table 1 in the Section 1.3 we find θ = 3π2

. (Also we could plot the complex number(− i

2) as a point in the complex plane and find the angle from the picture.) Therefore arg z = 3π

2+

2πk. Since 3π2 is not from the interval (−π, π] we can subtract 2π and get that Arg z = 3π

2 −2π = −π2 .

So, the polar representation is

− i

2=

12

(cos(−π

2

)+ i sin

(−π

2

)).

13. We have z = x + i y = 13 + i 2. Since x > 0 we compute

Arg z = tan−1(y

x) = tan−1(

213

) ≈ 0.153.

Hence we express arg z ≈ 0.153 + 2πk for all integer k.

17. First, we need to express the number z = −√

3+ i in the polar form. We compute r =√

3 + 1 =2. And since −

√3 < 0, and 1 > 0 we see that Arg z = tan−1

(1

−√

3

)= tan−1

(−

√3

3

)= 5π

6 . Sincewe need to find the cube of the number z we can use the De Moivre’s Identity to get the polarrepresentation:

(−√

3 + i)3 =(

2(

cos(

5π

6

)+ i sin

(5π

6

)))3

= 23

(cos(

5π

6

)+ i sin

(5π

6

))3

= 8(

cos(

15π

6

)+ i sin

(15π

6

))

= 8(

cos(

3π

6

)+ i sin

(3π

6

)).

Section 1.3 Polar form 9

Where in the last identity we used the fact that Arg (−√

3 + i)3 = 15π6 − 2π = 3π

6 should be in theinterval (−π, π].

21. First, we find the modulus and the argument of the number z = 1+ i. We have r =√

12 + 12 =√2. And, since for z = x + iy = 1 + i1 we get x > 0 we compute Arg z = tan−1

(yx

)= tan−1

(11

)=

tan−1(1) = π4 . Hence z can be expressed in the polar form

1 + i =√

2(cos(π

4

)+ i sin

(π

4

)).

After this we can use De Moivre’s identity to get

(1 + i)30 =(√

2(cos(π

4

)+ i sin

(π

4

)))30

= (√

2)30

(cos(

30π

4

)+ i sin

(30π

4

))

= 215 (0 − i · 1) = 215 i.

Thus Re((1 + i)30

)= 0, and Im

((1 + i)30

)= 215.

25. By De Moivre’s Identity we have

in = (cos(π/2) + i sin(π/2))n = cos(nπ/2) + i sin(nπ/2).

Now since the functions cos(θ), and sin(θ) both have period 2π we notice that the values cos(nπ/2),and sin(nπ/2) depend only on what remainder gives n when is divided by 4. Therefore, we concludethat

in =

1, if n = 4k, for some integer k,i, if n = 4k + 1, for some integer k,−1, if n = 4k + 2, for some integer k,−i, if n = 4k + 3, for some integer k.

29. We know that for any complex numbers z1 = r1(cos θ1 + i sin θ1), and z2 = r2(cos θ2 + i sin θ2)we have

z1 z2 = r1[cos(θ1) + i sin(θ1)]r2[cos(θ2) + i sin(θ2)]= r1 r2[cos(θ1 + θ2) + i sin(θ1 + θ2)].

We can use this property several times to get the identity above step by step.

z1z2 · · ·zn = (z1 z2)z3 · · ·zn

= (r1(cos(θ1) + i sin(θ1)) r2(cos(θ2) + i sin(θ2)))z3 · · · zn

= (r1 r2)(cos(θ1 + θ2) + i sin(θ1 + θ2))z3z4 · · ·zn

= (r1 r2)(cos(θ1 + θ2) + i sin(θ1 + θ2))r3(cos(θ3) + i sin(θ3))z4 · · · zn

= (r1 r2 r3)(cos(θ1 + θ2 + θ3) + i sin(θ1 + θ2 + θ3))z4z5 · · ·zn

= · · ·= (r1 r2 · · · rn−1)(cos(θ1 + θ2 + · · ·+ θn−1) + i sin(θ1 + θ2 + · · ·+ θn−1))zn

= (r1 r2 · · · rn−1)(cos(θ1 + θ2 + · · ·+ θn−1) + i sin(θ1 + θ2 + · · ·+ θn−1))·rn(cos(θn) + i sin(θn))

= (r1 r2 · · · rn−1 rn)·(cos(θ1 + θ2 + · · ·+ θn−1 + θn) + i sin(θ1 + θ2 + · · ·+ θn−1 + θn))


33. First, we express the number i in the polar form:

i = 1(cos(π/2) + i sin(π/2)).

Now the solutions of the equation z4 = i are the 4th roots of i. So, by the formula for the nth rootswith n = 4 we find the solutions

z1 = (cos(π/8) + i sin(π/8)), z2 = (cos(5π/8) + i sin(5π/8)),

z3 = (cos(9π/8) + i sin(9π/8)), and z4 = (cos(13π/8) + i sin(13π/8)).

37. In order to solve this equation we need to find the 7-th root of (−1). Thus we express (−1) inthe polar form. −1 = cos(π) + i sin(π). Hence by the formula for the n-th root with n = 7 we havethat the solutions are

z1 = cos(π/7) + i sin(π/7), z2 = cos(3π/7) + i sin(3π/7),z3 = cos(5π/7) + i sin(5π/7), z4 = cos(π) + i sin(π/7),z5 = cos(9π/7) + i sin(9π/7), z6 = cos(11π/7) + i sin(11π/7), andz7 = cos(13π/7) + i sin(13π/7).

41. First, we can make a substitution and put w = z − 5. Then we need to solve the equation

w3 = −125 = 125(cos(π) + i sin(π)).

