Exercises 4.

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Before we proceed, we recall that ez1 = ez2 if and only if z1 = z2 + 2kπi (k ∈ Z).

1. Find all values of z for which

(a) e3z = 1
Solution. We use the above-mentioned fact. For this purpose, we express the right-hand
side in the form of ew . Notice that 1 = e0 . Then it is easily checked that 3z = 0 + 2kπi
(k ∈ Z). Therefore z = 2kπi/3 (k ∈ Z). ////

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(b) ez = 1
Solution. Since 1 = e0 , it follows that z 2 = 2kπi (k ∈ Z). If k = 0 then z = 0. If k 6= 0
then we consider the polar representation for 2kπi:
( ¡ ¢
2kπ cos π2 + i sin π2 , k>0
2kπi = ¡ ¡ π¢ ¡ ¢¢ (1)
2|k|π cos − 2 + i sin − π2 , k < 0.

For k ∈ Z, we define zk as z0 = 0 and

( √
2kπ(cos(π/4) + i sin(π/4)), k>0
zk = p
2|k|π(cos(−π/4) + i sin(−π/4)), k < 0.

then zk is a root of z 2 = 2kπ. Therefore {zk | k ∈ Z} ∪ {−zk | k ∈ Z} is the set of all roots
of z 2 = 2kπi. ////

z
(c) ee = 1
Solution. Since ez =
6 0 for any z ∈ C, it follows that ez = 2kπi (k ∈ Z − {0}). Then we
can rewrite (1) as (
eln(2kπ)+πi/2 , k > 0
2kπi =
eln(2|k|π)−πi/2 , k < 0.

If k > 0, we obtain that

ez = 2kπi = eln(2kπ)+πi/2 ⇔ z = [ ln(2kπ) + πi/2] + 2nπi (n ∈ Z).

Moreover, if k < 0 then

ez = 2kπi = eln(2|k|π)−πi/2 ⇔ z = [ ln(2|k|π) − πi/2] + 2nπi (n ∈ Z).

Combining these two results, we conclude that

³ 1´
z = ln(2kπ) + i n + π (k ∈ N, n ∈ Z).
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2. Show that all the zeros of sin z and cos z are real.
Solution. (1) Note that sin z = 0 ⇔ eiz − e−iz = 0 ⇔ e2iz = 1 = e0 . Therefore every
zero of sin z takes the form
z = nπ (n ∈ Z).

(2) Note that cos z = 0 ⇔ eiz + e−iz = 0 ⇔ e2iz = −1 = eπi ⇔ 2iz = πi + 2nπ
(n ∈ Z). Therefore every zero of cos z takes the form
π
z = nπ + (n ∈ Z).
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3. (a) Show that both sin z and cos z are unbounded on the ray Arg z = θ, 0 < θ < π.
Solution. If z is located on the ray L : Arg z = θ (0 < θ < π) then z = |z|(cos θ + i sin θ).
Consequently

|eiz | = eRe (iz) = e−|z| sin θ and |e−iz | = e−Re (iz) = e|z| sin θ .

Since sin θ > 0, |e−iz | → ∞ and |eiz | → 0 as z → ∞ along the ray L. Therefore
1 iz 1
| sin z| = |e − e−iz | ≥ (|e−iz | − |eiz |) → ∞,
2 2
1 iz 1
| cos z| = |e + e | ≥ (|e−iz | − |eiz |) → ∞
−iz
2 2
as z → ∞ along the ray L. ////
(b) Show that sin z is bounded only on sets contained in a horizontal strip.
Solution. We first show that sin z is bounded on a horizontal strip. Let A = {z | c0 ≤
Im z ≤ c1 } be a horizontal strip, where c0 , c1 are real constants. Note that

|eiz | = eRe (iz) = e−Im z and |e−iz | = e−Re (iz) = eIm z .

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Then it follows that |eiz | ≤ e−c0 and |e−iz | ≤ ec1 , and consequently, | sin z| ≤ (|eiz | +
2
1
|e−iz |) ≤ (e−c0 + ec1 ) on A.
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On the other hand, choose a set B such that B is not contained in any horizontal strip. Then
we can choose a sequence {zn } in B such that |Im zn | → ∞. Without loss of generality,
we may assume that Im zn → ∞. Then follows that |eizn | = e−Im zn → 0 and |e−izn | =
eIm zn → ∞. Consequently | sin zn | ≥ 12 (|e−izn | − |eizn |) → ∞.
Therefore sin z is unbounded on a set B. ¤

4. For |z| = r, prove the following inequalities. When will equality hold?
Remark. Note that e−|w| ≤ |ew | = eRe w ≤ e|w| for w ∈ C. Here we used the fact that
−|w| ≤ Re w ≤ |w|.
(a) e−r ≤ |ez | ≤ er :

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 Solution. Since |ez | = eRe z and −|z| ≤ Re z ≤ |z|, the inequalities immediately follows.
Taking into account the proof of the inequality, we have that |ez | = er if and only if
Re z = |z|. Therefore |ez | = er if and only if z = r. Similarly, |ez | = e−r if and only if
z = −r. ¤
n n n
(b) e−r ≤ |ez | ≤ er , n ∈ N. (left as an exercise)

5. (left as exercises.)

