Solutions of Homework 2
Complex Analysis: Math 412
Professor Alain Bourget
1.3.2 (a) We have e3−i = e3 e−i = e3 cos(−1) + ie3 sin(−1) = e3 cos(1) − ie3 sin(1).
(b) We have
ei(2+3i) + e−i(2+3i)
2
1 −3
=
e cos(2) + ie−3 sin(2) + e3 cos(2) − ie3 sin(2)
2
e3 − e−3
e−3 + e3
cos(2) − i
sin(2)
=
2
2
= cosh(3) cos(2) − i sinh(3) sin(2)
cos(2 + 3i)
=
1.3.4 (a) We have
sin z =
eiz − e−iz
= 3/4 + i/4 =⇒ eiz − e−iz = −1/2 + 3/2i.
2i
Multiplying last equation by 2eiz , we obtain the quadratic equation
2e2iz + (1 − 3i)eiz − 2 = 0,
so
e
iz
=
−1 + 3i +
By Proposition 1.1.3, we obtain
p
(1 − 3i)2 + 16
−1 + 3i +
=
4
2
√
√
8 − 6i
.
8 − 6i = ±(3 − i).
Therefore,
eiz =
−1 + 3i ± (3 − i)
= 1 + i or − 2 + 2i.
2
Hence,
z = −i log(1 + i) = −i(log |1 + i| + i arg(1 + i) + 2kπi) = π/4 + 2kπ −
i
log 2
2
and
z = −i log(−2 + 2i) = −i(log | − 2 + 2i| + i arg(−2 + 2i) + 2kπi) = 3π/4 + 2kπ −
(b) We have
sin z =
eiz − e−iz
= 4 =⇒ eiz − e−iz = 8i.
2i
Multiplying last equation by eiz , we obtain the quadratic equation
e2iz − 8ieiz − 1 = 0,
(k ∈ Z),
3i
log 2.
2
so
e
iz
=
8i +
p
√
(−64 + 4)
= 4i ± i 15.
2
Hence,
√
√
√
π
z = −i log(4i + i 15) = −i(log |4 + 15| + i + 2kπi) = π/2 + 2kπ − i log(4 + 15)
2
and
(k ∈ Z),
√
√
√
z = −i log(4i − i 15) = −i(log |4 − 15| + πi/2 + 2kπi) = +π/2 + 2kπ − i log(4 − 15).
1.3.6 (a) We have
log(−i) = log | − i| + i arg(−i) + 2kπi = −iπ/2 + 2kπi
(k ∈ Z).
(b) We have
2i = ei log 2 = ei(log 2+i arg(2)+2kπi) = ei log 2−2kπ
1.3.12 Let w = tan z. We then have
w = −i
(k ∈ Z).
eiz − e−iz
.
eiz + e−iz
That is,
iw(eiz + e−iz ) = eiz − e−iz .
Multiplying by eiz , we get
iw(e2iz + 1) = e2iz − 1.
Solving for ebiz , we obtain
e2iz =
1 + iw
.
1 − iw
Hence,
1
z=
log
2i
1 + iw
1 − iw
1
= log
i
1 + iw
1 − iw
1/2
(mod pi).
Finally,
w = tan z = tan
1
log
i
1 + iw
1 − iw
1/2 !
since tan z is periodic of period π.
1.1.16 (a) We have
2
2
cosh z − sinh z
=
=
=
2
ez − e−z
−
2
1 2z
+ 2 + e−2z −
e − 2 + e−2z
4
ez − e−z
2
1 2z
e
4
1.
2
(b) We have
sinh z cosh w + cosh z sinh w
=
ez − e−z
2
ew + e−w
2
ez + e−z
2
ew − e−w
2
ez − e−z
2
ew − e−w
2
+
ez+w − e−(z+w)
2
= sinh(z + w)
=
(c) Similarly, we have
cosh z cosh w + sinh z sinh w
=
ez + e−z
2
ew + e−z
2
+
ez+w + e−(z+w)
2
= cosh(z + w)
=
1.1.20 (a) We have
ab ac = eb log a ec log a = e(b+c) log a = ab+c .
(b) Since we assumed that log(ab) = log a + log b, we have
(ab)c = ec log(ab) = ec log a+c log b = ec log a ec log b = ac bc .
1.3.22 (a) The map f (z) = z 3 = |z|3 e3i arg(z) will cube the length of z and triple its argument, therefore f
maps the first quadrant onto the first three quadrant.
(b) The map g(z) = z 1/3 = |z|1/3 ei3 arg(z) within a branch will take the cubic root of |z| and divide its
argument by 3, so it will map the first quadrant to the sector S = {z ∈ C : 0 ≤ z ≤ π/6}.
1.3.24 Let f (z) = cos z = u(x, yi v(x, y). Since
cos(z) = cos(x + iy) = cos x cos(iy) − sin x sin(iy) = cos x cosh y − i sin x sin y,
it follows that
u(x, y) = cos x cosh y and v(x, y) = − sin x sinh y.
For the horizontal line y = b, we obtain the ellipse
v2
u2
= 1.
2 +
cosh b sinh2 b
While for the vertical line x = a, we obtain the hyperbola
u2
v2
−
= 1.
2
cos a sin2 a
1.3.26 (a) We have f (z) = z 2 = (x + iy)2 = x2 − y 2 + 2ixy, so that u(x, y) = Re(f (z)) = x2 − y 2 and
v(x, y) = Im(f (z)) = 2xy. Hence, the horizontal line y = b is mapped to u = x2 − b2 and v = 2bx, i.e.
v
u2 = 2b
− b2 for b 6= 0 which is the equation of a parabola in the uv-plane.
(b) Note that f (z) =
√
i
h
arg(z)
+
i
sin
for some chosen branch,
z = |z|1/2 ei arg(z)/2 = |z|1/2 cos arg(z)
2
2
say θ0 ≤ arg(z) < θ0 + 2π. So,it follows that
(arg(z)
arg(z)
1/2
1/2
u(x, y) = |z| cos
and v(x, y) = |z| sin
.
2
2
Consequently, the horizontal line y = |z| sin(arg(z)) = b is mapped to the hyperbola
arg(z)
arg(z)
|z| sin(arg(z))
b
uv = |z| cos
sin
=
= .
2
2
2
2
1.3.29 Done in class.