The solution set is given by the n-th root formula:

w1 = 3√

125(cos(π/3) + i sin(π/3)) = 5(1/2 + i√

3/2),

w2 = 3√

125(cos(π) + i sin(π)) = 5(−1 + i 0), and

w3 = 3√

125(cos(5π/3) + i sin(5π/3)) = 5(1/2 − i√

3/2).

Recalling that z = w + 5 we conclude that the solutions for the original problem are

z1 = 5 + w1 =152

+ i5√

32

, z2 = 5 + w2 = 0, and z3 = 5 + w3 =152

− i5√

32

.

45. We can use the formula for the solution of the quadratic equation az2 + bz + c = 0 given by

z =−b ±

√b2 − 4ac

2a.

We have in our problem a = 1, b = 1, and c = i4 . Therefore the solution is

z =−1 ±

√1 − i

2.

Therefore we need to find all possible square roots of (1 − i). So, we express (1 − i) in the polarform:

1 − i = 21/2(cos(−π

4

)+ i sin

(−π

4

)).

Hence two square roots of this number are

w1 = 21/4(cos(−π

8

)+ i sin

(−π

8

)), and w2 = 21/4

(cos(

7π

8

)+ i sin

(7π

8

)).


And, of course w1 = −w2. Thus the solution for the original problem is

z1 =−1 + w1

2, and z2 =

−1 − w1

2.

Also since −π/2 < −π8

< 0 by the half-angle formulas we find

cos(−π

8

)=

√1 + cos(π/4)

2=

√12

+√

24

and

sin(−π

8

)= −

√1 + sin(π/4)

2= −

√12

+√

24

.

Finally we can write down the solutions for the original equation in the form with radicals:

z1 =12

+21/4

2

√12

+√

24

− i21/4

2

√12

+√

24

, and

z2 =12− 21/4

2

√12

+√

24

+ i21/4

2

√12

+√

24

.

49. By De Moivre’s Identity with n = 4 we have

(cos θ + i sinθ)4 = cos 4θ + i sin 4θ.

Therefore the real part of the number z = (cos θ + i sinθ)4 is equal to cos 4θ. So, we can directlyevaluate the number z:

(cos θ + i sinθ)4 =((cos θ + i sinθ)2

)2

= (cos2 θ + 2i cos θ sin θ + i2 sin2 θ)2

= (cos2 θ + 2i cos θ sin θ − sin2 θ)2

= (cos2 θ)2 + (2i cos θ sin θ)2 + (sin2 θ)2 + 2(cos2 θ)(2i cos θ sin θ)+2(2i cos θ sin θ)(− sin2 θ) + 2(− sin2 θ) cos2 θ

(In the last equality we used the formula (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca which is easyto verify by direct computation similar to (a + b)2 = a2 + 2ab + b2.)

= cos4 θ − 4 cos2 θ sin2 θ + sin4 θ + 4i cos3 θ sin θ − 4i cos θ sin3 θ − 2 sin2 θ cos2 θ

=(cos4 θ − 6 cos2 θ sin2 θ + sin4 θ

)+ i

(cos3 θ sin θ − 4 cos θ sin3 θ

).

Finally, we see that the real part of the number z is equal to(cos4 θ − 6 cos2 θ sin2 θ + sin4 θ

)and is

equal to cos 4θ by De Moivre’s Identity. That is why these two trigonometric expressions are equalto each other.

53. (a) We compute(ω0 ωj)n = ωn

0 ωnj = 1 · 1 = 1,

since both numbers ω0 and ωj are nth roots of unity. This verifies that z = ω0 ωj satisfies zn = 1and thus is an nth root of 1.

(b) We see that ω0 ωj − ω0 ωk = ω0(ωj − ωk) 6= 0 because ω0 6= 0 and ωj 6= ωk since they aredifferent nth roots of unity. This verifies that ω0 ωj 6= ω0 ωk.


Also since all the complex numbers ω0 ωj for 1 ≤ j ≤ n are different, are nth roots of unity bythe part (a), and there are exactly n nth roots of unity we conclude that the set (ω0 ωj)

nj=1 is the

same as the set of all n roots of unity.

(c) First, we choose some nth root of unity ω0 which is different from 1 (such one exists because allroots are different and there are at least 2 of them since n ≥ 2). Since by the part (b) the set of allnth roots of unity (ωj)

nj=1 is the same as the set (ω0 ωj)

nj=1 we see that the sum of all the numbers

from the first set will be the same as the sum of all the numbers from the second one. That is

ω1 + ω2 + · · ·+ ωn = ω0 ω1 + ω0 ω2 + · · ·+ ω0 ωn, or

ω1 + ω2 + · · ·+ ωn = ω0(ω1 + ω2 + · · ·+ ωn).

Now if we subtract the right hand side of the equation from both sides of it and factor out the sumof all nth roots of unity we find that

(1 − ω0)(ω1 + ω2 + · · ·+ ωn) = 0.

This means that either 1− ω0 = 0, or ω1 + ω2 + · · ·+ ωn = 0. But the first equation cannot be truebecause ω0 is different from 1. Therefore we conclude that

ω1 + ω2 + · · ·+ ωn = 0.

(d) Let us call z = 1 + ω0 + ω20 + · · ·+ ωn−1

0 . If z is not zero then the number z(1− ω0) is also notzero because 1 − ω0 6= 0 (we remember that ω0 is different from 1). So, we evaluate

z(1 − ω0) = (1 + ω0 + ω20 + · · ·+ ωn−1

0 )(1 − ω0)= 1(1 − ω0) + ω0(1 − ω0) + · · ·+ ωn−2

0 (1 − ω0) + ωn−10 (1 − ω0)

= 1=0︷︸︸︷

−ω0 + ω0

=0︷︸︸︷−ω2

0 + ω20 −ω3

0 + · · ·+ ωn−20

=0︷︸︸︷−ωn−1

0 + ωn−10 −ωn

0

= 1 − ωn0 = 0,

since ω0 is an nth root of unity. Hence we have

z(1 − ω0) = 0,

and we divide by (1 − ω0) 6= 0 to show that z = 0.

