Math 185 Complex Analysis
Fall 2018, Professor Klass
Midterm
Oct 5, 2018
Name:
Student ID:
Instructions
1. This exam contains 5 regular questions: each worth 25 points. Choose four questions
that you want graders to grade. Please specify your choice in the second column of the
following box. Otherwise, only the first 4 will be graded.
2. Please begin by writing your name on the first page. You are allowed 55 minutes for this
exam. These questions are not necessarily ordered in the increasing difficulty. Please
first pick the problems you feel most comfortable with and pace yourself accordingly.
3. You may not use any calculator, notes, or other assistance on this exam. In order to
receive full credit, you must show your work and carefully justify your answers. Partial
credit will be given if you show sufficient progress in solving a problem. If you need
more room, use the backs of the pages and indicate that you have done so. Please
write neatly.
Grade Table
Question Which to Grade
1
2
3
4
5
Total:
Score
Math 185 Complex Analysis
Midterm - Page 2 of ??
Oct 5, 2018
1. (a) (12.5 points) Find all the values of ii .
Solution:
By the definition of complex exponent,
ii = ei log i .
In the definition of logarithm, each branch is different up to 2πki for k ∈ Z. So
log i = log |i| + i arg i = i
Hence,
π
+ 2πki.
2
(
)
π
π
ii = ei log i = e− 2 −2πk = e− 2 +2πk
for k ∈ Z.
(b) (12.5 points) Simplify
(1 − i)5
√
(1 + i 3)8
into the form a + bi, where a, b ∈ R.
Solution:
To simplify computation, we convert the numerator and the denominator into polar
form:
√
√
π
π
1 − i = 2e−i 4 and 1 + i 3 = 2ei 3 .
Therefore,
(1 − i)5
2 2 e−i 4
− 11
−i 47π
− 11
iπ
2 e
12 = 2
2 e 12 .
√
=
=
2
8π
(1 + i 3)8
28 ei 3
5
5π
Hence, we can convert back to Cartesian coordinate
11
11
π
π
(1 − i)5
√
= 2− 2 cos
+ i2− 2 sin .
8
12
12
(1 + i 3)
π
π
Remark: you don’t have to calculate out cos 12
and sin 12
.
Math 185 Complex Analysis
Midterm - Page 3 of ??
Oct 5, 2018
2. (a) (18 points) Calculate the following integral
∫
dz
,
2
γ z +z −2
where γ is the circle centered at the origin with radius 3 traversed once in the
counterclockwise direction.
Solution 1: partial fraction + winding number
Observe that the two roots of z 2 + z − 2 are z1 = −2 and z2 = 1, both of which are
inside the circle. We can decompose the integrand by using partial fraction
(
)
1
1
1
1
=
−
+
.
z2 + z − 2
3
z+2 z−1
Observe that
∫
γ
1
dz
=
2
z +z−2
3
( ∫
−
γ
dz
+
z+2
∫
γ
dz
z−1
)
.
Observe that the first integral is the winding number 2πiI(γ; −2) = 2πi, and the
second is I2πi(γ; 1) = 2πi. (You can conclude the same thing by claiming they are
Cauchy integral formula with f ≡ 1.) Hence, the integral is zero.
Solution 2: generalized deformation theorem + Cauchy integral formula
By generalized deformation theorem (we covered in lecture),
∫
∫
∫
dz
dz
dz
=
+
,
2
2
2
γ z +z−2
γ1 z + z − 2
γ2 z + z − 2
where γ1 is a circle centered at −2 with radius, say, 0.5, and γ1 is a circle centered
at 1 with radius 0.5 (other small radius also works). The goal here is to construct
contours such that each circle contains only one singularity point. By Cauchy
integral formula,
∫
∫
dz
1/(z − 1)dz
2
=
=
−
πi.
2
z+2
3
γ1 z + z − 2
γ1
Here f =
1
z−1
is analytic inside γ1 and z0 = −2. Similarly,
∫
γ2
dz
=
2
z +z−2
∫
γ2
1/(z + 2)dz
2
= πi.
z−1
3
1
Here f = z+2
is analytic inside γ2 and z0 = 1. Notice that this type of application
appears in our homework and lecture before. Hence, the overall integral is zero.
(b) (7 points) Please briefly describe another way to calculate the integral above. (We
currently know there are at least three ways.)
Math 185 Complex Analysis
Midterm - Page 4 of ??
Oct 5, 2018
Solution 3: deformation theorem + integral bound
By deformation theorem,
∫
∫
dz
dz
=
,
2
2
γR z + z − 2
γ z +z −2
where γr is any circle centered at 0 and radisu R > 3. In particular, we can make
R as large as we want. For ξ on γR , notice that by reverse triangle inequality
|ξ 2 + ξ − 2| ⩾ |ξ|2 − |ξ| − 2 = R2 − R − 2.
