Solutions for Math 311 Assignment #3
(1) Compute the limits:
iz 3 − 1
;
(a) lim 2
z→i z + 1
4z 2
(b) lim
.
z→∞ (z − 1)2
Solution. (a)
i(z 3 − i3 )
iz 3 − 1
=
lim
z→i z 2 + 1
z→i z 2 + 1
i(z − i)(z 2 + iz + i2 )
= lim
z→i
(z − i)(z + i)
2
i(z + iz + i2 )
3
= lim
=−
z→i
z+i
2
or by complex L’Hospital,
lim
3
iz 3 − 1
3iz 2
=
lim
=−
2
z→i z + 1
z→i 2z
2
lim
(b)
4z −2
4z 2
=
lim
z→0 (z −1 − 1)2
z→∞ (z − 1)2
4
= lim
=4
z→0 (1 − z)2
lim
(2) Show that the limit of the function
z 2
f (z) =
z
as z tends to 0 does not exist.
Solution. Let z = x + yi. When x → 0 and y = 0,
z 2
x2
=
lim
=1
lim
x→0
x→0 x2
z
y=0
When x = y and x → 0,
z 2
(x + xi)2
lim
=
lim
= −1
x→0
x→0 (x − xi)2
z
x=y
Therefore, limz→0 f (z) does not exist.
1
2
(3) Let
az + b
cz + d
where ad − bc 6= 0. Show that
(a) lim T (z) = ∞ if c = 0;
z→∞
a
(b) lim T (z) = if c 6= 0 and lim T (z) = ∞ if c 6= 0.
z→∞
z→−d/c
c
T (z) =
Proof. Since
lim
z→∞
1
1
= lim
z→0
T (z)
T (z −1 )
cz −1 + d
= lim −1
z→0 az
+b
c
c + dz
=
= lim
z→0 a + bz
a
it follows that limz→∞ T (z) = ∞ when c = 0 and limz→∞ T (z) =
a/c if c 6= 0.
Since ad − bc 6= 0, a(−d/c) + b 6= 0. Therefore,
1
cz + d
= lim
=0
z→−d/c T (z)
z→−d/c az + b
lim
and hence limz→−d/c T (z) = ∞ when c 6= 0.
(4) Find f 0 (z) when
(a) f (z) = 3z 2 − 2z + 4;
(b) f (z) = (1 − 4z 2 )3 ;
z−1
1
(c) f (z) =
(z 6= − );
2z + 1
2
(1 + z 2 )4
(d) f (z) =
(z 6= 0).
z2
Answer. (a) 6z − 2 (b) −24z(1 − 4z 2 )2 (c) 3(2z + 1)−2 (d)
2(3z 2 − 1)(1 + z 2 )3 z −3
(5) Show that f 0 (z) does not exist at any point when
(a) f (z) = Im(z);
(b)
(
z 2 /z if z 6= 0
f (z) =
0
if z = 0
3
Proof. (a) Since
∂
∂
∂
∂
+i
f=
+i
(x − yi) = 2 6= 0
∂x
∂y
∂x
∂y
for all z = x + yi, f 0 (z) does not exist at any point.
(b) For z = x + yi 6= 0,
∂
∂
∂
∂ (x − yi)2
+i
f=
+i
∂x
∂y
∂x
∂y
x + yi
2(x − yi)(x + yi) − (x − yi)2
=
(x + yi)2
−2i(x − yi)(x + yi) − i(x − yi)2
+i
(x + yi)2
4(x − yi)
6= 0
=
x + yi
Therefore, f 0 (z) does not exist for z 6= 0.
At z = 0,
f (z) − f (0)
z2
lim
= lim 2
z→0
z→0 z
z−0
When x → 0 and y = 0,
x2
z2
=
lim
=1
lim
x→0 z 2
x→0 x2
y=0
When x = y and x → 0,
z2
(x − xi)2
lim
=
lim
= −1
x→0 z 2
x→0 (x + xi)2
x=y
Therefore, f 0 (0) does not exist. In conclusion, f 0 (z) does not
exist anywhere.
(6) Use Cauchy-Riemann equations to verify that f (z) is analytic
when
(a) f (z) = z 3 in C;
(b) f (z) = z −1 for z 6= 0;
2
(c) f (z) = e−z in C.
Proof. (a) Since
∂
∂
∂
∂
3
z =
(x + iy)3
+i
+i
∂x
∂y
∂x
∂y
= 3(x + yi)2 + i2 3(x + yi)2 = 0
4
and ∂f /∂x and ∂f /∂y are continuous everywhere, f (z) is entire.
(b) Since
∂
∂
∂
∂
−1
+i
z =
+i
(x + iy)−1
∂x
∂y
∂x
∂y
= −(x + yi)−2 − i2 (x + yi)−2 = 0
and ∂f /∂x and ∂f /∂y are continuous for z 6= 0, f (z) is analytic
for z 6= 0.
(c) Since
∂
∂
∂
∂
2
−z 2
+i
e
=
+i
e−(x+iy)
∂x
∂y
∂x
∂y
2
2
= −2(x + yi)e−(x+iy) − 2i2 (x + yi)e−(x+iy) = 0
and ∂f /∂x and ∂f /∂y are continuous everywhere, f (z) is entire.
(7) Show that if both f (z) and g(z) satisfy the Cauchy-Riemann
equations at z0 , so does f (z)g(z).
Proof. Since C-R equations hold for f (z) and g(z) at z0 ,
∂f
∂f
∂g
∂g
+i
=
+i
=0
∂x
∂y
∂x
∂y
at z0 . Therefore,
∂
∂
∂g
∂g
∂f
∂f
+i
g+f
g+f
(f g) =
+i
∂x
∂y
∂x
∂x
∂y
∂y
∂f
∂f
∂g
∂g
g+i g + f
+ if
=
∂x
∂y
∂x
∂y
∂f
∂f
∂g
∂g
=g
+i
+f
+i
=0
∂x
∂y
∂x
∂y
at z0 . Therefore, C-R equations hold for f (z)g(z) at z0 .
(8) Suppose that f (z) = u + iv is analytic at z0 . Show that
∂v
i ∂u
0
+i
f (z0 ) = −
z0 ∂θ
∂θ
at z = z0 , where (r, θ) are the polar coordinates.
5
Proof. By chain rule,
∂
∂
sin θ ∂
∂
∂
cos θ ∂
= cos θ −
and
= sin θ +
.
∂x
∂r
r ∂θ
∂y
∂r
r ∂θ
Since f (z) is analytic at z0 ,
∂
∂
+i
f =0
∂x
∂y
and hence
∂
i ∂
∂f
i ∂f
+
f =0⇒
=−
∂r r ∂θ
∂r
r ∂θ
in |z − z0 | < a for some a > 0. Therefore,
∂f
∂f
sin θ ∂f
f 0 (z) =
= cos θ
−
∂x
∂r
r ∂θ
i cos θ ∂f
sin θ ∂f
=−
−
r ∂θ
r ∂θ
−iθ
i ∂f
e ∂f
= − iθ
= −i
r ∂θ
re
∂θ
i ∂u
∂v
=−
+i
z ∂θ
∂θ