57. (a) We can use a formula from the problem 56. For n = 1, 2, . . .,

cos nθ =[n2 ]∑

k=0

(n2k

)(cos θ)n−2k(−1)k(sin θ)2k

Since for the trigonometric functions there is an identity sin2 θ = 1 − cos2 θ we can use it in theformula above to get

cos nθ =[n2 ]∑

k=0

(n2k

)(cos θ)n−2k(−1)k(sin2 θ)k

=[n2 ]∑

k=0

(n2k

)(cos θ)n−2k(−1)k(1 − cos2 θ)k


Now if we substitute x = cos θ into the formula above we find

Tn(x) = Tn(cos θ) = cos nθ =[n2 ]∑

k=0

(n2k

)(cos θ)n−2k(−1)k(1 − cos2 θ)k

=[n2 ]∑

k=0

(n2k

)xn−2k(−1)k(1 − x2)k.

(b) We can use the general formula for Chebyshev polynomials proved above for particular n. Wehave

T0(x) =[ 02 ]∑

k=0

(02k

)(−1)kx0−2k(1 − x2)k =

0!0!0!

x0(1 − x2)0 =1 · 11

· 1 · 1 = 1.

For n = 1 we have

T1(x) =[12 ]∑

k=0

(12k

)(−1)kx1−2k(1 − x2)k =

1!0!1!

x1−2·0(1 − x2)0 =1 · 11

· x · 1 = x.

For n = 2 we have

T2(x) =[ 22 ]∑

k=0

(22k

)(−1)kx2−2k(1 − x2)k

=2!

2!0!x2−2·0(1 − x2)0 + (−1)1

2!0!2!

x2−2·1(1 − x2)

= x2 − (1 − x2) = 2x2 − 1.

For n = 3 we get

T3(x) =[32 ]∑

k=0

(32k

)(−1)kx3−2k(1 − x2)k

=3!

3!0!x3(1 − x2)0 − 3!

1!2!x1(1 − x2)1

= x3 − 3x(1 − x2) = 4x3 − 3x.

For n = 4 we get

T4(x) =[42 ]∑

k=0

(42k

)(−1)kx4−2k(1 − x2)k

=4!

4!0!x4(1 − x2)0 − 4!

2!2!x2(1 − x2)1 +

4!0!4!

x0(1 − x2)2

= x4 − 6x2(1 − x2) + (1 − 2x2 + x4) = 8x4 − 8x2 + 1.

Finally for n = 5 we have

T5(x) =[52 ]∑

k=0

(52k

)(−1)kx5−2k(1 − x2)k

=5!

5!0!x5(1 − x2)0 − 5!

3!2!x3(1 − x2)1 +

5!1!4!

x1(1 − x2)2

= x5 − 10x3(1 − x2) + x(1 − 2x2 + x4) = 16x5 − 20x3 + 5x.



1. We computef(3 + 4i) = (3 + 4i) + i = 3 + 5i.

5. In order to find the principal root we need to express the value of the expression (z2 + 2) forz = i + 1 in the polar form

z2 + 2 = (i + 1)2 + 2 = i2 + 2i + 12 + 2 = 2i + 2 =√

8(cos(π

4

)+ i sin

(π

4

)).

Then using the formula for the principal root, that is

(r(cos θ + i sin θ)1/2 = r1/2(cos(θ/2) + i sin(θ/2))

we find the needed value:

f(1 + i) =(√

8(cos(π

4

)+ i sin

(π

4

)))1/2

= (√

8)1/2(cos(π

8

)+ i sin

(π

8

)).

The values of cos(

π8

)and sin

(π8

)can be found by the half-angle formulas:

cos2θ

2=

1 + cos θ

2and sin2 θ

2=

1− cos θ

2.

We can use them for θ = π/4.

cos2π

8=

1 +√

22

2and sin2 θ

2=

1 −√

22

2.

Finally we substitute these values to the expression found above to get

f(1 + i) = 81/4

(12

+√

24

+ i

(12−

√2

4

)).

9. We compute for z = x + iy

f(z) = z3 = (x + iy)3 = x3 + 3x2 i y + 3x(iy)2 + (iy)3 = x3 − 3xy2 + i(3x2y − y3).

Hence we find u(x, y) = x3 − 3xy2 and v(x, y) = 3x2y − y3.

13. We know that for z = r(cos θ + i sin θ) 6= 0, we have

1z

=1r(cos(−θ) + i sin(−θ)).

So, we see that the formula makes sense for any complex number z except when z = 0. Hence thelargest subset of C where this function makes is all complex plane C except z = 0.

17. We again use the formula for the principal root. For the complex number z = r(cos θ + i sin θ)we have

z12 = r

12

(cos(

θ

2

)+ i sin

(θ

2

)).

Since r ≥ 0 and the trigonometric functions cos θ and sin θ make sense for any θ we see that theprincipal root is defined for any z. And it follows that f(z) is defined for all complex numbers z.

Section 1.4 Complex Functions 15

21. We see that S is the disc of radius 1 with center at the origin. Now if we multiply anynumber z from S by 4 we get a point inside the disc of radius 4 centered at the origin which isS1 = {z : |z| < 4}. This means that the image f(S) is contained in S1. On the other hand if z1 isin S1 then z1

4 is in the original disc S and f(

z14

)= 4 z

4 = z. And this means that S1 is in the imagef(S). So, S1 is contained in f(S) and contains f(S). This is only possible if

f(S) = S1 = {z : |z| < 4}.

25. (a) We compute

f(g(z)) = ag(z) + b = a(cz + d) + b = acz + ad + b = (ac)z + (ad + b).

