United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 1 { SOLUTION
Section 1 { Section 5
Complex Analysis I
MATH 315 SECTION 01 CRN 23516
9:30 { 10:45 on Monday & Wednesday
Due Date: Monday, September 13, 2009
Complex Analysis I
HOMEWORK 1 { SOLUTION
Fall, 2009
Section 1. Sums and Products and Section 2 Basic Algebraic Properties
1. Verify that
(1.1)
p
2
i
1
i
p
i
2 = 2i
p
Proof.
2
i
i
p
p
+ i2 2 = 2
i
i
p
+ ( 1) 2 = 2i:
(1.2) (2; 3)( 2; 1) = ( 1; 8)
Proof.
(2; 3)( 2; 1) = ( 4 + 3; 2 + 6) = ( 1; 8):
(1.3) (3; 1)(3;
1 1
1) ;
= (2; 1)
5 10
Proof.
(3; 1)(3; 1) = (9 + 1; 3 + 3) = (10; 0);
(3; 1)(3;
1 1
1 1
1) ;
= (10; 0) ;
= (2
5 10
5 10
0; 1 + 0) = (2; 1):
2. Show that
(2.1) Re(iz ) =
Proof.
Im(z )
For z = x + yi,
iz
= i(x + yi) =
y
+ xi;
Re(iz ) =
y
=
Im(z );
Re(iz ) =
Im(z ):
(2.2) Im(iz ) = Re(z )
Proof.
For z = x + yi,
iz
= i(x + yi) =
y
+ xi;
Im(iz ) = x = Re(z );
Page 1 of 7
Im(iz ) = Re(z ):
Complex Analysis I
HOMEWORK 1 { SOLUTION
Fall, 2009
3. Solve the equation z 2 + z + 1 = 0 for z = (x; y ) by writing
(x; y )(x; y ) + (x; y ) + (1; 0) = (0; 0)
and then solving a pair of simultaneous equations in
Answer.
and y .
x
We expand the equation,
(0; 0) = (x; y )(x; y ) + (x; y ) + (1; 0)
= (x2 y 2 ; 2xy ) + (x; y ) + (1; 0) = (x2 y 2 + x + 1; 2xy + y + 0);
i:e:;
0 = x2 y 2 + x + 1;
and
0 = 2xy + y = y (2x + 1):
(1)
From the second equation in (1), we get y = 0 or 2x + 1 = 0, i.e., x = 1=2.
Case 1. y = 0: Putting y = 0 into the rst equation in (1), we get
0 = x2 + x + 1;
which cannot be solved for a real x. That is, y = 0 cannot be the solution of the given equation.
Case 2. x = 1=2: Putting x = 1=2 into the rst equation in (1), we get
1
0=
4
y
1
3
+1=
2
4
2
It implies the solutions to the given equation
z
z
2
2
y ;
y
=
p
3
:
2
+ z + 1 = 0 are
p
3
1
;
2
2
= (x; y ) =
!
:
Section 3. Further Properties
4. Reduce each of these quantities to a real number:
(4.1)
1 + 2i 2 i
+
3 4i
5i
Answer.
(4.2)
(1
)(2
i
!
2 i
1 + 2i 2 i 1 + 2i 3 + 4i
+
=
+
3 4i
5i
3 4i 3 + 4 i
5i
5i
)(3
i
2
:
5
)
i
Answer.
(1
(4.3) (1
!
5i
=
5i
)(2
5i
i
1
:
2
)2 = [ 2i]2 = 4:
)(3
)
i
i
=
5i
=
10i
)4
i
Answer.
(1
h
)4 = (1
i
i
i2
Page 2 of 7
Complex Analysis I
HOMEWORK 1 { SOLUTION
Fall, 2009
5. Prove that if z1 z2 z3 = 0, then at least one of the three factors is zero.
We recall that for complex numbers z and w, if zw = 0, then z = 0 or w = 0 or possibly
both z and w equal zero. It implies that if (z1 z2 )z3 = z1 z2 z3 = 0, then z1 z2 = 0 or z3 = 0 or
possibly both z1 z2 and z3 equal zero. Again if z1 z2 = 0, then z1 = 0 or z2 = 0 or possibly both
z1 and z2 equal zero. Therefore, combining those two statements, we can deduce that z1 = 0 or
z2 = 0 or z3 = 0, i.e., at least one of the three factors is zero.
Proof.
6. Use identity
z1 z2
z3 z4
=
to derive the cancelation law:
z1
z3
z1 z
z2 z
Proof.
z2
z4
=
z1
=
z1
z2
;
;
z3
6= 0 6= 0
6= 0 6=
z2
z4 ;
z:
By the identity,
z1 z
z2 z
=
z1
z2
z
z
zz
z2
1
=
z1
z2
;
z2
6= 0 6=
z:
Section 4. Moduli
7. Locate the numbers z1 + z2 and z1
(7.1)
z1
z1
2
i
3
z1 = 2i = (0; 2), z2 = 2=3
=(
p
z1
p
=(
p
z2
= ( 2=3; 3):
p
3; 1), z2 = ( 3; 0).
+ z2 = (0; 1);
z1
z2
p
= ( 2 3; 1):
= ( 4; 3):
= ( 3; 1), z2 = (1; 4).
z1
z1
z1
= ( 3; 1) and z2 = (1; 4)
Answer. z1
(7.4)
= (2=3; 1).
+ z2 = (2=3; 1);
z1
z1
i
3; 1) and z2 = ( 3; 0)
Answer. z1
(7.3)
vectorially when
= 2i and z2 =
Answer.
(7.2)
z2
= x1 + y1 i and z2 = x1
Answer. z1
+ z2 = ( 2; 5);
z2
y1 i
= x1 + y1 i = (x1 ; y1 ), z2 = x1
z1
z1
y1 i
+ z2 = (2x1 ; 0);
Page 3 of 7
= (x1 ;
z1
z2
y1
).
= (0; 2y1 ):
Complex Analysis I
HOMEWORK 1 { SOLUTION
p
8. Verify that
Proof.
Fall, 2009
2jz j j Re(z )j + j Im(z )j:
Let z = x + yi. Then
Re(z ) = x;
and
Im(z ) = y;
and
j j2 =
z
2
x
+ y 2 = jxj2 + jy j2 :
We observe
0 (jxj
j j)2 = j j2 + j j2
y
x
2jxjjy j;
y
2jxjjy j jxj2 + jy j2 :
i:e:;
Adding jxj2 + jy j2 to both sides, we get
j j2 + j j2 + 2j jj j 2 j j2 + j j2
x
y
x y
x
y
;
(jxj + jy j)2 2
i:e:;
j j2 + j j2
x
y
:
Now we take both sides to be squared. The inequality does not change the order, because both
sides are nonnegative.
q
p
q
(jxj + jy j)2 2 (jxj2 + jy j2 );
j j+j j
i:e:;
x
p
y
p
q
2 (jxj2 + jy j2 );
which is j Re(z )j + j Im(z )j 2jz j.
9. In each case, sketch the set of points determined by the given condition:
(9.1)
j
1 + ij = 1
z
Answer.
(9.2)
j
z
i
with radius 1.
+ ij 3
Answer.
3.
