Midterm Exam
Summer 2013
Math 1560
Solutions
100 points total
1. [10 points] Evaluate Im ((−i)−1+i ). Find all values. Simplify your answer.
Solution:
(−i)−1+i = e(−1+i) ln(−i) = e(−1+i)(0+i(−π/2−2πk)) = e(1+i)(π/2+2πk)) = eπ/2+2πk ei(π/2+2πk)
= eπ/2+2πk (cos(π/2 + 2πk) + i sin(π/2 + 2πk)) = eπ/2+2πk i, k = 0, ±1, ±2...
Hence Im ((−i)−1+i ) = eπ/2+2πk , k = 0, ±1, ±2...
1
2. [10 points] Find Re sin
.
z̄
z
x + iy
x
y
1
=
= 2
= 2
+ 2
i.
Solution:
2
2
z̄
z z̄
x +y
x +y
x + y2
1
x
y
Then Re sin
= sin
cosh
.
z̄
x2 + y 2
x2 + y 2
3. [10 points] Prove or disprove that the function

2

 z + iz + 6 ,
2i − z
f (z) =

 −5i,
z 6= 2i,
z = 2i.
is continuous at z = 2i.
Solution:
A function f (z) is continuous at a point z0 if lim f (z) = f (z0 ).
z→z0
In the definition of f it is given that f (2i) = −5i.
2z + i
z 2 + iz + 6 0 H
” 0 ” = lim
= −5i.
z→2i −1
z→2i
2i − z
lim f (z) = lim
z→2i
So, we have shown that lim f (z) = f (2i). Hence, f (z) is continuous at z = 2i.
z→2i
z2 + 2
. Express the result in rectangular form.
z+i
2z(z + i) − z 2 − 2
z 2 + 2zi − 2
f 0 (z) =
=
(z + i)2
(z + i)2
4. [10 points] Find f 0 (2) if f (z) =
Solution:
f 0 (2) =
2 + 4i
2 + 4i
2 + 4i 3 − 4i
22 + 4i
22
4
=
=
·
=
=
+ i
2
(2 + i)
3 + 4i
3 + 4i 3 − 4i
25
25 25
1
5. [10 points] Use CREs to show that the function f (z) = cos z̄ is non-analytic everywhere.
Solution:
f (z) = cos z̄ = cos(x − iy) = cos x cosh y + i sin x sinh y,
v = sin x sinh y,
u = cos x cosh y,
ux = − sin x cosh y, vx = cos x sinh y, uy = cos x sinh y, vy = sin x cosh y.
If we assume that CREs hold. Then ux = vy gives sin x = 0 and uy = −vx gives cos x sinh y = 0.
The system has solutions x = πk, y = 0. Hence, CREs can hold only at the isolated points
z = πk. But a function cannot be analytic at an isolated point.
Hence, f (z) is non-analytic everywhere.
6. [10 points] Let u(x, y) = 2xy−y be a real part of an analytic function f (z) = u(x, y)+v(x, y)i.
Find f (z).
Solution:
From CREs we have vy = ux = 2y, vx = −uy = −2x + 1.
From vy = 2y we get v = y 2 + φ(x). To find φ(x) we use vx = −2x + 1 or φ0 (x) = −2x + 1.
Then φ(x) = −x2 +x and v = −x2 +y 2 +x. Hence, f (z) = 2xy −y +(−x2 +y 2 +x)i = −iz 2 +iz.
7. [10 points] Evaluate the limit using the L’Hospital’s rule
lim (sin z)z .
z→0
Define and write the types of all indeterminate forms occurred during the process of obtaining
the solution.
Solution:
z
lim z ln sin z
lim (sin z)z = lim eln(sin z) = ez→0
z→0
z→0
cos z/ sin z
z 2 cos z
ln sin z ∞ H
”
”
=
lim
=
−
lim
∞
z→0
z→0 sin z
z→0 z −1
−z −2
lim z ln sin z = lim
z→0
2z cos z − z 2 sin z
=0
z→0
cos z
= − lim
Hence, lim (sin z)z = e0 = 1.
z→0
Z
8. Evaluate I =
3z 2 dz if
C
(a) [10 points] C is the curve in the complex plane given by y = x3 − 3x − 2 and joining points
(0, −2) and (2, 0).
2
Solution:
The function 3z 2 is analytic in the whole complex plane and the integral is
independent of path. Hence,
Z2
I=
2
3z dz = z 2
= 8 − (−2i)3 = 8 − 8i
3
−2i
−2i
(b) [10 points] C is the circle |z + 2| = 3.
Solution: The function 3z 2 is analytic in the whole complex plane. The path C is closed. By
Cauchy’s theorem I = 0.
I
5z̄ dz where C is the circle |z + i| = 2.
9. [10 points] Evaluate I =
C
Solution:
By the complex form of Green’s theorem we have
Z
Z
∂
I = 2i
(5z̄) dA = 10i
dA
∂ z̄
R
R
Z
dA represents the area of the circle |z + i| = 2 with radius 2. Its area is
The integral
R
π · 22 = 4π.
Hence, I = 10i · 4π = 40πi.
I
bonus problem [15 points extra] Evaluate I =
dz
.
z−i
|z−i|=1
Solution:
On the circle |z − i| = 1 we have z = i + eiθ , 0 ≤ θ < 2π, dz = ieiθ dθ.
Then
Z2π
I=
ieiθ dθ
=
eiθ
0
Z2π
i dθ = 2πi
0
3