This means that f(g(z)) is also linear.(b) First, we express the number a in the polar form a = r(cos θ + i sin θ). If we substitute thisvalue to the function f(z) we obtain

f(z) = r(cos θ + i sin θ)z + b = ((cos θ + i sin θ)(rz)) + b.

So, if we take g1(z) = z + b, g2(z) = (cos θ + i sin θ)z, and g3(z) = rz we can find a representationfor f(z):

f(z) = g1((cos θ + i sin θ)(rz)) = g1(g2(rz)) = g1(g2(g3(z))).

And we notice that g1 is a translation, g2 is a rotation, and g3 is a dilation.

29. If a real number z is positive then we can express it in the polar form z = r = r(cos(0)+i sin(0))for r > 0. And it follows that Argz = 0. If a real number z is negative then we can express it inthe polar form z = −r = r(cos(π) + i sin(π)) for r > 0. And it follows that Argz = π. So, it followsthat the image of the set S is two numbers 0 and π.

33. If we write z = r(cos θ + i sin θ), then f(z) = 1z = 1

r (cos(−θ) + i sin(−θ)). Hence the polarcoordinates of w = f(z) = ρ(cos φ+i sin φ) are 1

3 < ρ < ∞, and −2π3 ≤ Arg w ≤ −π

3 . As r increasesfrom 0 to 3, ρ decreases from ∞ to 1

3 ; and as θ goes from π3 up to 2π

3 , φ decreases from (−π3 ) to

(−2π3

).

37. (a) From the equation for f(z) we get f(z) · z = 1. So, since z 6= 0 and 1 6= 0 it follows thatf(z) 6= 0 (otherwise the product of f(z) and z would be 0, not 1).(b) We know from the part (a) that for any z from S we have f(z) 6= 0. Since f [S] is simply theunion of all possible values of f(z) when z is from S it follows that f [S] cannot include zero.(c) We compute

f(f(z)) =1

f(z)=

11z

=1 · z1z· z

=z

1= z.

(d) We know that f [f [S]] is the set of all complex numbers w such that w = f(u) when u is fromf [S]. But then u = f(z) for some complex number z from S. And, if we use the identity from thepart (c) we get w = f(u) = f(f(z)) = z. So, it follows that f [f [S]] is contained in S. On the otherhand if z is in S then f(z) is in f [S] and f(f(z)) is in f [f [S]]. So, it follows that S is contained inf [f [S]]. Combining these two facts we conclude that f [f [S]] = S.

41. We need to solve the equation 1z

= z which is equivalent to z2 = 1. The solutions of the lastequation are the square roots of 1 which are +1 and −1. Hence the fixed points are ±1.

45. (a) We know that w is in f [L]. This means that there is z = m + i n with integers m and nsuch that z2 = w. By the way we can compute

w = z2 = (m + i n)2 = m2 + 2i mn + (i n)2 = m2 − n2 + 2i mn


Now take m1 = −n and n1 = m. Then for z1 = m1 + i n1 we compute

z21 = (m1 + i n1)2 = m2

1 − n21 + 2i m1n1 = (−n)2 − m2 + 2i (−m)n

= −(m2 − n2 + 2i mn) = −w.

Hence −w is also from f [L].

If we take z2 = z = m − i n then we find

z22 = (z)2 = z2 = w.

Hence w is from f [L].

Finally, we can compute −Re w + i Im w = −( Re w − i Imw) = −w. Since we already provedu = w is from f [L]. And, for u from f [L] we have (−u) is from f [L] as well. We can conclude that−Re w + i Imw = −u is from f [L].(b) This part follows from the formula given in the proof of the part (a). That is for z = m + i n

w = z2 = (m + i n)2 = m2 + 2i mn + (i n)2 = (m2 − n2) + i(2 mn).

We notice that (m2 − n2) is integer and (2mn) is integer. So, w = z2 is also from L.(c) By the part (b) it follows that w is in L. Hence the function f maps the number w to f(w)which is already in f [L]. And this is what we needed to prove.

Section 1.5 The Complex Exponential 17


1. We compute

eiπ = e0+iπ = e0(cos π + i sin π) = 1(−1 + i 0) = −1.

5. We compute

ei 3π4 = e0+i 3π

4 = e0

(cos

3π

4+ i sin

3π

4

)

= 1

(−√

22

+ i

√2

2

)= −

√2

2+ i

√2

2.

9. We evaluate

3e3+i π2 = 3e3

(cos

π

2+ i sin

π

2

)= 3e3 (0 + i 1) = 3e3 i.

13. For z = −1 −√

3 i we have that the radius

r = |z| =√

(−1)2 + (√

3)2 =√

4 = 2.

Since z is in the third quadrant, we have Arg z = tan−1(√

31

)+ π = 4π

3.

17. First, we can find the standard representation for the number z = 1+i1−i by dividing and multi-

plying by the conjugate of the denominator

1 + i

1 − i=

(1 + i)(1 + i)(1 − i)(1i)

=1 + 2 i + i2

12 + 12=

2i

2= i.

Then we find |z| =√

02 + 12 = 1. And, since the number z = i is pure imaginary we find thatArg z = π

2 . Therefore

z = i = 1(cos

π

2+ i sin

π

2

)= 1 · e

π2 .

21. We can notice that we need to find only the radius of the given complex number z = e−i π12 .

And, if we write it in the polar form we have

z = e−i π12 = e0−i π

12 = 1(cos

π

12− i sin

π

12

).

From this it follows that∣∣e−i π

12∣∣ = |z| = 1.

25. (a) For z = x + i y we have

Re e3z = Re e3(x+i y) = Re(e3x (cos 3y + i sin 3y)

)= e3x cos 3y.

Similar we have

Im e3z = Im(e3x (cos 3y + i sin 3y)

)= e3x sin 3y.