(9.3)
It is a circle centered at the point 1
j
It is a disc inside and including the circle centered at the point
4ij 4
z
Answer.
4.
i
with radius
It is a region outside and including the circle centered at the point 4i with radius
Section 5. Complex Conjugates
10. Use properties of conjugates and moduli to show that
(10.1) z + 3i = z
Proof.
3i
+ 3i = z + 3i = z
z
(10.2)
iz
=
3i:
iz
Proof.
iz
= iz =
Page 4 of 7
iz :
Complex Analysis I
HOMEWORK 1 { SOLUTION
(10.3) (2 + i)2 = 3
Fall, 2009
4i
Since (2 + i)2 = 3 + 4i, so
Proof.
(2 + i)2 = 3 + 4i = 3
p
j = 3j2
Recall j j = j j.
p
(10.4) j (2
z + 5)
Proof.
p
2
i
z
4i:
+ 5j
z
z
(2
z + 5)
i 2
p
i
= j2
z + 5j 2
p
p
p
= 3j2
z + 5j
p
= 3j2
z + 5j =
3j2z + 5j = 3j2z + 5j:
11. Sketch the set of points determined by the condition
z
(11.1) Re(
)=2
i
Answer.
We recall Re(z ) = Re(
z ). So
Re (
z
) = Re z
i
i
= Re z i = Re (z + i) :
That is, the given equation is equivalent to Re (z + i) = 2. Let z = x + yi and put into the
equation, then
x
= Re(x + (y + 1)i) = Re(x + yi + i) = Re(z + i) = 2;
i:e:;
x
It implies that for any y , z = 2 + yi satises the equation. That is, the line
complex plane satises the given equation.
(11.2) j2z
= 2:
x
= 2 in the
j=4
i
Answer.
We simplify the given equation
2
z
i
2
= 4;
i:e:;
z
i
2
= 2;
which represents a circle centered at the point i=2 with radius 2.
Page 5 of 7
Complex Analysis I
HOMEWORK 1 { SOLUTION
Fall, 2009
12. Use established properties of moduli to show that when jz3 j 6= jz4 j,
z1
z3
Proof.
The Triangle
j
z1
+ z2 + z4 j 1j + j 2j
jj 3j j 4jj
z
z
z
z
:
implies
Inequality
+ z2 j jz1 j + jz2 j;
z1
z3
and so
+ z2 =
+ z4 j
j
+ z2 j
z3 + z4 j
z1
jj 1j ++ j 2jj
3
4
z
z
z
z
:
Now we recall, from the triangle inequality,
jj 3j j 4jj j
z
z
z3
+ z4 j;
and so
0<
1
jz3 + z4j
jj j 1 j jj
3
4
z
z
:
Therefore, we conclude
z1
z3
+ z2 + z4 j 1j + j 2j j 1j + j 2j
j 3 + 4j jj 3j j 4jj
z
z
z
z
13. Show that
Proof.
z
z
z
z
;
i:e:;
Re(2 + z + z 3 ) 4;
Recall for a complex number w, j Re(w
z1
z3
+ z2 + z4 j 1j + j 2j
jj 3j j 4jj
z
z
z
z
:
j j1
)j j j. So for j j 1, we have
when
w
z
:
z
Re(2 + z + z 3 ) 2 + z + z 3 j2j + jzj + jz 3 j = 2 + jz j + jz j3 2 + 1 + 1 = 4:
14. Prove that
(14.1)
z
is real if and only if z = z .
Let z = x + yi.
If z is real, then y = 0 and so z = x, i.e., z = z .
If z = z , then x yi = x + yi, 2yi = 0, i.e., y = 0 and so z = x + yi becomes a real x.
Proof.
(14.2)
z
is either real or pure imaginary if and only if z2 = z 2 .
Let z = x + yi.
If z is real, i.e., y = 0, then from the problem above, z = z and so z2 = z 2 .
If z is pure imaginary, i.e., x = 0 and so z = yi, then z = yi and z2 = y 2 = z 2 , i.e., z2 = z 2 .
If z2 = z 2 , then z2 z 2 = 0, i.e.,
Proof.
0 = z2
z
2
= (x
yi
)2
(x + yi)2 = x2
y
2
2xyi
(x2
i.e., y = 0 or x = 0, which means z is either real or pure imaginary.
Page 6 of 7
y
2
+ 2xyi) = 4xyi;
Complex Analysis I
HOMEWORK 1 { SOLUTION
15. Using expressions,
+ z
;
Im(z ) =
2
2
y = 1 can be written
Re(z ) =
show that the hyperbola
2
x
z
z
Proof.
2
Fall, 2009
z
z
2i
;
+ z2 = 2:
Let z = x + yi. Then by the given expressions,
x
=
z
+ z
;
2
Putting them into the hyperbola equation
+ z 2
2
2
2
x
z
z
y
=
2
= 1, we have
y
2i
:
(z + z)2 (z z)2
+
2i
4
4
1
1
= z 2 + z2 + 2z z + z 2 + z2 2z z = z 2 + z2 ;
4
2
1=
z
z
z
=
Page 7 of 7
i:e:;
z
2
+ z2 = 2:
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 2 – SOLUTION
Complex Analysis I
MATH 315 SECTION 01 CRN 23516
9:30 – 10:45 on Monday & Wednesday
Due Date: Monday, September 28, 2009
Complex Analysis I
Fall, 2009
HOMEWORK 2 – SOLUTION
Section 6. Exponential Form
Section 7. Products and Quotients in Exponential Form
1. Find the principal argument Arg(z) when
(1.1) z =
i
−2 − 2i
Answer.
z=
(1.2) z =
√
i
1
= − (1 + i) ,
−2 − 2i
4
tan (Arg(z)) =
−1/4
= 1,
−1/4
Arg(z) = −
3π
.
4
6
3−i
Answer. Let w =
√
3 − i. Then z = w6 and
−1
π
tan (Arg(w)) = √ = − ,
6
3
π
Arg(w) = − ,
6
Arg(z) = Arg(w6 ) = 6 Arg(w) = −π,
where Arg(z1 z2 ) = Arg(z1 ) + Arg(z2 ) is used. Since −π < Arg(z) ≤ π, hence we conclude
Arg(z) = π.
In fact, z =
√
6
3 − i = −64 and so Arg(z) = π.
2. Show that
(2.1) |eiθ | = 1
Proof. By Euler’s formula,
eiθ = cos θ + i sin θ,
|eiθ | = | cos θ + i sin θ| =
p
cos2 θ + sin2 θ = 1.
(2.2) eiθ = e−iθ .
Proof. By Euler’s formula,
eiθ = cos θ + i sin θ = cos θ − i sin θ = cos(−θ) + i sin(−θ) = e−iθ .
3. Using the fact that the modulus eiθ − 1 is the distance between the points eiθ and 1, give a
geometric
iθ
argument to find a value of θ in the interval 0 ≤ θ < 2π that satisfies the equation
e − 1 = 2.
Answer. Since eiθ , 0 ≤ θ < 2π, has the modulus |eiθ | = 1, so the points eiθ , 0 ≤ θ < 2π, are all
on the unit circle centered at the origin. (The unit circle means a circle with radius 1.) If eiθ ,
0 ≤ θ < 2π, satisfies the equation |eiθ − 1| = 2, then the point eiθ is 2 units away from the point
(1, 0) in the plane. Hence the point eiθ should be at (−1, 0) and it means θ = π.
