Solutions for Math 311 Assignment #4
(1) Let a function f be analytic everywhere in a domain D. Prove
that if f (z) is real-valued for all z in D, then f (z) must be
constant throughout D.
Proof. Let f (z) = u(x, y) + iv(x, y) where u(x, y) = Re(f (z))
and v(x, y) = Im(f (z)). Since f (z) is real-valued for all z in D,
v(x, y) ≡ 0 in D. Since f (z) is analytic in D, u and v satisfy
Cauchy-Riemann equations, i.e., ux = vy and uy = −vx . And
since v ≡ 0, ux = uy = vx = vy = 0 for all z = x + yi in D.
Therefore, f (z) is constant in D.
2 + πi
(2) Compute (a) exp(2 ± 3πi) (b) exp
4
√
√
Answer. (a) −e2 (b) e1/2 ( 2/2 + i 2/2)
(3) Explain why the function f (z) = 2z 2 − 3 − zez + e−z is entire.
Proof. Since 2 and z are entire and products of entire functions
are entire, 2z 2 is entire. Since z and ez are entire, −zez is
entire. Since ez and −z are entire, their composition e−z is
entire. Finally, since 2z 2 , −3, −zez and e−z are entire, their
sum f (z) is entire.
(4) Show that the function f (z) = exp(z) is not analytic anywhere.
Proof. Since
∂
∂
∂
∂
+i
+i
f (z) =
ex−yi
∂x
∂y
∂x
∂y
x−yi
=e
+ i(−i)ex−yi = 2ex−yi 6= 0
for all z ∈ C, f (z) = exp(z) is not analytic anywhere.
(5) Write | exp(2z + i)| and | exp(iz 2 )| in terms of x and y. Then
show that
| exp(2z + i) + exp(iz 2 )| ≤ e2x + e−2xy .
1
2
Proof. We have
exp(2z + i) = exp(2(x + yi) + i) = e2x e(2y+i)i
= e2x (cos(2y + 1) + i sin(2y + 1))
and
exp(iz 2 ) = exp(i(x + yi)2 ) = exp(i(x2 − y 2 + 2xyi))
= exp(−2xy + i(x2 − y 2 ))
= e−2xy (cos(x2 − y 2 ) + i sin(x2 − y 2 )).
Therefore,
| exp(2z + i) + exp(iz 2 )| ≤ | exp(2z + i)| + | exp(iz 2 )|
= e2x + e−2xy .
(6) Find the complex derivatives of (a) sinh(z) cosh(z) (b) (tanh(z))2
Answer. (a) cosh2 (z) + sinh2 (z) = cosh(2z)
(b) 2 tanh(z) cosh−2 (z)
(7) Show that | sinh x| ≤ | cosh z| ≤ cosh x for all z ∈ C, where
x = Re(z).
Proof. Since cosh z = (ez + e−z )/2, it follows that
| cosh z| =
1
1
|ez + e−z |
≤ (|ez | + |e−z |) = (ex + e−x ) = | cosh x|
2
2
2
and
| cosh z| =
1
|ez + e−z |
1
≥ |ez | − |e−z | ≥ |ex − e−x | = | sinh x|.
2
2
2
(8) Show that | sin z| ≥ | sin x| and | cos z| ≥ | cos x| for all z ∈ C,
where x = Re(z).
Proof. Since
1
1
sin z = (eiz − e−iz ) = (eix−y − ey−ix )
2i
2i
1 −y
=
e (cos x + i sin x) − ey (cos x − i sin x)
2i
e−y + ey
ey − e−y
=
sin x + i
cos x
2
2
3
it yields
| sin z| ≥
e−y + ey
| sin x|.
2
And since
e−y + ey
(ey/2 − e−y/2 )2
e−y + ey
−1=
⇒
≥1
2
2
2
it follows that | sin z| ≥ | sin x|. That is, | sin z| ≥ | sin(Re(z))|
for all z ∈ C. Replacing z by z + π/2, we obtain
π π sin z +
≥ sin Re z +
2
2π π ⇒ sin z +
≥ sin x +
2
2
⇒ | cos(z)| ≥ | cos(x)|.