(b) For z = x + i y we first compute z2 = (x + i y)2 = x2 + 2i xy + (i y)2 = x2 − y2 + 2xy i. Thenwe find

Re ez2= Re ex2−y2+2xy i

= Re(ex2−y2

(cos 2xy + i sin 2xy))

= ex2−y2cos 2xy.

Similar we have

Im ez2= Im

(ex2−y2

(cos 2xy + i sin 2xy))

= ex2−y2sin 2xy.

(c) For z = x + i y we compute

Re ez = Re ex−i y

= Re ex (cos(−y) + i sin(−y))= ex cos(−y).

Similar we compute

Im ez = Im ex−i y

= Im ex (cos(−y) + i sin(−y))= ex sin(−y).

(d) For z = x + i y we compute

Re eiz = Re ei x−y

= Re e−y (cos(x) + i sin(x))= e−y cos(x).

Similar we compute

Im eiz = Im ei x−y

= Im e−y (cos(x) + i sin(x))= e−y sin(x).

29. Any complex number z inside the square S can be expressed in the form z = x + i y wherex1 < x < x2 and α1 < y < α2. We can compute

f(z) = ez = ex+i y = ex (cos(y) + i sin(y))

This means that in the polar form the radius of the number f(z) is ex. And, since x is a real numberbetween x1 and x2 then ex is any number between ex1 and ex2 . The argument of the complexnumber f(z) is equal to y. And, y is any number between α1 and α2. Therefore, f [S] is a sectorof all complex numbers having in the polar form the radius between ex1 and ex2 and the argumentbetween α1 and α2.

33.(a) Since we know that

sin θ =eiθ − e−iθ

2i

Section 1.5 The Complex Exponential 19

we compute

sin4 θ =(

eiθ − e−iθ

2i

)4

=116(e4iθ − 4e3iθe−iθ + 6e2iθe−2iθ − 4eiθe−3iθ + e−4iθ

)

=18

(e4iθ + e−4iθ

2− 4

e2iθ + e−2iθ

2+ 6)

=18

(cos 4θ − 4 cos 2θ + 6) .

(b) We compute the integral using the linearization we received in (a).∫

sin4 θ dθ =∫

18

(cos 4θ − 4 cos 2θ + 6) dθ =132

sin 4θ − 14

sin 2θ +34θ + C.

37. For a complex number z = x + i y we can compute |ez| and e|z| separately. We have

|ez| = |ex+i y| = |exeiy| = |ex| = ex.

Also we have

e|z| = e|x+i y| = e√

x2+y2

Then we use a property of the exponential function. For real numbers x1 ≤ x2 we have ex1 ≤ ex2 .Now combining with the fact that x ≤

√x2 ≤

√x2 + y2 we conclude

|ez| = ex ≤ e√

x2+y2= e|z|

and the equality happens only if x =√

x2 + y2. And this is true only if y = 0 and x ≥ 0. Or inother words when z is a non-negative real number.

Complex Numbers and Functions

Solutions to Exercises 1.61. We have z = i. Then we evaluate using the definition of cos(z) and sin(z)

cos(z) =eiz + e−iz

2=

ei2 + e−i2

2=

e−1 + e1

2=

12e

+e

2= cosh 1,

which verifies (2) and (15); and

sin(z) =eiz − e−iz

2i=

ei2 − e−i2

2i=

e−1 − e1

2i

=

(e−1 − e

)(−i)

2i(−i)=

i e

2− i

2ei sinh 1,

which verifies (3) and (16).

5. (a) For z = 1 + i we evaluate using the definition of cos(z)

cos(z) =eiz + e−iz

2

=ei(1+i) + e−i(1+i)

2=

ei−1 + e1−i

2

=e−1 (cos(1) + i sin(1)) + e1 (cos(−1) + i sin(−1))

2

=e−1 (cos(1) + i sin(1)) + e1 (cos(1) − i sin(1))

2

= cos(1)e−1 + e1

2− i sin(1)

e1 − e−1

2= cos(1) cosh(1) − i sin(1) sinh(1).

Similar we evaluate sin(z)

sin(z) =eiz − e−iz

2i

=ei(1+i) − e−i(1+i)

2i=

ei−1 − e1−i

2i

=e−1 (cos(1) + i sin(1)) − e1 (cos(−1) + i sin(−1))

2i

=1i· e−1 (cos(1) + i sin(1)) − e1 (cos(1) − i sin(1))

2

= (−i) ·(

cos(1)e−1 − e1

2i+ i sin(1)

e1 + e−1

2

)

= sin(1) sinh(1) + i cos(1) cosh(1),

and tan(z)

tan(1 + i) = tan(z) =sin(z)cos(z)

=sin(1) sinh(1) + i cos(1) cosh(1)cos(1) cosh(1) − i sin(1) sinh(1)

=(sin(1) sinh(1) + i cos(1) cosh(1)) (cos(1) cosh(1) + i sin(1) sinh(1))(cos(1) cosh(1) − i sin(1) sinh(1)) (cos(1) cosh(1) + i sin(1) sinh(1))

= icos2(1) cosh2(1) + sin2(1) sinh2(1)cos2(1) cosh2(1) + sin2(1) sinh2(1)

Section 1.6 Trigonometric and Hyperbolic Functions 21

Where we used the computed values of cos(1 + i) and sin(1 + i) for evaluating tan(1 + i).(b) In this part we use the formulas REF(17) and (18) for the values of | cos z| and | sin z|. We

have for z = x + i y with x = 1 and y = 1

| cos(1 + i)| = | cos z| =√

cos2 x + sinh2 y =√

cos2(1) + sinh2(1),

and

| sin(1 + i)| = | cos z| =√

sin2 x + sinh2 y =√

sin2(1) + sinh2(1).