4. Use de Movire’s formula to derive the following trigonometric identities:
(4.1) cos(3θ) = cos3 θ − 3 cos θ sin2 θ
(4.2) sin(3θ) = 3 cos2 θ sin θ − sin3 θ
Page 1 of 9
Complex Analysis I
Fall, 2009
HOMEWORK 2 – SOLUTION
Proof.
eiθ = cos θ + i sin θ,
cos(3θ) + i sin(3θ) = ei3θ = (eiθ )3 = (cos θ + i sin θ)3
= cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ
= cos3 θ − 3 cos θ sin2 θ + i 3 cos2 θ sin θ − sin3 θ .
By equating the real parts and then the imaginary parts, we arrive at the desired trigonometric
identities,
cos(3θ) = cos3 θ − 3 cos θ sin2 θ,
sin(3θ) = 3 cos2 θ sin θ − sin3 θ.
5. By writing the individual factors on the left in exponential form, performing the needed operators,
and finally changing back to rectangular coordinates, show that
√
√ √
(5.1) i(1 − i 3)( 3 + i) = 2(1 + i 3)
√
πi √
πi
πi
Answer. It is straightforward to see i = e 2 , 1 − i 3 = 2e− 3 , 3 + i = 2e 6 . So,
√ √ √
πi
πi
πi
πi
π
π
=2 1+i 3 .
i(1 − i 3)( 3 + i) = e 2 2e− 3 2e 6 = 4e 3 = 4 cos + i sin
3
3
(5.2)
5i
= 1 + 2i
2+i
πi
Answer. It is straightforward to see 5i = 5e 2 , 2 + i =
√
−1
5ei tan
1
2.
So,
πi
√
√ i π −tan−1 1
π
5i
5e 2
π
−1 1
−1 1
)
(
2
2
= 5 cos
=√
= 5e
− tan
− tan
+ i sin
−1 1
2+i
2
2
2
2
5ei tan 2
√
1
2
= 5 √ + i √ = 1 + 2i.
5
5
(5.3) (−1 + i)7 = −8 (1 + i)
Answer. It is straightforward to see −1 + i =
√ 3πi
2e 4 . So,
√ 7 3πi 7
21πi
21π
21π
7/2
7/2
(−1 + i) =
2
e 4
=2 e 4 =2
cos
+ i sin
4
4
1
1
= 27/2 − √ − i √
= −8 (1 + i) .
2
2
7
(5.4)
√ −10
√ −11
=2
−1 + i 3
1+i 3
√
πi
Answer. It is straightforward to see 1 + i 3 = 2e 3 . So,
−10
√ −10
πi
10πi
10π
10π
−
−10
−10
−10
(1 + i 3)
=2
e3
=2 e 3 =2
cos
− i sin
3
3
Page 2 of 9
Complex Analysis I
Fall, 2009
HOMEWORK 2 – SOLUTION
√ !
√ 1
3
−11
− +i
=2
−1 + i 3 .
2
2
−10
=2
6. Prove that two nonzero complex numbers z1 and z2 have the same moduli if and only if there are
complex numbers c1 and c2 such that z1 = c1 c2 and z2 = c1 c̄2 .
Proof. Suppose there are complex numbers c1 and c2 such that z1 = c1 c2 and z2 = c1 c̄2 . Then, by
the properties on moduli, we have
|z1 | = |c1 c2 | = |c1 ||c2 | = |c1 ||c̄2 | = |c1 c̄2 | = |z2 |,
|z1 | = |z2 |.
i.e.,
Suppose that two nonzero complex numbers z1 and z2 have the same moduli. Let |z1 | = r1 = |z2 |
and Arg(z1 ) = r1 and Arg(z2 ) = θ2 . Then we may write
z2 = r2 eiθ2 .
z1 = r1 eiθ1 ,
If we introduce the numbers
c1 = r1 ei
θ1 +θ2
2
,
and
c2 = ei
θ1 −θ2
2
,
we find that
c1 c2 = r1 ei
θ1 +θ2
2
ei
θ1 −θ2
2
= r1 eiθ1 = z1 ,
and
c1 c̄2 = r1 ei
z1 = c1 c2 ,
and
z2 = c1 c̄2 .
θ1 +θ2
2
e−i
θ1 −θ2
2
= r1 eiθ2 = z2 .
That is,
Hence, there are complex numbers c1 and c2 desired in the problem.
7. Establish the identity
1 − z n+1
,
1 + z + z + ··· + z =
1−z
2
n
z 6= 1,
and then use it to derive Lagrange’s trigonometric identity:
1 + cos θ + cos(2θ) + · · · + cos(nθ) =
Proof. Let S =
Pn
k=0
1 sin[(2n + 1)θ/2]
+
,
2
2 sin(θ/2)
0 < θ < 2π.
z k . Then we observe for z 6= 1,
(1 − z)S = S − zS =
n
X
k=0
k
z −
n
X
z
k+1
=
k=0
n
X
z k − z k+1 = 1 − z n+1 ,
k=0
Substituting z = eiθ into the equation above, we get
n
X
k=0
ikθ
e
1 − ei(n+1)θ
=
.
1 − eiθ
Page 3 of 9
S=
1 − z n+1
.
1−z
Complex Analysis I
HOMEWORK 2 – SOLUTION
Fall, 2009
Euler’s formula implies
n
X
cos(kθ) + i
k=0
n
X
n
X
1 − ei(n+1)θ
1 − eiθ
k=0
1 − ei(n+1)θ e−iθ/2
e−iθ/2 − ei(2n+1)θ/2
=
=
,
1 − eiθ
e−iθ/2
e−iθ/2 − eiθ/2
sin(kθ) =
k=0
eikθ =
Again Euler’s formula implies
n
X
n
X
e−iθ/2 − ei(2n+1)θ/2
cos(kθ) + i
sin(kθ) =
e−iθ/2 − eiθ/2
k=0
k=0
cos(θ/2) − i sin(θ/2) − [cos((2n + 1)θ/2) + i sin((2n + 1)θ/2)]
−2i sin(θ/2)
cos(θ/2) − i sin(θ/2) − [cos((2n + 1)θ/2) + i sin((2n + 1)θ/2)] i
=
−2i sin(θ/2)
i
sin(θ/2) + sin((2n + 1)θ/2) + i [cos(θ/2) − cos((2n + 1)θ/2)]
.
=
2 sin(θ/2)
=
By equating the real parts and the imaginary parts, we find
n
X
k=0
n
X
cos(kθ) =
sin(θ/2) + sin((2n + 1)θ/2)
1 sin((2n + 1)θ/2)
= +
,
2 sin(θ/2)
2
sin(θ/2)
sin(kθ) =
1
cos((2n + 1)θ/2)
cos(θ/2) − cos((2n + 1)θ/2)
= cot(θ/2) −
,
2 sin(θ/2)
2
sin(θ/2)
k=0
where 0 < θ < 2π.
