SOLUTION OF TEST 4
MINGFENG ZHAO
December 07, 2012
1. [10 Points] Sketch the level curves Re
1
1
1
1
= ± and Im = ± . Explain your reasoning.
z
2
z
2
Proof. a. Let z = x + iy ∈ C such that Re
±
1
2
1
1
= ± , then z 6= 0 and
z
2
=
Re
1
z
=
Re
1
x + iy
=
Re
x − iy
x2 + y 2
=
x
x2 + y 2
So we know that
x2
1
x
=±
2
+y
2
That is, we have
x2 + y 2 ± 2x = 0
but x2 + y 2 6= 0
(x ± 1)2 + y 2 = 1
but x2 + y 2 6= 0
Hence we know that
b. Let z = x + iy ∈ C such that Im
1
1
= ± , then z 6= 0 and
z
2
±
1
2
=
Im
1
z
=
Im
1
x + iy
=
Im
x − iy
x2 + y 2
1
2
MINGFENG ZHAO
=
−y
x2 + y 2
So we know that
1
−y
=±
2
+y
2
x2
That is, we have
x2 + y 2 ± 2y = 0
but x2 + y 2 6= 0
x2 + (y ± 1)2 = 1
but x2 + y 2 6= 0
Hence we know that
SOLUTION OF TEST 4
3
2. [10 Points] Find a linear fractional transformation that maps the real axis onto itself and the
imaginary axis onto the circle with center 2 and radius 2. Explain your reasoning.
Proof. Let f (z) =
4z
, then it easy to see that f (z) ∈ R for all z ∈ R. Also we know that
z−2
f (0) = 0 and f (∞) = 4, since f is a linear fractional transformation, then f maps the imaginary
axis onto a circle or straight line which must pass 0 and 4 and be perpendicular to the real axis, but
f (i) =
4i
6∈ R, then f maps the imaginary axis onto the circle with center 2 and radius 2.
i−2
4
MINGFENG ZHAO
3. [10 Points] Let f (z) =
z2
. Find the maximum and minimum values of |f (z)| for |z| ≤ 1 and
z−3
find all points z at which the max/min occur. Show all steps.
Proof. Let g(t) =
t2
for all 0 ≤ t ≤ 1, then
t+3
g 0 (t)
=
2t(t + 3) − t2
(t + 3)2
=
t2 + 6t
(t + 3)2
≥0
Then g is increasing on [0, 1]. For all |z| ≤ 1, then
2 z |f (z)| = z − 3
=
|z|2
|z − 3|
≤
|z|2
3 − |z|
≤
1
3−1
1
2
2 z |f (z)| = z − 3
=
=
|z|2
|z − 3|
≥
|z|2
|z| + 3
≤
1
1+3
=
1
4
Since g(t) =
t2
is increasing
t+3
So we know that
1
1
≤ |f (z)| ≤ ,
4
2
∀|z| ≤ 1
SOLUTION OF TEST 4
5
On the other hand, we know that
f (1)
f (−1)
=
1
|3 − 1|
=
1
2
=
1
| − 1 − 3|
=
1
4
If |f (z)| = 12 , then we know that |z| = 1 and |z − 3| = 3 − |z| = 2. Hence z = 1.
If |f (z) = 41 , then |z| = 1 and |z − 3| = 3 + |z| = 4. Hence z = −1.
6
MINGFENG ZHAO
4. [10 Points] Answer the same question for f (z) =
Proof. Let g(t) =
z3
.
z2 − 3
t3
for all 0 ≤ t ≤ 1, then
t2 + 3
g 0 (t)
=
3t2 (t2 + 3) − t3 · 2t
(t2 + 3)2
=
t4 + 9t2
(t2 + 3)2
≥0
Then g is increasing on [0, 1]. For all |z| ≤ 1, then
|f (z)| =
3 z z2 − 3 =
|z|3
|z 2 − 3|
≤
|z|3
3 − |z|2
≤
1
3−1
=
|f (z)| =
1
2
3 z z2 − 3 =
|z|3
|z 2 − 3|
≥
|z|3
|z|2 + 3
≤
1
1+3
=
1
4
Since g(t) =
t2
t3
is increasing
+3
So we know that
1
1
≤ |f (z)| ≤ ,
4
2
∀|z| ≤ 1
SOLUTION OF TEST 4
7
On the other hand, we know that
f (1)
f (i)
=
1
|3 − 1|
=
1
2
=
1
| − 1 − 3|
=
1
4
If |f (z)| = 12 , then |z| = 1 and |z 2 − 3| = 3 − |z|2 = 2. Hence z 2 = 1, which implies that z = ±1.
If |f (z)| = 41 , then |z| = 1 and |z 2 − 3| = 3 + |z|2 = 4. Hence z 2 = −1, which implies that z = ±i.
8
MINGFENG ZHAO
5. [10 Points] Find the number of zeros of the polynomial z 4 + 4 that lie in the right half plane.
Explain your reasoning.
Proof. Since −4 = 4 · eiπ , then
z1 =
√
π
2ei 4 ,
z2 =
√
π
2e−i 4 ,
z3 =
√
2ei
3π
4
,
and z4 =
√
2e−i
3π
4
If z lies in the right half plane, then only z2 and z4 are our choices.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu