1. Find the real and imaginary parts of the following, where z = x + iy with x, y ∈ R.
(a)
ez
2
2 −y 2
Thus Re ez = ex
2
2
2
2
2
2
= e(x+iy) = ex −y +i2xy = ex −y ei2xy
2
2
= ex −y (cos 2xy + i sin 2xy) .
2
2 −y 2
cos 2xy and Im ez = ex
sin 2xy.
(b)
µ
¶
3x + i(3y + 1)
3x + i(3y + 1) x + 4 − iy
3z + i
=
=
z+4
x + 4 + iy
x + 4 + iy
x + 4 − iy
3x(x + 4) + y(3y + 1) + i ((x + 4) (3y + 1) − 3xy)
=
(x + 4)2 + y 2
3x2 + 12x + 3y 2 + y + i (x + 12y + 4)
=
.
(x + 4)2 + y 2
¡ 3z+i ¢
¢
¡
2
2
2
2
+
12x
+
3y
+
y)/((x
+
4)
+
y
)
and
Im
=
(3x
= (x + 12y +
Thus Re 3z+i
z+4
z+4
2
2
4)/((x + 4) + y ).
2. Find the norm and the argument arg(z) of the following complex numbers if we assume that
the branch of the argument is 0 ≤ arg(z) < 2π.
22π
(a) z = 2e 5 i .
|z| = 2 and arg z = 2π/5 since 22π/5 = 20π/5 + 2π/5 = 4π + 2π/5.
(b) z = 3 + 3i.
√
√
√
|z| = 32 + 32 = 18 = 3 2, and arg z = arctan (1) = π/4.
√
(c) z = −4 + 4 3i.
q
¡ √ ¢2 √
¡ √
¢
¡ √ ¢
z = (−4)2 + 4 3 = 16 + 48 = 8, and arg z = arctan 4 3/ − 4 +π = arctan − 3 +
π = −π/3 + π = 2π/3 (since z is in the 2nd quadrant)
3. Prove or give a counterexample:
The intersection of an arbitrary number of open subsets of C is an open set.
The statement is false. Let Un = D(0;
T 1/n) = {z ∈ C : |z| < 1/n}, as we saw in class all of
these sets are open. Note that z ∈ ∞
n=1 Un means that |z| < 1/n for every n natural number,
so the only possibility is z = 0. Thus
∞
\
n=1
Un = {0}.
However, f or every ε > 0, the disk D(0, ε) is not contained in {0}. Therefore {0} is not open.
1
4. Find the set of points where the function f (z) =
(z 3
the derivative of f .
2
is analytic and compute
− 1) (z 2 + 2)
The function (z 3 − 1)(z 2 + 2) is entire and so is the constant function 2. Thus f (z) is analytic
and only
in C except where the denominator vanishes. Note that (z 3 − 1) (z 2 + 2) = 0 if √
√ if
3
2
2πi/3 4πi/3
z = 1 or z = 2. The solutions to these equations are z = 1,√e
,√
e
, and z = 2i, − 2i,
respectively. Therefore f is analytic in C \ {1, e2πi/3 , e4πi/3 , 2i, − 2i}.
Differentiating we get,
¡
¢−2 ¡ 2 ¢ ¡ 2
¢−1
¡
¢−1 ¡ 2
¢−2
f 0 (z) = −2 z 3 − 1
− 2 z3 − 1
(2z)
3z
z +2
z +2
¢−2 ¡ 2
¢−2 ¡ 2 ¡ 2
¢
¡ 3
¢¢
¡ 3
z +2
3z z + 2 + 2z z − 1
= −2 z − 1
3
3
−2z (5z + 6z − 2)
=
.
(z 3 − 1)2 (z 2 + 2)2
5. Evaluate the following integral over the path γ shown in the figure. Recall that z = x + iy.
Z
γ
(2x − iy) dz
The path γ can be divided and parametrized according to the following two curves, γ = γ 1 +γ 2 :
γ 1 (t) = t + it, 0 ≤ t ≤ 1
γ 2 (t) = 1 + 2t + i (1 − t) , 0 ≤ t ≤ 1.
(other parametrizations are possible)
Then we have that
γ 01 (t) = 1 + i and γ 02 (t) = 2 − i.
Then
Z
γ
(2x − iy) dz =
Z
γ
Z 11
(2x − iy) dz +
Z
γ2
(2x − iy) dz
Z
1
(2t − it) (1 + i) dt +
(2 (1 + 2t) − i (1 − t)) (2 − i) dt
0
Z 1
Z 1
= (1 + i) (2 − i)
tdt + (2 − i)
(2 − i + t (4 + i)) dt
0
0
¶¯t=1
µ
µ 2 ¶¯t=1
t2 ¯¯
t ¯¯
+ (2 − i) (2 − i) t + (4 + i)
= (3 + i)
2 ¯t=0
2 ¯t=0
µ
¶
µ ¶
1
9
1
+ (2 − i) (2 − i) + (4 + i)
= 9 − i.
= (3 + i)
2
2
2
=
0
2
6. Show that the function f : C → C given by f (z) = z + z̄ is continuous but is not analytic.
Let ε > 0 and set δ = ε/2. Assume that |z − z0 | < δ, then
|f (z) − f (z0 )| = |z + z̄ − (z0 + z0 )| = |z − z0 + z̄ − z0 | ≤ |z − z0 | + |z − z0 |
≤ |z − z0 | + |z − z0 | = 2 |z − z0 | < 2δ = ε.
Therefore limz→z0 f (z) = f (z0 ), i.e., f is continuous.
Note that f (z) = 2x, if z = x + iy. Thus u = 2x and v = 0, consequently ∂u/∂x = 2 and
∂v/∂y = 0. Since these are different f does not satisfy the Cauchy-Riemann equations, thus
f is not analytic.
7. (Extra) Let f (z) be an entire function such that f (2z) = 2f (z) for every z ∈ C. Prove that
there is a constant complex number c such that f (z) = cz for all z ∈ C.
Solution 1: Taking derivatives on both sides we get that 2f 0 (2z) = 2f 0 (z), that is f 0 (2z) =
f 0 (z) for all z ∈ C. In particular
³ ´
³ ´
³ ´
0
0 z0
0 z0
0 z0
=f
= ... = f
.
f (z0 ) = f
2
22
2n
Because f 0 is continuous we can take the limit as n → ∞. Thus f 0 (z0 ) = f 0 (0) = c for all
z0 ∈ C. That is, f 0 is constant. Let g(z) = f (z) − cz, taking derivatives yields g 0 (z) =
f 0 (z) − c = c − c = 0 for all z. Thus, as we saw in class, g is a constant function in C,
that is g(z) = c0 , and thus f (z) = cz + c0 . However, because f (2z) = 2f (z), it follows that
2cz + c0 = 2cz + 2c0 . Therefore c0 = 0 and f (z) = cz.
Solution 2: First, f (0) = f (2·0) = 2f (0) and so f (0) = 0. By repeatedly using the condition,
we get
f (2n z) = 2f (2n−1 z) = 22 f (2n−2 z) = · · · = 2n f (z).
Let z0 be an arbitrary complex number. Because f is differentiable at zero, by letting z =
z0 /2n , it follows that
f (z) − f (0)
f (z)
= lim
z→0
z→0 z
z−0
f (z0 /2n )
2n f (z0 /2n )
f (z0 )
f (z0 )
=
lim
=
lim
=
.
= lim
n→∞ z0 /2n
n→∞
n→∞ z0
z0
z0
f 0 (0) = lim
Therefore f (z0 ) = f 0 (0)z0 for all z0 .
3