Another Proof. Euler’s formula implies
n
X
cos(kθ) + i
k=0
n
X
k=0
n
X
1 − ei(n+1)θ
1 − eiθ
k=0
1 − ei(n+1)θ 1 − e−iθ
1 − e−iθ + einθ − ei(n+1)θ
=
=
1 − eiθ
1 − e−iθ
2 − e−iθ − eiθ
1 − cos θ + cos(nθ) − cos(n + 1)θ + i [sin θ + sin(nθ) − sin(n + 1)θ]
=
2(1 − cos θ)
sin(kθ) =
eikθ =
That is, we deduce
n
X
k=0
n
X
k=0
cos(kθ) =
1 − cos θ + cos(nθ) − cos(n + 1)θ
1 cos(nθ) − cos(n + 1)θ
= +
,
2(1 − cos θ)
2
2(1 − cos θ)
sin(kθ) =
sin θ + sin(nθ) − sin(n + 1)θ
,
2(1 − cos θ)
where we will simplify the first one.
Page 4 of 9
Complex Analysis I
HOMEWORK 2 – SOLUTION
Fall, 2009
We recall from Calculus
1 − cos A = 2 sin2 (A/2),
cos(A + B) = cos A cos B − sin A sin B,
sin(2A) = 2 sin A cos A,
sin(A + B) = sin A cos B + cos A sin B.
Using them, we have
n
X
cos(kθ) =
k=0
1 cos(nθ) − cos(n + 1)θ
+
2
2(1 − cos θ)
=
1 cos(nθ) − cos(nθ) cos θ + sin(nθ) sin θ
+
2
4 sin2 (θ/2)
=
1 cos(nθ) (1 − cos θ) + 2 sin(nθ) sin(θ/2) cos(θ/2)
+
2
4 sin2 (θ/2)
=
1 2 cos(nθ) sin2 (θ/2) + 2 sin(nθ) sin(θ/2) cos(θ/2)
+
2
4 sin2 (θ/2)
1 cos(nθ) sin(θ/2) + sin(nθ) cos(θ/2)
+
2
2 sin(θ/2)
1 sin(nθ + θ/2)
1 sin[(2n + 1)θ/2]
= +
= +
.
2
2 sin(θ/2)
2
2 sin(θ/2)
=
Page 5 of 9
Complex Analysis I
Fall, 2009
HOMEWORK 2 – SOLUTION
Section 8. Roots of Complex Numbers and Section 9. Examples
8. Find the square roots of the followings and express them in rectangular coordinates.
(8.1) 2i
Answer. We look for z such that z 2 = 2i, i.e., z = (2i)1/2 . The complex number 2i has the
modulus |2i| = 2 and the principal argument Arg(2i) = π/2. From the information, 2i has
the exponential form,
π
2i = |2i|ei arg(2i) = 2ei(Arg(2i)+2kπ) = 2ei( 2 +2kπ) ,
√
π
π
k = 0, ±1, ±2, . . . .
(2i)1/2 = 21/2 ei( 4 +kπ) = 2ei( 4 +kπ) ,
Hence, we deduce all two square roots by putting k = 0, 1,
√
√
1
1
c0 = 2e
= 2 √ + √ i = 1 + i,
2
2
√ i π +π
√
1
1
(
)
4
c1 = 2e
= 2 − √ − √ i = −1 − i.
2
2
i( π4 )
√
(8.2) 1 − i 3
√
√ 1/2
2
Answer.
We
look
for
z
such
that
z
=
1
−
i
3,
i.e.,
z
=
(1
−
i
3) . The complex
number
√
√
√
1 − i 3 has the modulus |1√− i 3| = 2 and the principal argument Arg(1 − i 3) = −π/3.
From the information, 1 − i 3 has the exponential form,
√
√
√
√
π
1 − i 3 = |1 − i 3|ei arg(1−i 3) = 2ei(Arg(1−i 3)+2kπ) = 2ei(− 3 +2kπ) ,
√
√
π
π
(1 − i 3)1/2 = 21/2 ei(− 6 +kπ) = 2ei(− 6 +kπ) ,
k = 0, ±1, ±2, . . . .
Hence, we deduce all two square roots by putting k = 0, 1,
! √
√
1
3 1
3
c0 = 2e
= 2
− i = √ − √ i,
2
2
2
2
!
√
√
√
√
π
1
1
3
3
c1 = 2ei(− 6 +π) = 2 − √ + √ i = − √ + √ i.
2
2
2
2
√
i(− π6 )
√
9. In each case, find all of the roots in rectangular coordinates, exhibit them as vertices of certain
squares, and point out which is the principal root.
(9.1) (−16)1/4
Answer. We look for z such that z = (−16)1/4 . The complex number −16 has the modulus
|−16| = 16 and the principal argument Arg(−16) = π. From the information, −16 has the
exponential form,
−16 = |−16| ei arg(−16) = 16ei(Arg(−16)+2kπ) = 16ei(π+2kπ) ,
(−16)1/4 = 161/4 e
(2k+1)
πi
4
= 2e
(2k+1)
πi
4
Page 6 of 9
,
k = 0, ±1, ±2, . . . .
Complex Analysis I
Fall, 2009
HOMEWORK 2 – SOLUTION
Hence, we deduce all two square roots by putting k = 0, 1, 2, 3,
√
√
3πi
1
1
1
1
c0 = 2e = 2 √ + √ i = 2 (1 + i) ,
c1 = 2e 4 = 2 − √ + √ i = 2 (−1 + i) ,
2
2
2
2
√
√
7πi
5πi
1
1
1
1
c3 = 2e 4 = 2 √ − √ i = 2 (1 − i) .
c2 = 2e 4 = 2 − √ − √ i = − 2 (1 + i) ,
2
2
2
2
πi
4
√ √
√ √
The
numbers
2), (− 2, 2),
0 , c1 , c2 and c3 are respectively located at ( 2,
√ complex
√
√ c√
(− 2, − 2), and
√ (√ 2, − 2) in the complex plane and forms a square. The principal root is
c0 located at ( 2, 2).
√
(9.2) (−8 − 8i 3)1/4
√ 1/4
√
Answer. We look for √
z such
−8 − 8i 3 has
that z = (−8 − 8i 3) . The complex number
√
the modulus −8 − 8i 3 = 16 and the principal argument Arg(−8 − 8i 3) = −2π/3. From
√
the information, −8 − 8i 3 has the exponential form,
√
√
√ i arg(−8−8i√3)
2π
−8 − 8i 3 = −8 − 8i 3 e
= 16ei(Arg(−8−8i 3)+2kπ) = 16ei(− 3 +2kπ) ,
√
(3k−1)
π
kπ
(−8 − 8i 3)1/4 = 161/4 ei(− 6 + 2 ) = 2e 6 πi ,
k = 0, ±1, ±2, . . . .
Hence, we deduce all two square roots by putting k = 0, 1, 2, 3,
− πi
6
c0 = 2e
c2 = 2e
5πi
6
!
√
√ !
√
√
πi
1
3 1
3
=2
− i = 3 − i,
c1 = 2e 3 = 2
+
i = 1 + i 3,
2
2
2
2
!
√
√ !
√
√
4πi
3 1
1
3
=2 −
+ i = − 3 + i,
c3 = 2e 3 = 2 − −
i = −1 − i 3.