(9) Find all the roots of the equation cos z = 2.
Solution. Let u = eiz . We have
cos z = 2 ⇒ eiz + e−iz = 4 ⇒ u2 − 4u + 1 = 0.
Solving the equation u2 − 4u + 1 = 0, we obtain u = 2 ±
Therefore,
√
√
iz = log(2 ± 3) = ln(2 ± 3) + 2kπi
√
⇒ z = 2kπ + i ln(2 ± 3)
for k ∈ Z.
√
3.
4
Solutions for Math 311 Assignment #5
(1) Compute (a) Log(−ei) (b) Log(1 − i).
Answer. (a) 1 − πi/2 (b) (1/2) ln 2 − πi/4
(2) Show that
(a) Log(1 + i)2 = 2 Log(1 + i);
(b) Log(−1 + i)2 6= 2 Log(−1 + i).
Proof. Since
Log(1 + i)2 = Log(2i) = ln 2 +
πi
2
and
√
1
πi
πi
Log(1 + i) = ln 2 +
= ln 2 +
4
2
4
we have Log(1 + i)2 = 2 Log(1 + i).
Since
πi
Log(−1 + i)2 = Log(−2i) = ln 2 −
2
and
√
1
3πi
3πi
= ln 2 +
Log(−1 + i) = ln 2 +
4
2
4
2
we have Log(−1 + i) 6= 2 Log(−1 + i).
(3) Find the domain D where the function
f (z) =
Log(z + 4)
z2 + i
is analytic.
Solution. √
f (z) is analytic when z 2 +i 6= 0 and z +4 6∈ (−∞, 0],
i.e., z 6= ± 2(1 − i)/2 and z 6∈ (−∞, −4]. It is analytic in the
domain
( √
)
!
2(1 − i)
C\
±
∪ (−∞, −4] .
2
(4) Show that if Re(z1 ) > 0 and Re(z2 ) > 0, then
Log(z1 z2 ) = Log z1 + Log z2 .
5
Proof. We know that
Log(z1 z2 ) = Log z1 + Log z2 + 2kπi
for some integer k. Since Re(z1 ) > 0 and Re(z2 ) > 0,
π
π
π
π
− < Arg(z1 ) < and − < Arg(z2 ) <
2
2
2
2
and hence −π < Arg(z1 ) + Arg(z2 ) < π. And since −π <
Arg(z1 z2 ) ≤ π,
−2π < 2kπ = Arg(z1 z2 ) − Arg(z1 ) − Arg(z2 ) < 2π ⇒ −1 < k < 1.
Since k is an integer, we must have k = 0.
(5) Show that for any two nonzero complex numbers z1 and z2 ,
Log(z1 z2 ) = Log z1 + Log z2 + 2N πi
where N has one of values 0, ±1.
Proof. We know that
Log(z1 z2 ) = Log z1 + Log z2 + 2N πi
for some integer N . Since
π < Arg(z1 ) ≤ π and − π < Arg(z2 ) ≤ π,
−2π < Arg(z1 )+Arg(z2 ) ≤ 2π. And since −π < Arg(z1 z2 ) ≤ π,
−3π < 2N π = Arg(z1 z2 ) − Arg(z1 ) − Arg(z2 ) < 3π ⇒ −3 < 2N < 3.
Since N is an integer, we must have N = −1, 0 or 1.
(6) Compute (a) (1 + i)i (b) (−1)1/π and also find their principal
values.
√
√
Answer. (a) exp(2kπ−π/4)(cos(ln 2)+i sin(ln 2)) for k ∈ Z
and its principal value is taken when k = 0.
(b) cos(2k + 1) + i sin(2k + 1) for k ∈ Z and its principal value
is taken when k = 0.
(7) Assuming that the derivative f 0 (z) exists, find the derivative of
3f (z) .
Solution.
(3
f (z) 0
) = (exp(f (z) ln 3))0 = f 0 (z)(ln 3) exp(f (z) ln 3) = f 0 (z)3f (z) ln 3.