United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
EXAM 1 – SOLUTION
Complex Analysis I
MATH 315 SECTION 01 CRN 23516
9:30 – 10:45 on Monday & Wednesday
Due Date: Monday, October 19, 2009
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These are the notes of Exam of Complex Analysis and its key important points are: Express, Compute, Real, Exponential Form, Formula, Integer, Clearly, Condition, Satisfies, Unit Circle
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United Arab Emirates University College of Sciences Department of Mathematical Sciences
EXAM 1 – SOLUTION
Complex Analysis I
MATH 315 SECTION 01 CRN 23516
9:30 – 10:45 on Monday & Wednesday
Due Date: Monday, October 19, 2009
Complex Analysis I EXAM 1 – SOLUTION Fall, 2009
1. (Total 10 points) Let z =
3 + i and w = 1 + i
(1.1) (5 points) Compute w − z, zw, and | z |. Express w/z in the form x + iy with real x and y.
Answer.
w − z = 1 + i
3 + i
3 + i
zw =
3 + i
1 + i
3 + i 2
3 + i
[ (
]
3 + i ( 3 + 1 ) = 4i
| z | =
3 + i =
+ 1^2 =
w
z
1 + i
3 + i
1 + i
3 + i
3 − i √ 3 − i
1 + i
3 − i
3 + 2i
4
3 + i
2
.
(1.2) (5 points) Express z =
3 + i in the exponential form reiθ.
Answer. For the given z, we have
| z | = 2, −π < Arg(z) = tan
π
6
≤ π, z = | z | e i Arg(z) = 2e i π 6
.
2. (5 points) Write down Euler’s formula and de Moivre’s formula.
Answer. Euler’s formula and de Movire’s formula are, respectively, as follows
e iθ = cos θ + i sin θ, [ cos θ + i sin θ ]
n = cos ( nθ ) + i sin ( nθ ) ,
where n is an integer.
3. (Total 5 points) Determine whether the following is true (T) or false (F).
(3.1) (3 points) If Im(z) > 0, then | z − i | > | z + i |......................................... F
Answer. z = i satisfies the condition Im(z) = 1 > 0. But, clearly z = i does not satisfy the inequality | z − i | = 0 < 2 = | z + i |.
(3.2) (2 points) If z ̸= 0 lies inside the unit circle centered at the origin, then 1/¯z lies outside the
circle.................................................................................. T
Answer. With z = x + iy ̸= 0,
¯z
x − iy
x + iy
x^2 + y^2
, and
z¯
x^2 + y^2
If z = x + iy ̸= 0 lies inside the unit circle, then | z¯ |
2 = | z |
2 = x 2
- y 2 < 1 which implies | 1 /z¯ | > 1. That is, 1/¯z should be outside the unit circle.
Complex Analysis I EXAM 1 – SOLUTION Fall, 2009
7. (5 points) Find all 3rd^ roots of 1 + i, i.e., (1 + i)
1 / 3 .
Answer. We are asked to find all z’s satisfying z^3 = 1 + i. We start with the exponential forms of z, and 1 + i,
z = | z | e i arg( z ) , z 3 = | z |
3 e i3 arg( z ) , 1 + i = | 1 + i | e i arg( 1+i ) =
2 e i( π 4 +2nπ (^) )
where n = 0, ± 1 , ± 2 ,.... From z 3 = 1 + i, we obtain
| z |
3 e i3 arg( z ) = z 3 = 1 + i =
2 e i( π 4 +2nπ (^) ) , i.e., | z |
3
2 , 3 arg ( z ) =
π
4
- 2nπ.
It gives us
| z | =
1 / 6 , arg ( z ) =
π
12
2 nπ
3
, i.e., z = 2 1 / 6 e i( 12 π + 2 nπ 3 ) ,
where n = 0, ± 1 , ± 2 ,.... Putting n = − 1 , 0 , 1, we have all the 3rd^ roots:
c 2 = c− 1 = 2 1 / 6 e i( 12 π − 23 π ) =
− 21 /^6
− i
c 0 = 2 1 / 6 e i( 12 π ) =
21 /^6
− i
c 1 = 2 1 / 6 e
i( 12 π + 23 π ) = −
1 / 6 − i √ 2
.
8. (Total 10 points) Solve the following conversion problems.
(8.1) (5 points) Write f (z) = xy + iy 2 in terms of z and ¯z, where z = x + iy.
Answer. With z = x + iy, we have
x = Re ( z ) =
z + ¯z
2
, y = Im ( z ) =
z − ¯z
2 i
Before putting them into the function, we modify the given function
f (z) = xy + iy 2 = y ( x + iy ) = ( Im ( z ) ) z,
which gives us
f (z) = ( Im ( z ) ) z =
z − z¯
2 i
z = −i
z 2 − z z¯
2
= −i
z 2 − | z |
2
| z |
2 − z 2
i.