6. Remark. By definition, |ew | = eRe w and arg (ew ) = Im w + 2nπ (n ∈ Z). Moreover, note
1 z ³ 1 ´ Re z ³1´ Im z
that = 2 , and consequently, Re = 2
and Im =− 2.
z |z| z |z| z |z|
(a) Separate e1/z , z 6= 0, into its real and imaginary parts.
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Solution. Since |e1/z | = eRe(1/z) = e(Re z)/|z| and arg(e1/z ) = −(Im z)/|z|2 + 2nπ (n ∈ N),
it follows from the polar representation that
2
³ Im z ´ 2
³ Im z ´
Re(e1/z ) = e(Re z)/|z| cos and Im(e1/z ) = e(Re z)/|z| sin .
|z|2 |z|2

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(b) Show that |e1/z | is bounded in the region |z| ≥ ² (² > 0).
Solution. Note that
¯ Re z ¯ |Re z| 1 1
¯ ¯
¯ 2¯= ≤ ≤ in the region |z| ≥ ² > 0.
|z| |z|2 |z| ²
2
Therefore |e1/z | = e(Re z)/|z| ≤ e1/² in that region. ¤

7. (a) Prove that eiz is periodic with period 2π.

Solution. Recall that ez1 = ez2 if and only if z1 = z2 + 2nπi (n ∈ Z). Consequently
eiz1 = eiz2 if and only if z1 = z2 + 2nπ (n ∈ Z). ////
(b) For an arbitrary nonzero complex number a, show that eaz is periodic, and find its period.
Solution. eaz1 = eaz2 ⇔ az1 = az2 + 2nπi (n ∈ Z) ⇔ z1 = z2 + 2nπi/a. The period
is 2πi/a. ¤

8. Prove the following identities. (z = x + iy)

2 2
−y 2
(a) |ez + ez | ≤ ex + ex
2 2 2
Solution. |ez + ez | ≤ |ez | + |ez | ≤ eRe z + eRe(z ) . ////

2
(b) |eiz + eiz | ≤ e−y + e−2xy
2 2 2 2
Solution. |eiz + eiz | ≤ |eiz | + |eiz | ≤ eRe(iz) + eRe(iz )
= e−Im z + e−Im(z ) . ////

(c) | sin z|2 + | cos z|2 ≥ 1

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 Solution. Recall that z − z = 2iIm z and z + z = 2Re z. Moreover, we note that ew = ew
for any w ∈ C (why?). Since iz = iz = −iz and −iz = −iz = iz,

|eiz ± e−iz |2 = (eiz ± e−iz )(eiz ± e−iz ) = (eiz ± e−iz )(eiz ± e−iz )
= (eiz ± e−iz )(e−iz ± eiz )
= ei(z−z) + e−i(z−z) ± e−i(z+z) ± ei(z+z)
= e−2Im z + e2Im z ± e−2iRe z ± e2iRe z .

Therefore we conclude that

1 iz 1 1
| sin z|2 + | cos z|2 = |e − e−iz |2 + |eiz + e−iz |2 = (e−2Im z + e2Im z )
4 4 2
1³ 1 2Im z
´
= +e ≥ 1.
2 e2Im z
Remark. Equality holds ⇔ e2Im z = 1 ⇔ Im z = 0 (∵ Im z is real). ¤

Exercises 4.2 (#3 may be skipped.)

Recall that ez = ex (cos y +i sin y), z = x+iy. Namely, |ez | = ex and arg (ez ) = y +2nπ (n ∈ Z).
If −π < y ≤ π then Arg(ez ) = y.