2
2
2
2
√
√
√
The complex
numbers
c
,
c
,
c
and
c
are
respectively
located
at
(
3,
−1),
(1,
3),
(−
3, 1),
0
1
2
3
√
and
(−1,
−
3)
in
the
complex
plane
and
forms
a
square.
The
principal
root
is
c
located
at
0
√
( 3, −1).
10. The three cube roots of a nonzero complex number z0 can be written c0 , c0 ω3 , c0 ω32 , where c0 is
the principal cube root of z0 and
ω3 = exp
2πi
3
√
−1 + i 3
=
.
2
√
√
√
Show that if z0 = −4 2+4i 2, then c0 = 2(1+i) and the other two cube roots are in rectangular
form, the numbers
√
√
−( 3 + 1) + ( 3 − 1)i
√
c0 ω3 =
,
2
c0 ω32
and
√
√
( 3 − 1) − ( 3 + 1)i
√
=
.
2
√
Proof. √
We find all cube roots of z0 = −4 2 (1 − i). The given complex number z0 has the modulus
|z0 | = 32 + 32 = 8 and the principal argument Arg(z0 ) = 3π/4. With this information, z0 has
the exponential form,
3π
z0 = |z0 |ei arg(z0 ) = |z0 |ei(Arg(z0 )+2kπ) = 8ei( 4 +2kπ)
Page 7 of 9
Complex Analysis I
1/3
z0
Fall, 2009
HOMEWORK 2 – SOLUTION
π
= 81/3 ei( 4 +
2kπ
3
h 2π ik
π
π
) = 2ei π4 ei 2kπ
3
= 2ei 4 ei 3
= 2ei 4 ω3k ,
k = 0, ±1, ±2, . . . .
Hence, we deduce all three cube roots by putting k = 0, 1, 2,
√
1
1
= 2e = 2 √ + √ i = 2 (1 + i) ,
c0 = 2e
2
2
"
√ #
√
√
√
π
π 2πi
−1
+
i
3
−(
3
+
1)
+
(
3 − 1)i
√
c1 = 2ei 4 ω31 (= c0 ω3 ) = 2ei 4 e 3 = 2 (1 + i)
=
,
2
2
"
√ #2
√
√
h 2πi i2 √
π
π
−1
+
i
3
( 3 − 1) − ( 3 + 1)i
i4 2
2
i4
√
c2 = 2e ω3 (= c0 ω3 ) = 2e
e 3
= 2 (1 + i)
=
.
2
2
i π4
i π4
ω30
11. Find the four roots of the equation z 4 + 4 = 0 and use them to factor z 4 + 4 into quadratic factors
with real coefficients.
Answer. The equation z 4 + 4 = 0 is equivalent to z 4 = −4, i.e., z = (−4)1/4 . We express the
complex numbers −4 and (−4)1/4 in the exponential form. Since −4 has the modulus | − 4| = 4
and the principal argument Arg(−4) = π, because we use a principal argument in (−π, π]. Using
the information, we have
−4 = | − 4|ei arg(−4) = 4ei(Arg(−4)+2kπ) = 4ei(π+2kπ) = 4e(2k+1)πi ,
√ (2k+1)πi
(2k+1)πi
= 2e 4 ,
k = 0, ±1, ±2, . . . .
(−4)1/4 = 41/4 e 4
Hence, we deduce all four roots by putting k = 0, 1, 2, 3,
√ 3πi √
1
1
1
1
c1 = 2e 4 = 2 − √ + √ i = −1 + i,
c0 = 2e = 2 √ + √ i = 1 + i,
2
2
2
2
√ 5πi √
√ 7πi √
1
1
1
1
c2 = 2e 4 = 2 − √ − √ i = −1 − i,
c3 = 2e 4 = 2 √ − √ i = 1 − i.
2
2
2
2
√
πi
4
√
Since c0 , c1 , c2 , and c3 are roots of the equation z 4 + 4 = 0, so we can express z 4 + 4 as follows:
z 4 + 4 = [z − c0 ] [z − c1 ] [z − c2 ] [z − c3 ]
= [z − (1 + i)] [z − (−1 + i)] [z − (−1 − i)] [z − (1 − i)]
= [(z − 1) − i] [(z + 1) − i] [(z + 1) + i] [(z − 1) + i]
= (z − 1)2 − i2 [(z + 1) − i] [(z + 1) + i]
= (z − 1)2 − i2 (z + 1)2 − i2 = z 2 − 2z + 2 z 2 + 2z + 2 .
12. Show that if c is any nth root of unity other than unity itself, then
1 + c + c2 + · · · + cn−1 = 0.
Page 8 of 9
Complex Analysis I
Fall, 2009
HOMEWORK 2 – SOLUTION
Proof. We observe that if zw = 0 for complex numbers z 6= 0 and w, then w = 0. For this reason,
(1 − c) 1 + c + c2 + · · · + cn−1 = 0
implies
1 + c + c2 + · · · + cn−1 = 0,
where c 6= 1. Hence, it is enough to prove that
(1 − c) 1 + c + c2 + · · · + cn−1 = 0.
Expanding the left–hand side, we get
(1 − c) 1 + c + c2 + · · · + cn−1 = 1 + c + c2 + · · · + cn−1 − c 1 + c + c2 + · · · + cn−1
= 1 + c + c2 + · · · + cn−1 − c + c2 + c3 + · · · + cn−1 + cn
= 1 + c + c2 + · · · + cn−1 − c + c2 + c3 + · · · + cn−1 + 1 = 0,
where cn = 1 is used.
Another Proof. Using the identity in the problem 7 above, we have
1 + c + c2 + · · · + cn−1 =
1 − cn
1−1
=
= 0,
1−c
1−c
because c is the nth root of unity, i.e., cn = 1.
Page 9 of 9
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 3 – SOLUTION
Complex Analysis I
MATH 315 SECTION 01 CRN 23516
9:30 – 10:45 on Monday & Wednesday
Due Date: Wednesday, October 7, 2009
Complex Analysis I
HOMEWORK 3 – SOLUTION
Fall, 2009
1. Sketch the following sets and determine which are domains.
(1.1) | z − 2 + i | ≤ 1.
Answer. With z = x + iy, the given inequality implies
1 ≤ | x − 2 + i ( 1 + y ) | = ( x − 2 )2 + ( 1 + y )2 ,
i.e., it represents the region outside (including boundary) the circle centered at (2, −1) with the
radius 1. See the figure below.
(1.2) | 2z + 3 | > 4.
Answer. With z = x + iy, the given inequality implies
(
4 < | 2x + 3 + 2yi | = ( 2x + 3 ) + ( 2y ) ,
2
2
1<
3
x+
2
)2
+ y2,
i.e., it represents the region outside (excluding boundary) the circle centered at (−3/2, 0) with
the radius 1. See the figure below.
H1.2L È2z+3È>4
H1.1L Èz-2+iÈ£1
0.0
2
-0.5
1
Y
0
Y -1.0
-1
-1.5
-2
-2.0
1.0
1.5
2.0
2.5
3.0
-4
X
-3
-2
-1
0
1
X
(1.3) Im(z) > 1.
Answer. With z = x + iy, the given inequality implies
1 < Im(z) = x
i.e., it represents the region on the left–hand side (excluding boundary) the line x = 1. See the
figure below.