(c) From the part (a) we see that cos(1 + i) is represented by the point with the coordinates(cos(1) cosh(1),− sin(1) sinh(1)), sin(1+i) is represented by the point with the coordinates (sin(1) sinh(1), cos(1) sinh(1)),and tan(1 + i) has coordinates

(0,

cos2(1) cosh2(1) + sin2(1) sinh2(1)cos2(1) cosh2(1) + sin2(1) sinh2(1)

).

9. We use the definition of sin z. We have for z = x + i y

sin(2z) =ei2z − e−i2z

2i

=ei2(x+i y) − e−i2(x+i y)

2inow we use polar representation

=e−2y(cos(2x) + i sin(2x)) − e2y(cos(−2x) + i sin(−2x))

2i

=(e−2y cos(2x) − e2y cos(−2x)) + i (e−2y sin(2x) − e2y sin(−2x))

2i

(−i)(−i)

= −i cos(2x)e−2y − e2y

2+ sin(2x)

e−2y + e2y

2

Therefore we have sin(2z) = u(x, y)+i v(x, y) for u(x, y) = sin(2x) cosh(2y) and v(x, y) = cos(2x) sinh(2y).

13. For z = x + i y we evaluate using the formulas REF(15)-(16) for sin z and cos z as well as theformula REF(17) for | cos z|

tan z =sin z

cos z=

sin zcos z

cos zcos z=

sin zcos z

| cos z|2[by formulas REF(15)-(17)]

=(sin x cosh y + i cos x sinh y)(cos x cosh y + i sinx sinh y)

cos2 x + sinh2 y

=cos x sinx(cosh2 y − sinh2 y) + i sinh y cosh y(cos2 x + sin2 x)

cos2 x + sinh2 y

[by properties REF(8) and (27)]

=cos x sinx + i sinh y cosh y

cos2 x + sinh2 y

Therefore we have tan z = u(x, y)+i v(x, y) where u(x, y) = cosx sin xcos2 x+sinh2 y

and v(x, y) = sinh y coshycos2 x+sinh2 y

.

17. As in the Example 4, we will first find the image under f of a simple curve in the domain ofdefinition, often a line segment or line. Then we will sweep the domain of definition with this curveand keep track of the area that is swept by the image. Fix α < y0 < β and consider the horizontal


line segment EF defined by: y = y0, −π2≤ x ≤ π

2. Let u+i v denote the image of a point z = x+i y0

on EF . Using (??), we get

u + i v = sin(x + i y0) = sin x cosh y0 + i cos x sinh y0.

Henceu = sin x cosh y0 and v = cos x sinh y0.

Therefore

(1)u

cosh y0= sin x and

v

sinh y0= cos x.

Note that v ≥ 0 because cos x ≥ 0 for −π2 ≤ x ≤ π

2 . Squaring both equations in (1) then addingthem, we get (

u

cosh y0

)2

+(

v

sinh y0

)2

= sin2 x + cos2 x = 1.

Hence as x varies in the interval −π2 ≤ x ≤ π

2 , the point (u, v) traces the upper semi-ellipse

(u

cosh y0

)2

+(

v

sinh y0

)2

= 1, v ≥ 0.

The u-intercepts of the ellipse are at u = ± cosh y0 and the v-intercept is at v = sinh y0. As y0 variesfrom α to β the upper semi-ellipses will vary between two upper semi-ellipse given by y0 = α andy0 = β whose u- and v-intercepts are at u = ± cosh α, v = sinh α and u = ± cosh β, v = sinh β.When y0 changes from α to β continously the corresponding upper semi-ellipses will continouslychange from the ellipse corresponding to y0 = α to the ellipse corresponding to y0 = β and thusfill-in the shadowed area on Figure 4.

21. To prove (??), we appeal to (??) and (??)–(??) and write

sin z = sin(x + i y)= sin x cos(iy) + cos x sin(iy)= sin x cosh y + i cos x sinh y.

To prove (??), we use (??) and the definition of the modulus of a complex number ((??), Section1.2). We also use the identity cosh2 y − sinh2 y = 1 for real hyperbolic functions. We get

| sin z|2 = sin2 x cosh2 y + cos2 x sinh2 y

= sin2 x (1 + sinh2 y) + cos2 x sinh2 y

= sin2 x + sinh2 y(cos2 x + sin2 x)= sin2 x + sinh2 y.

Computing the square roots of both sides of the received identity we get (??) proved.

25. To establish the identity we use the definition of sin z.

sin(−z) =e−iz − eiz

2i= −eiz − e−iz

2i= − sin z.

29. sin(z1 + z2) = sin z1 cos z2 + cos z1 sin z2. Expanding the right-hand side of the identity above


we have

(eiz1 − e−iz1)(eiz2 + e−iz2)4i

+(eiz1 + e−iz1)(eiz2 − e−iz2)

(4i

[expanding the numerators we have]

=ei(z1+z2) + ei(z1−z2) − ei(−z1+z2) − ei(−z1+z2)

4i

+ei(z1+z2) + ei(z2−z1) − ei(z1−z2) − e−i(z1+z2)

4i[adding the fractions and canceling the middle terms we receive]

=2ei(z1+z2) − 2e−i(z1+z2)

4i

=ei(z1+z2) − e−i(z1+z2)

2i= sin(z1 + z2).

33. Using already proved properties (??) and (??) of cos z we expand the right-hand side of theidentity above

(cos z1 cos−z2 − sin z1 sin−z2) − (cos z1 cos z2 − sin z1 sin−z2)= cos z1 cos z2 + sin z1 sin z2 − cos z1 cos z2 + sin z1 sin z2

= 2 sin z1 sin z2.