1. Find the image of the following sets under the transformation w = ez , and sketch.
(a) −5 ≤ x ≤ 5, y = π/4
Solution. e−5 ≤ ex = |w| ≤ e5 and Arg w = y = π/4. The image is a line segment
{(x, y) | y = x and e−10 ≤ x2 + y 2 ≤ e10 }. ////

(b) x = 3, −π/2 < y < π/2

Solution. |w| = ex = e3 and −π/2 < Arg w = y < π/2. The image is a semicircle
{(x, y) | x2 + y 2 = e6 and x > 0}. ////

(c) −2 < x < 1, 0 < y < π

Solution. e−2 < |w| = ex < e and 0 < Arg w = y < π. Therefore the image is the set
1
{(x, y) | 4 < x2 + y 2 < e2 and y > 0}. ////
e

(d) x < 1, −π/3 < y < 2π/3

Solution. 0 < |w| = ex < e and −π/3 < Arg w = y < 2π/3. Moreover lim ex = 0.
x→−∞
√
Therefore the image is the set {(x, y) | 0 < x2 + y 2 < e2 and y > − 3 x}. ¤

2. Find the image of the region 0 ≤ x ≤ π, y ≥ 0, for the transformation (z = x + iy)

(a) w = eiz

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 Solution. Note that eiz = e−y+ix = e−y (cos x + i sin x). Then it follows that 0 < e−y ≤ 1
and lim e−y = 0. Moreover, 0 ≤ Arg(eiz ) = x ≤ π. Consequently the image is {w |
y→∞
Im w ≥ 0, w 6= 0}. ////
Remark. The line segment {z | Im z = y, 0 ≤ x ≤ π} parallel to the x-axis, is mapped
onto the semicircle {w | |w| = e−y , Im w ≥ 0}.

(b) w = ieiz
Solution. Let f1 (z) = eiz and f2 (z) = iz. Then w = ieiz = (f2 ◦ f1 )(z) and f2 denotes a
rotation by π/2.
By the result in (a), we conclude that the image is {w | Re w ≤ 0, w 6= 0}. ////

(c) w = ie−iz
Solution. Note that ie−iz = eπi/2−iz = ey+i(π/2−x) , in other words,
³ ³π ´ ³π ´´
ie−iz = ey cos − x + i sin −x .
2 2
π π π
Since 0 ≤ x ≤ π, it follows that − ≤ −x ≤ . Moreover, |w| = ey > 0 and lim ey = ∞.
2 2 2 y→∞
Therefore the image is {w | Re w ≥ 0, w 6= 0}. ¤
Remark. The line segment {z | Im z = y, 0 ≤ x ≤ π} parallel to the x-axis, is mapped
onto the semicircle {w | |w| = ey , Re w ≥ 0}.

3. (Optional) Find the images of straight lines for the transformation w = ecz , c a complex
constant.
Solution. Let L be a straight line. Then L can be described as the following (vector)
equation L : z0 + tz1 , where t ∈ R is a parameter (see section 7.1). The image of L under
w = ecz is described as the equation ec(z0 +tz1 ) = ecz0 +tcz1 (t ∈ R).
We consider three cases:
(1) Case 1: cz1 = 0. In this case ecz0 +tcz1 = ecz0 is a constant.
(2) Case 2: cz1 6= 0 and Im(cz1 ) = 0. In this case cz1 is real, and hence

ecz0 +tcz1 = r(cos θ0 + i sin θ0 ), r = r(t) = eRe(cz0 )+tcz1 , θ0 = Im(cz0 ).

Thus the image is a ray emitting from the origin (the origin is not contained in the image).
(3) Case 3: Im(cz1 ) 6= 0. In this case cz0 + tcz1 = Re(cz0 ) + tRe(cz1 ) + iIm(cz0 ) + itIm(cz1 ).
Then it follows that
ecz0 +tcz1 = r(cos θ + i sin θ),

where r = r(t) = eRe(cz0 )+tRe(cz1 ) and θ = θ(t) = Im(cz0 ) + tIm(cz1 ). Since Im(cz1 ) 6= 0, it
follows that
θ − Im(cz0 )
t= .
Im(cz1 )

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 Consequently, we obtain that
£ ¤ h θ − Im(cz0 ) i
r = exp Re(cz0 ) + tRe(cz1 ) = exp Re(cz0 ) + Re(cz1 ) = Reaθ ,
Im(cz1 )
h i Re(cz1 )
[Re(cz1 )][Im(cz0 )]
where R = exp Re(cz0 ) − Im(cz1 ) and a = . Therefore the image is a
Im(cz1 )
logarithmic spiral. ¤

4. Show that the image of the disk |z| ≤ 1 under the transformation w = ez is contained in the
annulus 1/e ≤ |w| ≤ e.
Solution. Recall that −|z| ≤ Re z ≤ |z| and |w| = |ez | = eRe z . If |z| ≤ 1 then
1
= e−1 ≤ e−|z| ≤ |w| = eRe z ≤ e|z| ≤ e.
e
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