(1.4) | z − 4 | ≥ | z |.
Answer. With z = x + iy, the given inequality implies
x2 + y 2 = | x + iy | ≤ | x − 4 + iy | = ( x − 4 )2 + y 2 = x2 − 8x + 16 + y 2 ,
8x ≤ 16,
x ≤ 2,
i.e., it represents the region on the left–hand side (including boundary) the line x = 2. See the
figure below.
Page 1 of 5
Complex Analysis I
HOMEWORK 3 – SOLUTION
H1.3L ImHzL>1
Fall, 2009
H1.4L Èz-4ȳÈzÈ
6
3
5
2
4
1
Y 3
Y
0
2
-1
1
-2
0
-3
-2
-1
0
1
2
-3
-3
3
-2
-1
X
0
1
2
3
X
2. In each case, sketch the closure of the set.
(2.1) | Re(z) | < | z |.
Answer. With z = x + iy, we observe
√
| x | = | Re(z) | < | z | = x2 + y 2 ,
i.e.,
x2 = | x |2 < x2 + y 2 ,
0 < |y|,
0 < y2,
which represents the whole complex plane except the real axis (y = 0). Hence, the closure of the
set will be the whole complex plane.
(2.2) Re(z 2 ) > 0.
Answer. With z = x + iy, we have
z 2 = x2 − y 2 + 2xyi,
Re(z 2 ) = x2 − y 2 .
So the region, 0 < x2 − y 2 = (x − y)(x + y), is the region satisfying (1) x − y > 0 and x + y > 0
and (2) x − y < 0 and x + y < 0. Hence, the closure of the set will be the region including the
boundaries y = ±x. See the figure below.
H2.2L ReHz2 L>0
Y
H2.2L ReHz2 L³0
3
3
2
2
1
1
0
Y
0
-1
-1
-2
-2
-3
-3
-2
-1
0
1
2
3
X
-3
-3
-2
-1
0
X
Page 2 of 5
1
2
3
Complex Analysis I
HOMEWORK 3 – SOLUTION
Fall, 2009
3. For each of the functions below, describe the domain of definition that is understood.
(
(3.1) f (z) = Arg
1
z
)
.
Answer. Since 1/z is defined at any z, except at z = 0, thus f (z) is defined everywhere except
at z = 0.
(3.2) f (z) =
z
.
z + z̄
Answer. The denominator of f (z) vanishes at the point satisfying
z + z̄ = 0,
z = −z̄,
x + iy = − ( x − iy ) = −x + iy,
i.e.,
i.e.,
x = 0,
i.e., f (z) is defined everywhere except at the points on the imaginary axis.
(3.3) f (z) =
1
.
1 − | z |2
Answer. The denominator of f (z) vanishes at the point satisfying
1 − | z |2 = 0,
| z |2 = 1,
i.e.,
x2 + y 2 = 1,
i.e., f (z) is defined everywhere except at the points on the unit circle centered at the origin.
4. Suppose that f (z) = x2 − y 2 − 2y + i ( 2x − 2xy ), where z = x + yi. Use the expressions,
x=
z + z̄
,
2
and
y=
z − z̄
,
2i
to write f (z) in terms of z and simplify the result.
Answer. We recall that
z 2 = ( x + iy )2 = x2 − y 2 + 2ixy,
( )
Re z 2 = x2 − y 2 ,
( )
Im z 2 = 2xy,
( )
( )
z 2 = Re z 2 − i Im z 2 .
Using them with the given expressions above, we have
f (z) = x2 − y 2 − 2y + i ( 2x − 2xy )
( )
( )
z − z̄
z + z̄
= Re z 2 − 2
+ 2i
− i Im z 2
2i
2
( )
( 2)
= Re z + i ( z − z̄ ) + i ( z + z̄ ) − i Im z 2
( )
( )
= Re z 2 − i Im z 2 + 2iz = z 2 + 2iz.
Page 3 of 5
Complex Analysis I
HOMEWORK 3 – SOLUTION
5. Write the function
1
f (z) = z + ,
z
Fall, 2009
z ̸= 0,
in the form f (z) = u(r, θ) + iv(r, θ).
Answer. Using the exponential form z = reiθ , we have
1
e−iθ
iθ
=
re
+
reiθ
r
1
= r ( cos θ + i sin θ ) + ( cos θ − i sin θ )
(
)r (
) (
)
(
)
1
1
1
1
= r cos θ + cos θ + i r sin θ − sin θ = r +
cos θ + i r −
sin θ.
r
r
r
r
f (z) = reiθ +
Hence, in the form f (z) = u(r, θ) + iv(r, θ), we have
(
)
(
)
1
1
u(r, θ) = r +
cos θ,
v(r, θ) = r −
sin θ.
r
r
6. Find and sketch the image S ′ of the semi–infinite strip S = { z = (x, y) | 0 ≤ x, 0 ≤ y ≤ π } under
the given transformation.
(6.1) w = z 2 .
Answer. A simple computation shows with z = x + iy,
u(x, y) + iv(x, y) = w = z 2 = x2 − y 2 + 2ixy,
u(x, y) = x2 − y 2 ,
v(x, y) = 2xy.
(1) The line y = 0 corresponds to u(x, 0) = x2 and v(x, 0) = 0, and (u, v) = (x2 , 0) makes the
positive real axis in the w–plane. That is, y = 0 corresponds to the positive real axis in the
w–plane. As x decreases on the line y = 0, u = x2 also decreases on the positive real axis.
(2) The line x = 0 corresponds to u(0, y) = −y 2 and v(0, y) = 0, and (u, v) = (−y 2 , 0) makes
the negative real axis in the w–plane. That is, x = 0 corresponds to the negative real axis in the
w–plane. As y increases on the line x = 0, u = −y 2 decreases on the negative real axis.
(3) The line y = π corresponds to u(x, π) = x2 − π 2 and v(x, π) = 2πx and (u, v) = (x2 − π 2 , 2πx)
makes a parabolic curve of equation
( v )2
√
)
(
v = ±2π u + π 2 .
− π2,
4π 2 u + π 2 = v 2 ,
u=
2π
As x increases
on the line y = π, both u = x2 − π 2 and v = 2πx increases on the parabolic curve
√
v = ±2π u + π 2 .
(4) Since 0 ≤ x and 0 ≤ y ≤ π, so v = 2xy ≥ 0.
Combining (1) through (4), the image S ′ should be bounded by the real axis and the parabolic
curve, i.e., we have the image S ′ as follows:
}
{
√
′
2
.
S = w = (u, v) | 0 ≤ v ≤ 2π u + π
Page 4 of 5
Complex Analysis I
HOMEWORK 3 – SOLUTION
Fall, 2009
H6.1L S=8Hx,yLÈ 0£x, 0£y£Π <
u + Π2 <
H6.1L S'=8Hu,vLÈ 0£v£2Π
30
4
3
20
2
y
v
1
10
0
0
-1
-1
0
1
2
-10
-15
3
-10
x
0
-5
5
10
u
(6.2) w = ez .
Answer. A simple computation shows with z = x + iy,
ρeiϕ = w = ez = ex+iy = ex eiy ,
ρ = ex ,
ϕ = y.