Complex Analysis I EXAM 1 – SOLUTION Fall, 2009
(8.2) (5 points) Write f (z) = z 2 − ¯z 2 in the form u(r, θ) + iv(r, θ), where r and θ are the modulus and the principal argument of z, respectively.
Answer. With z = re iθ and ¯z = re −iθ , we have
f (z) = z 2 − z¯ 2 = r 2 e 2 iθ − r 2 e − 2 iθ = r 2
e 2 iθ − e − 2 iθ
= r 2 [ cos ( 2θ ) + i sin ( 2θ ) − ( cos ( 2θ ) − i sin ( 2θ ) ) ]
= 0 + 2r 2 i sin ( 2θ ) = u(r, θ) + iv(r, θ).
That is, u(r, θ) = 0 and v(r, θ) = 2r^2 sin(2θ).
9. (5 points) Sketch and find the image S
′ of the semi–infinite strip S given below under the trans- formation w = f (z) = ez^ :
S = { z = (x, y) ∈ C | − 1 ≤ x ≤ 1 , 0 ≤ y ≤ π }.
Answer. Using Exponential Form. For z = x + iy ∈ S, we use the exponential form,
ρe iϕ = w = f (z) = e x+iy = e x e iy , i.e., ρ = e x , ϕ = y.
For − 1 ≤ x ≤ 1 and 0 ≤ y ≤ π, we have e − 1 ≤ ρ = e x ≤ e and 0 ≤ ϕ = y ≤ π. It implies
S
′ = f (S) =
ρe iϕ ∈ C | e − 1 ≤ ρ ≤ e, 0 ≤ ϕ ≤ π
The inequality e − 1 ≤ ρ ≤ e represents the annulus between two disks centered at the origin with radii e−^2 and e^2. The other inequality 0 ≤ ϕ ≤ π represents the upper–half space in the w–plane. Therefore, the region S ′ is the part of annulus lying in the upper–half space.
Answer. Using Cartesian Form. For z = x + iy ∈ S, we have
u(x, y) + iv(x, y) = f (z) = e x+iy = e x e iy = e x ( cos y + i sin y ) = e x cos y + ie x sin y,
i.e., u(x, y) = e x cos y and v(x, y) = e x sin y. For − 1 ≤ x ≤ 1 and 0 ≤ y ≤ π, we have e − 1 ≤ e x ≤ e and − 1 ≤ cos y ≤ 1 and 0 ≤ sin y ≤ 1. It implies
−e ≤ u(x, y) = e x cos y ≤ e, 0 ≤ v(x, y) = e x sin y ≤ e.
That is, for z ∈ S, the real and imaginary parts of f (z) = u + iv, Re(f (z)) = u(x, y) = ex^ cos y and Im(f (z)) = v(x, y) = ex^ sin y, should be in the intervals, −e ≤ u ≤ e and 0 ≤ v ≤ e, respectively.
Finally we observe that for z ∈ S,
u 2
- v 2 = ( e x cos y )
2
- ( e x sin y )
2 = e 2 x (^ cos 2 y + sin 2 y
= e 2 x , e − 2 ≤ u 2
- v 2 ≤ e 2 ,
because − 1 ≤ x ≤ 1. Therefore we deduce
S
′ = f (S) =
u + iv = (u, v) ∈ C | e − 1 ≤ u ≤ e, 0 ≤ v ≤ e, e − 2 ≤ u 2
- v 2 ≤ e 2
The graphs of u^2 + v^2 = e−^2 and u^2 + v^2 = e^2 are circles centered at the origin with the radius e−^2 and e^2 , respectively. Thus, the region S′^ is the part of annulus between two disks centered at the origin with radii e−^2 and e^2 and (u, v) ∈ [ e−^1 , e ] × [ 0, e ].
United Arab Emirates University College of Sciences Department of Mathematical Sciences
EXAM 2 { SOLUTION
Complex Analysis I MATH 315 SECTION 01 CRN 23516 9:30 { 10:45 on Monday & Wednesday Due Date: Wednesday, November 25, 2009
ID No: Solution
Name: Solution
Score: 50/
Complex Analysis I EXAM 2 { SOLUTION Fall, 2009
1. (5 points) Show that f^0 (z) does not exist at any point for f(z) = e
z .
Proof. With z = x + iy, e z = e x iy = e x cos y ie x sin y:
Let u(x; y) = e x cos y and v(x; y) = e x sin y. Then we observe
ux = e x cos y = u; uy = e x sin y = v; vx = e x sin y = v; vy = e x cos y = u:
Only when u = 0 and v = 0, u and v satisfy the Cauchy{Riemann equations, i.e.,
ux = u = vy; uy = v = vx:
However, ex^ 0 for any x and there is no y satisfying both cos y = 0 and sin y = 0 simultane- ously. Therefore, f(z) is not di erentiable at any point.
2. (5 points) Show that the function f(z) = e
cos (ln r) + ie sin (ln r) is di erentiable in the domain of de nition r > 0, 0 < < 2 , and also to nd f^0 (z).