37. We use the definitions of cosh z and sinh z to get

cosh(z + πi) =ez+πi + e−z−πi

2=

ezeπi + e−ze−πi

2

=ez(−1) + e−z(−1)

2= −ez + e−z

2= − cosh z, and

sinh(z + πi) =ez+πi − e−z−πi

2=

ezeπi − e−ze−πi

2

=ez(−1) − e−z(−1)

2= −ez − e−z

2= − sinh z.

41. First, we use the property (??) or the problem 40 twice to show that the second, the third,and the fourth sides of the formula are equal to each other. We have

cosh2 z + sinh2 z = cosh2 z + (cosh2 z − 1) = 2 cosh2 z − 1= 2(1 + sinh2 z) − 1 = 1 + 2 sinh2 z.

Finally, we show that cosh2 z + sinh2 z is equal to the left-hand side. We have

cosh2 z + sinh2 z =(ez + e−z)2

4+

(ez − e−z)2

4[adding the fractions]

=(e2z + 2 + e−2z) + (e2z − 2 + e−2z)

4

=e2z + e−2z

2= cosh 2z.


45. We use the definitions of the hyperbolic functions to prove this identity. We start from theright-hand side to show it is equal to the left-hand side.

cosh z1 cosh z2 + sinh z1 sinh z2

=ez1 + e−z1

2ez2 + e−z2

2+

ez1 − e−z1

2ez2 − e−z2

2[expanding the numerators we get]

=ez1+z2 + ez1−z2 + e−z1+z2 + e−z1−z2

4+

ez1+z2 − ez1−z2 − e−z1+z2 + e−z1−z2

4[adding the fractions and simplifying we get]

=2ez1+z2 + 2e−z1−z2

4=

ez1+z2 + e−z1−z2

2= cosh(z1 + z2).

49. We show that the left-hand side is equal to the right-hand side by using the result of theproblem 46 and the facts that sinh(−z) = − sinh z and cosh(−z) = cosh z which can proved bydirect computation (see the solution of a similar problem 25). By problem 46 the right-hand side isequal to

(sinh z1 cosh z2 + cosh z1 sinh z2) + (sinh z1 cosh(−z2) + cosh z1 sinh(−z2))= (sinh z1 cosh z2 + cosh z1 sinh z2) + (sinh z1 cosh(z2) − cosh z1 sinh z2)= 2 sinh z1 cosh z2.

53. If tan z = i had a solution then this would have meant that sin z = i cos z. Now we use (??)and (??) to show that the assumption leads to a contradiction. Really, we would have for z = x+ i y

sin x cosh y + i cos x sinh y = i cosx cosh y + sin x sinh y,

or equivalently

sin x(cosh y − sinh y) = i cos x(cosh y − sinh y).

We have a real number in the left-hand side and an imaginary number in the right-hand side. Theycan be equal only if they are both zeroes. Now we use the fact that the expressions in parenthesisnever are equal to zero. Really, we compute

cosh y − sinh y =ey + e−y

2− ey − e−y

2

=ey + e−y − ey + e−y

2= −e−y 6= 0

because the exponential function is never equal to zero.At the same time it is impossible that sin x and cos x are both equal to 0 because of the trigono-

metric identity cos2 x + sin2 x = 1. Therefore, we conclude that our assumption was wrong andtan z = i does not have any solutions.

Similar we show that the equation tan z = −i does not have solutions. We use again (??) and(??) to show that if tan z = −i for some z = x + i y then

sin x cosh y + i cos x sinh y = −i cosx cosh y − sin x sinh y,


or equivalently

sin x(cosh y + sinh y) = −i cos x(cosh y + sinh y)

which leads us to

(sin x + i cos x)(cosh y + sinh y) = 0.

Now by direct computation we show that cosh y + sinh y = ey 6= 0 and sin x + i cos x 6= 0 sincewe always have sin2 x + cos2x = 1. Thus, the assumption tan z = −i is wrong and the equationtan z = −i also does not have any solutions.



1. We will use the formula (??) for evaluating log(2 i). We have

log(2 i) = ln |2 i| + i arg (2 i) = ln 2 + i (π

2+ 2kπ)

where k is an integer. We remember that log z is a multiple-valued function.

5. Now we will use the formula (??) for evaluating Log (3 + i√

3). We have

Log (3 + i√

3) = ln |3 + i√

3|+ i Arg (3 + i√

3)

Since for the number z = 3 + i√

3 = x + i y we have x > 0 we compute

Arg (3 + i√

3) = Arg z = tan−1(y

x) = tan−1(

√3

3) =

π

6.

Now we can substitute the found value to the formula (1) to find

Log (3 + i√

3) = ln√

32 + 3 + iπ

6=

ln 122

+ iπ

6= .

9. We will use the formula (??) to evaluate logπ 1. We have

logπ 1 = ln |1|+ i arg π1.

So, we need to compute arg π1. Since arg 1 = 2kπ for all integer k we have π < arg π1 = 2π < π+2π.We substitute the found value to the formula (1) to get

logπ 1 = 0 + i 2π = i 2π.

13. We know that

ez = 3 ⇔ z = log 3.

Therefore the solutions of the equations are

z = log3 = ln |3|+ i 2kπ = ln3 + i 2kπ

for all integer k.

17. The equation is equivalent to e2z = −5. We know that

e2z = −5 ⇔ 2z = log(−5).

Therefore the solutions of the equations are

z =12

log(−5) =12(ln | − 5| + i (2k + 1)π) =

12

ln 5 + i(2k + 1)π

2,

for all integer k.

21. We will use the formula (??) for evaluating all three numbers

Log1 = ln |1|+ i Arg (1) = ln1 + i 0 = 0,

Section 1.7 Logarithms and Powers 27

Log (i) = ln |i|+ i Arg (i) = ln1 + iπ

2= i

π

2, and

Log (−i) = ln | − i| + i Arg (−i) = ln 1 + i3π

2= i

3π

2.