(1) The line y = 0 corresponds to ϕ = 0 and ρ = ex , and so in polar coordinates system, it
corresponds to the positive real axis in the w–plane. As x decreases on the line y = 0, ρ = ex
also decreases on the positive real axis.
(2) The line x = 0 corresponds to ρ = e0 = 1 and ϕ = y, and so in polar coordinates system, it
corresponds to the circle with radius 1. As y increases on the line x = 0, ϕ = y increases on the
circle, i.e., the points on the circles moves counterclockwise.
(3) The line y = π corresponds to ϕ = π and ρ = ex , and so in polar coordinates system, it
corresponds to the negative real axis. As x increases on the line y = π, the point on the negative
real axis ρ = ex goes away from the origin, i.e., decreases.
Combining (1) through (3), the image S ′ should be above the real axis and outside the unit circle
centered at the origin, i.e., we have the image S ′ as follows:
{
}
S ′ = w = ρeiϕ | 1 ≤ ρ, 0 ≤ ϕ ≤ π .
H6.2L S'=8ΡãiΦ È 1£Ρ, 0£Φ£Π <
H6.1L S=8Hx,yLÈ 0£x, 0£y£Π <
4
3
3
2
2
y
v
1
1
0
0
-1
-1
0
1
2
3
x
-1
-2
-1
0
u
Page 5 of 5
1
2
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 4 – SOLUTION
•
•
•
•
Section
Section
Section
Section
15
16
17
18
Limits
Theorem on Limits
Limits Involving The Point At Infinity
Continuity
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
HOMEWORK 4 – SOLUTION
Fall, 2009
1. Compute the given complex limit.
(1.1) lim
z → 2i
z 2 − z̄
Answer. Since z 2 and z̄ are continuous everywhere in the complex plane, so we have
lim z 2 − z̄ = ( 2i )2 − 2i = −4 − ( −2i ) = −4 + 2i.
z → 2i
(1.2)
z − z̄
z → 1+i z + z̄
lim
Answer. Since z − z̄ and z + z̄ are continuous everywhere in the complex plane and z + z̄ 6= 0
at z = 1 + i, so we have
z − z̄
limz → 1+i ( z − z̄ )
1 + i − (1 − i)
2i
=
=
=
= i.
z → 1+i z + z̄
limz → 1+i ( z + z̄ )
1+i+1−i
2
lim
(1.3) lim ez
z → πi
Answer. Since ez is continuous everywhere in the complex plane, so we have
lim ez = elimz → πi z = eπi = cos π + i sin π = −1.
z → πi
(1.4)
lim ( ez + z )
z → 2+i
Answer. Since ez + z is continuous everywhere in the complex plane, so we have
lim ( ez + z ) = e2+i + 2 + i = e2 ei + 2 + i
z → 2+i
= e2 ( cos 1 + i sin 1 ) + 2 + i = 2 + e2 cos 1 + i 1 + e2 sin 1 .
(1.5)
z2 + 1
z → 1+i z 2 − 1
lim
Answer. Since z 2 + 1 and z 2 − 1 are continuous everywhere in the complex plane and z 2 − 1 6= 0
at z = 1 + i, so we have
limz → 1+i ( z 2 + 1 )
( 1 + i )2 + 1
3 + 4i
z2 + 1
2i + 1
lim 2
=
=
=
=
.
2
2
z → 1+i z − 1
limz → 1+i ( z − 1 )
2i − 1
5
(1 + i) − 1
z4 − 1
z → −i z + i
(1.6) lim
Answer. We observe
z4 − 1
(z 2 + 1)(z 2 − 1)
(z + i)(z − i)(z 2 − 1)
=
=
= (z − i)(z 2 − 1),
z+i
z+i
z+i
which is a polynomial. So we have
z4 − 1
= lim (z − i)(z 2 − 1) = ( −2i ) ( −i )2 − 1 = 4i.
z → −i z + i
z → −i
lim
Page 1 of 5
Complex Analysis I
HOMEWORK 4 – SOLUTION
Fall, 2009
Re(z)
.
Im(z)
(2.1) What value does the limit approach as z approaches 0 along the line y = x?
2. Consider the limit zlim
→0
Answer. As z = (x, y) moves along the line y = x, we have Re(z) = x = Im(z), which implies
Re(z)
x
= lim = lim 1 = 1.
z → 0 Im(z)
x→0 x
x→0
lim
(2.2) What value does the limit approach as z approaches 0 along the imaginary axis?
Answer. As z = (x, y) moves along the imaginary axis, i.e., the line x = 0, we have Re(z) = 0
and Im(z) = y, which implies
Re(z)
0
= lim = lim 0 = 0.
z → 0 Im(z)
y→0 y
y→0
lim
Re(z)
?
z → 0 Im(z)
(2.3) Based on your answers for (2.1) and (2.2), what can you say about lim
Answer. By the uniqueness of the limit, the answers for (2.1) and (2.2) imply that the limit
does not exist.
3. Consider the limit lim ( | z | + i Arg ( iz ) ).
z→i
(3.1) What value does the limit approach as z approaches i along the unit circle | z | = 1 in the first
quadrant?
Answer. For z = eiθ on the unit circle h z i = 1, we observe
π
π
iz = ei 2 eiθ = ei( 2 +θ ) .
π
So when z is in the first quadrant, i.e., 0 ≤ θ ≤ , we have
2
π
π
≤ + θ ≤ π,
2
2
i.e., iz is in the second quadrant and by the same argument, when z is in the second quadrant,
iz is in the third quadrant.
Now, as z approaches i along the unit circle | z | = 1 in the first quadrant, iz approaches −1 in
the second quadrant. So we have
lim ( | z | + i Arg ( iz ) ) = | i | + iπ = 1 + iπ.
z→i
(3.2) What value does the limit approach as z approaches i along the unit circle | z | = 1 in the second
quadrant?
Answer. As z approaches i along the unit circle | z | = 1 in the second quadrant, iz approaches
−1 in the third quadrant. So we have
lim ( | z | + i Arg ( iz ) ) = | i | + i ( −π ) = 1 − iπ.
z→i
(3.3) Based on your answers for (3.1) and (3.2), what can you say about lim ( | z | + i Arg ( iz ) )?
z→i
Answer. By the uniqueness of the limit, the answers for (3.1) and (3.2) imply imply that the
limit does not exist.
Page 2 of 5
Complex Analysis I
HOMEWORK 4 – SOLUTION
Fall, 2009
4. Compute the given limit.
z 2 + iz − 2
z → ∞ (1 + 2i)z 2
(4.1) lim
z 2 + iz − 2
, we compute f (1/z):
(1 + 2i)z 2
(1/z)2 + i/z − 2
1
1 + iz − 2z 2
=
,
f
=
z
(1 + 2i)/z 2
1 + 2i
1
1 + iz − 2z 2
1
1 − 2i
lim f
= lim
=
=
.
z→0
z→0
z
1 + 2i
1 + 2i
5
Answer. Letting f (z) =
Hence, by the Theorem, we deduce
z 2 + iz − 2
1 − 2i
.