Proof. Let u(r; ) = e cos (ln r) and v(r; ) = e sin (ln r). Then we observe
ur =
r
e sin (ln r) =
r
v; u = e cos (ln r) = u;
vr =
r
e cos (ln r) =
r
u; v = e sin (ln r) = v
At any z = re i in the given domain, u and v satisfy the Cauchy{Riemann equations, i.e.,
rur = v = v; u = u = rvr:
Thus, f(z) is di erentiable in the given domain and has the derivative
f 0 (z) = e i (ur + ivr) = e i
r
v + i
r
u
= ir 1 e i (u + iv) = iz 1 e [cos (ln r) + i sin (ln r)] :
Remark: We observe
f(z) = e cos (ln r) + ie sin (ln r) = e e i ln r = e i(ln r+i) = e i log z = z i ;
where log z is a branch of the logarithm. So when we di erentiate it, we get f^0 (z) = iz^1 i. In fact, the answer above implies
f 0 (z) = iz 1 e [cos (ln r) + i sin (ln r)] = iz 1 f(z) = iz 1 z i = iz 1 i :
3. (5 points) Show that u(x; y) = 2x x
3
- 3xy 2 is harmonic in some domain and nd a harmonic conjugate v(x; y).
Proof. A simple computation shows that for all x and y,
ux = 2 3 x 2
- 3y 2 ; uxx = 6 x; uy = 6xy; uyy = 6x; uxx + uyy = 6 x + 6x = 0:
Thus, the function u is harmonic everywhere. A harmonic conjugate v should satisfy ux = vy and uy = vx and so we compute
v =
Z vy dy =
Z ux dy =
Z (^) 2 3 x^2 + 3y^2
dy = 2y 3 x^2 y + y^3 + (x);
vx =
@x
2 y 3 x 2 y + y 3
- (x)
= 6 xy = uy;
6 xy + 0 (x) = 6 xy; 0 (x) = 0; (x) = C:
Therefore, we deduce the harmonic conjugate v(x; y) = 2y 3 x 2 y + y 3
- C, where C is any constant.
Complex Analysis I EXAM 2 { SOLUTION Fall, 2009
7. (5 points) Evaluate log
p 3 + i
.
Answer. By the de nition,
log
p 3 + i
= ln
p 3 + i + i arg
p 3 + i
= ln 2 + i
6
- 2n
; n = 0; 1 ; 2 ; : : : :
8. (5 points) Find all values of z such that Log
z 2 i
i
2
, where the Log represents the prin-
cipal value of log.
Answer. Because the principal value Log z is one of the values of the logarithm log z, so we have e Log z = z; z 6 = 0:
It implies
z 2 i = e Log(z^2 i) = e
2 i^ = cos
2
- i sin
2
= i; i:e:; z 2 = 2i:
So the problem is converted to nd the values of z such that z 2 = 2i.
With z = rei^ and the exponential form of 2i, we have
r 2 e 2 i = z 2 = 2i = 2e
i( 2 +2n) ; r 2 = 2; i:e:; r =
p 2 ; 2 =
- 2n; =
- n:
Therefore, we deduce
z =
p 2 e i( 4 +n) ; i:e:; z 1 =
p 2 e i 4 = 1+i; and z 2 =
p 2 e i( 4 +) = (1 + i) :
Complex Analysis I EXAM 2 { SOLUTION Fall, 2009
9. (5 points) Find the principal value of the complex exponent
e
2
1 i
p 3
^3 i , i.e.,
P: V:
e
2
1 i
p 3
3 i :
Answer. By the de nition,
P: V:
e
2
1 i
p 3
^3 i = e 3 i Log( e 2 ( 1 i
p (^3) )) :
A simple computation shows
Log
e
2
1 i
p 3
= ln
e
2
1 i
p 3
- i
= ln e
i = 1
i;
because < Arg
e
2
1 i
p 3
=
< . Thus, we deduce
P: V:
e
2
1 i
p 3
3 i = e
3 i Log( 2 e ( 1 i
p (^3) )) = e
3 i( 1 23 i) = e 3 i+2^2 = e 3 i e 2 ^2 = e 2 ^2 :
10. (5 points) Use the de nition zc^ = ec^ log^ z^ to show that
1 + i
p 3
3 = 2 = 2
p
Proof. By the de nition,
1 + i
p 3
3 = 2 = e (3=2) log( 1+i
p (^3) ) :
A simple computation shows
log
1 + i
p 3
= ln 1 + i
p 3 + i arg
1 + i
p 3
= ln 2 + i
2
3
- 2n
;
log
1 + i
p 3
ln 2 + i
2
3
- 2n
ln 2 + i ( + 3n) =
ln 2 + i (3n + 1) ;
e (3=2) log( 1+i
p (^3) ) = e
3 2 ln 2+i(3n+1)^ = e
3 2 ln 2ei(3n+1)^ = 2^3 =^2 (1) = 2
p 2 :