Hence,
∫
γ
dz
⩽
z2 + z − 2
∫
γ
1
2πR
|dz|
⩽
→0
z2 + z − 2
R2 − R − 2
as R → ∞. Hence, the integral must be zero.
Solution 4: partial fraction + deformation theorem + parametrization
Decompose by partial fraction. For the two integrals, deform γ into γ1 and γ2 ,
respectively. By deformation theorem, you can conclude
∫
∫
∫
dz
dz
dz
1
1
=−
+
.
2
3 γ1 z + 2 3 γ2 z − 1
γ z +z −2
Now you can parametrize these contours and get
∫
∫
dz
dz
= 2πi and
= 2πi.
γ1 z + 2
γ2 z − 1
(We have done such parametrization multiple times in homework.) Hence, the
overall integral is zero.
Remark: Solution 1-4 are equivalently valid for both (a) and (b). Many of you claim
that you can parametrize the original contour and carry out the computation. This
is theoretically correct, but the actual calculation is far from obvious. If you choose
to parametrize the original contour, you need to provide more details about your
calculation to earn full credit.
Math 185 Complex Analysis
Midterm - Page 5 of ??
Oct 5, 2018
3. (a) (10 points) Please state the Cauchy-Riemann equations.
Solution:
Suppose that f is an analytic function on G with f = u + iv. Then
∂u
∂v
=
∂x
∂y
and
∂u
∂v
=− ,
∂y
∂x
or in another notation
u x = vy
and uy = −vx
(b) (15 points) Let f be analytic on a connected open region A. Suppose that |f (z)|
is constant on A. Prove that f (z) is constant on A. (Hint: consider |f (z)|2 .)
Solution:
Write f = u + iv. Notice that |f (z)| constant implies that |f (z)|2 = u2 + v 2 is
constant. Taking partial derivative with respect to x and y yields
{
2uux + 2vvx = 0
2uuy + 2vvy = 0.
By Cauchy Riemann equations, we have ux = vy and uy = −vx . With these, we
can rewrite the above system as
{
2uux + 2vvx = 0
−2uvx + 2vux = 0.
Consider the above system with (ux , vx ) as unknown, which we intend to show is
(0, 0). View the coefficients of the system above as a matrix
(
)
2u 2v
.
2v −2u
Notice that the determinant is −4u2 − 4v 2 . It is zero if and only if u = 0 = v,
in which case f = u + iv = 0 is constant. When the determinant is non-zero,
ux = 0 = vx is the only solution to the system because the matrix is inevitable,
implying that u, v are constant. Hence, in either case, f = u + iv is constant.
Remark: Since part (a) asks to write down Cauchy Riemann equations, it should
give you a hint of what you are expected to do. You can also use more elementary
algebraic method to solve the equation above.
Math 185 Complex Analysis
Midterm - Page 6 of ??
Oct 5, 2018
4. (a) (10 points) Please state Liouville’s Theorem.
Solution:
Any bounded entire function is constant.
(b) (15 points) Suppose that f : C → C is non-constant and entire. Prove that for any
w0 ∈ C and any ϵ > 0, there exists z0 ∈ C such that |f (z0 ) − w0 | < ϵ.
Solution:
Suppose, towards a contraction, that there exists a w0 ∈ C and an ϵ > 0 such that
for all z ∈ C we have |f (z) − w0 | ⩾ ϵ. Then, consider the function g = 1/(f − w0 ).
Notice that
1
1
|g| =
⩽ ,
|f − w0 |
ϵ
so g is bounded. Also, g is well-defined on C and entire, since |f (z) − w0 | ⩾ ϵ.
Hence, g must be a constant. Consequently, f − w0 is a constant and f is a constant. But this is a contradiction.
Math 185 Complex Analysis
5. (25 points) Please evaluate
Midterm - Page 7 of ??
∫
2π
Oct 5, 2018
e−iθ ee dθ.
iθ
0
Solution:
To simplify the calculation, we observe that the integral above is a parametrized complex
contour integral. Let γ = eiθ for θ ∈ [0, 2π]. Then, dz = ieiθ dθ and dθ = −iz −1 dz. Hence
∫ 2π
∫ 2π
iθ
−iθ eiθ
e e dθ =
e−iθ ee (−ie−iθ )(ieiθ dθ)
0
∫0
= z −1 ez (−iz −1 )dz
γ
∫ z
e
= −i
dz
2
γ z
= −i2πie0
= 2π.
In the second to last step, we used Cauchy integral formula for derivative.