We see that for z1 = i and z2 = −i we have

Log z1 + Log z2 = iπ

2+ i

3π

2= i 2π.

On the other hand Log (z1z2) = Log (1) = 0. So, in general we see Log (z1z2) is not equal toLog z1 + Log z2.

25. We use the formula (9) for computing the principal value of zα. We have

(−5)1−i = e(1−i)Log (−5).

To evaluate this expression we need to compute the principal value of the logarithm of (−5). Weget

Log (−5) = ln | − 5|+ i Arg (−5) = ln 5 + i π.

Therefore, we conclude using the properties

(−5)1−i = e(1−i)(ln 5+i π)

= eln 5+π+i (− ln 5+π))

= eln 5+π(cos(− ln 5 + π) + i sin(− ln 5 + π))[since cos(x + π) = − cos(x) and sin(x + π) = − sin(x)]

= eln 5eπ(− cos(− ln 5) − i sin(− ln 5))= −5eπ(cos(− ln 5) + i sin(− ln 5))

[since cos(−x) = − cos(x) and sin(−x) = − sin(x)]= −5eπ(cos(ln 5) − i sin(ln 5)).

29. We use the definition of the complex power (??)to find

ii = ei log i = ei(ln |i|+i arg i)

= ei(0+i ( π2 +2kπ)) = e−

π2 −2kπ

for integer k. So, we see that ii has infinitely-many values and all of them are real numbers.

33. (a) Using the definition (??) in Section 1.6 we evaluate

1 + i tan w = 1 + isin w

cos w= 1 + i

eiw−e−iw

2ieiw+e−iw

2

= 1 +eiw − e−iw

eiw + e−iw=

eiw + e−iw

eiw + e−iw+

eiw − e−iw

eiw + e−iw

[adding two fractions and simplifying we get]

=2eiw

eiw + e−iw.


(b) Acting similar to the solution of the part (a) we evaluate

1 − i tan w = 1 − isin w

cos w= 1 − i

eiw−e−iw

2ieiw+e−iw

2

= 1 − eiw − e−iw

eiw + e−iw=

eiw + e−iw

eiw + e−iw− eiw − e−iw

eiw + e−iw

[subtracting two fractions and simplifying we get]

=2e−iw

eiw + e−iw.

(c) In order to prove this formula for finding the yet unknown value w of tan−1 z we will use thedefinition

tan−1 z = w ⇔ z = tan w.

To find the value of w we will explore the following expression

1 − i z

1 + i z=

1 − i tanw

1 + i tanw[now we use the formula from part (a) and (b)]

=2e−iw

eiw+e−iw

2eiw

eiw+e−iw

=e−iw

eiw= e−2iw.

So, we see that1 − i z

1 + i z= e−2iw

and this equation is equivalent to

−2iw = log1 − i z

1 + i z,

or if we divide both sides of the equation by (−2i) = 2i we get an equivalent equation

w =i

2log

1 − i z

1 + i z,

and since w = tan−1 z we have proved the needed formula.

37. (a) We use the definition of log z. We have

log(z−1) ={w : w = Log z−1 + 2kπi for some integer k

}

={

w : w = ln |z|−1 + i Arg1z

+ 2kπi for some integer k

}

={

w : w = − ln |z| + i Arg1z

+ 2kπi for some integer k

}.

Now we only need to notice that the principal value of the argument of z−1 is equal to the negativeprincipal argument of z in the case if Arg z 6= π. Really, if z = reiθ then z−1 = 1

r e−iθ for real r > 0and θ. So, if −π < θ < π then also −π < −θ < π. And, if Arg z = π then similar we computeArg z−1 = π. In either case we have that either Arg z−1 = −Arg z or Arg z−1 = −Arg z +2π. Andif follows that

log z−1 = {w : w = − ln z − i Arg z − 2kπi for some integer k}= {−w : w = ln z + i Arg z + 2kπi for some integer k}

Section 1.7 Logarithms and Powers 29

because if we add or subtract 2π from any number from the set above it is still in the same set.(b) We proceed similar to the solution of the part (a). We have

logz1

z2=

{w : w = Log

z1

z2+ 2kπi for some integer k

}

={

w : w = ln∣∣∣∣z1

z2

∣∣∣∣+ i Argz1


}

={

w : w = ln |z1| − ln |z2| + i Argz1


}.

Now we notice that the difference between two numbers Arg z1z2

and ( Arg z1 − Arg z2) is an integermultiple of 2π. Really, if z1 = r1e

iθ1 and z2 = r2eiθ2 for r1 > 0, r2 > 0, −π < θ1 ≤ π, and

−π < θ2 ≤ π then we also have z1z2

= r1r2

ei(θ1−θ2). And, since the exponential function is 2πi-periodicwe see that θ1 − θ2 differs from Arg z1

z2by only an integer multiple of 2π. Thus,

logz1

z2= {w : w = ln |z1| − ln |z2|+ i Arg z1 − i Arg z2 + 2kπi for some integer k} .

(c) To show that it is not true that log(z2) = 2 log z we choose z = 1. Then we evaluate

2 log z = 2 log1 = 2 {w : w = 2kπi for some integer k}= {w : w = 4kπi for some integer k} .

On the other hand

log(z2) = log(12) = log1= {w : w = 2kπi for some integer k} .

In particular, we see that 2πi is log(z2) but not in 2 log z since 2π is not an integer multiple of 4π.To show that log(z2) ⊃ 2 log z we choose some complex number w from the set 2 log z. By

definition of the multiple-valued function f(z) = log z it follows that

ew2 = z.

If we square this identity we get

e2 w2 = z2,

or equivalently

ew = z2.

And this means that w is from the set log(z2).

Solutions to Exercises 1 - University of...

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