=
z → ∞ (1 + 2i)z 2
5
lim f (z) = lim
z→∞
z2 − 1
z → i z2 + 1
(4.2) lim
Answer. Letting f (z) =
z2 − 1
, we have
z2 + 1
z2 + 1
1
= 2
,
f (z)
z −1
0
1
z2 + 1
= lim 2
=
= 0.
z → i f (z)
z→i z − 1
−2
lim
Hence, by the Theorem, we deduce
z2 − 1
= ∞.
z → i z2 + 1
lim f (z) = lim
z→i
z 2 − (2 + 3i)z + 1
z→∞
iz − 3
(4.3) lim
Answer. Letting f (z) =
z 2 − (2 + 3i)z + 1
, we have
iz − 3
1
(1/z)2 − (2 + 3i)/z + 1
1 − (2 + 3i)z + z 2
f
=
=
,
z
i/z − 3
iz − 3z 2
0
1
iz − 3z 2
lim
= lim
= = 0.
2
z → 0 f (1/z)
z → 0 1 − (2 + 3i)z + z
1
1
iz − 3z 2
=
,
f (1/z)
1 − (2 + 3i)z + z 2
Hence, by the Theorem, we deduce
z 2 − (2 + 3i)z + 1
= ∞.
z→∞
iz − 3
lim f (z) = lim
z→∞
Page 3 of 5
Complex Analysis I
HOMEWORK 4 – SOLUTION
Fall, 2009
5. Show that the function f is continuous at the given point.
1
(5.1) f (z) = z 3 − ; z0 = 3i
z
Proof. With z = x + iy, we observe
1
x + iy
x
y
3
2
2
3
= x − 3xy − 2
= u(x, y) + iv(x, y).
+ i 3x y − y + 2
x + y2
x + y2
f (x + iy) = ( x + iy )3 −
Since u(x, y) and v(x, y) are continuous at (x, y) = (0, 3) from Calculus/Real Analysis, hence, by
Theorem, we conclude f (z) is continuous at z0 = 3i. One may prove in a different way.

 z3 − 1
,
(5.2) f (z) =
 3,z − 1
|z| =
6 1
; z0 = 1
|z| = 1
Proof. We compute the limit
z3 − 1
(z − 1)(z 2 + z + 1)
= lim
= lim z 2 + z + 1 = 3 = f (1) = f (z0 ).
z→1 z − 1
z→1
z→1
z−1
lim f (z) = lim
z → z0
Hence, by the definition, f (z) is continuous at z0 = 1.
6. Show that the function f is discontinuous at the given point.
(6.1) f (z) = Arg ( iz ); z0 = i
Proof. (1) As z approaches z0 = i along the unit circle | z | = 1 in the first quadrant, iz approaches
−1 along the unit circle | z | = 1 in the second quadrant and so Arg ( iz ) moves to π.
(2) As z approaches z0 = i along the unit circle | z | = 1 in the second quadrant, iz approaches
−1 along the unit circle | z | = 1 in the third quadrant and so Arg ( iz ) moves to −π.
From (1) and (2), the limit is not unique and thus the function f is not continuous at z0 = i.

 z3 − 1
,
(6.2) f (z) =
 3,z − 1
|z| =
6 1
; z0 = i
|z| = 1
Proof. We compute the limit
z3 − 1
(z − 1)(z 2 + z + 1)
= lim
= lim z 2 + z + 1 = i 6= 3 = f (i) = f (z0 ).
z→i z − 1
z→i
z→i
z−1
lim f (z) = lim
z → z0
Hence, by the definition, f (z) is not continuous at z0 = i.
Page 4 of 5
Complex Analysis I
HOMEWORK 4 – SOLUTION
Fall, 2009
7. (7.1) Is it true that zlim
f (z) = lim f ( z̄ ) for any complex function f ? If so, give a brief justification.
→z
z→z
0
0
If not, find a counterexample.
z
Answer. No. We recall from the Example 2 on page 47 in the textbook that f (z) = is not
z̄
continuous at z = 0, because the limit does not exist there. With this function and z0 = 0, we
observe
z z̄
z̄
f (z) =
= = = f ( z̄ ) .
z̄
z̄¯ z
z̄
However, since the limit lim does not exist (by the similar argument as in the Example 2 on
z→0 z
page 47), so we cannot say lim f (z) = lim f ( z̄ ).
z→0
z→0
(7.2) If f (z) is a continuous function at z0 , then is it true that f (z) is continuous at z0 ?
Answer. Yes. We recall the theorem saying that a composition of two continuous functions is also
continuous. Since g(z) = z̄ is continuous everywhere, for a function f which is continuous at z0 , the
composition ( g ◦ f ) (z) = g(f (z)) = f (z) should be continuous at z0 .
8. If f satisfies xlim
f (x + i0) = 0 and lim f (0 + iy) = 0, then can you conclude that lim f (z) = 0? Ex→0
y→0
z→0
plain.
Answer. No. Consider the function f (z) =
2
x + iy
− 1,
f (x + iy) =
x − iy
x 2
f (x + i0) =
− 1 = 0,
x
z 2
z̄
− 1. We observe
f (0 + iy) =
iy
−iy
2
− 1 = ( −1 )2 − 1 = 0,
which implies
lim f (x + i0) = lim 0 = 0,
x→0
x→0
lim f (0 + iy) = lim 0 = 0.
y→0
y→0
However, as z = (x, y) approaches 0 along the line y = x, we observe
#
"
2
x + ix
−1
lim f (z) =
lim f (x + ix) =
lim
z→0
(x,x) → (0,0)
(x,x) → (0,0)
x − ix
"
# 2
2
1+i
1+i
=
lim
−1 =
− 1 = −2 6= 0.
(x,x) → (0,0)
1−i
1−i
Page 5 of 5
United Arab Emirates University
College of Sciences
Department of Mathematical Sciences
HOMEWORK 5 { SOLUTION
Section 19 Derivatives
Section 20 Dierentiation Formulas
Complex Analysis I
MATH 315 SECTION 01 CRN 23516
9:30 { 10:45 on Monday & Wednesday
Due Date: Monday, November 2, 2009
ID No:
Solution
Name:
Solution
Score:
Solution
Complex Analysis I
HOMEWORK 5 { SOLUTION
Fall, 2009
1. (#1 on Page 62) Use results in Sec. 20 to nd f (z ) when
0
(1.1) (b) f (z ) = 1
4z 2
3
Answer.
f
(1.2) (c) f (z ) =
Answer.
1
,
2z + 1
z
z
0
6=
(z ) = 3 1
4z 2
2 1
4z 2 = 24z 1
0
4z 2
2
:
1=2
f
0
(z ) =
2z + 1 (z 1) 2
3
=
:
2
(2z + 1)
(2z + 1)2
Page 1 of 3
Complex Analysis I
HOMEWORK 5 { SOLUTION
Fall, 2009
2. (#8 on Page 63) Use the method in Example 2, Sec. 19, to show that f (z ) does not exist at any
point z when
0
(2.1) (a) f (z ) = Re(z )
Proof.
4 = Re ( + 4 ) Re ( ) = + 4
4
=
4
4
4 +4
4 +4
where 4 = 4 + 4 and = + .
As 4 approaches (0 0) horizontally through the points (4 0) on the real axis, we have
4 = 4
4 =1
=
4 4 +4 4 +0
As 4 approaches (0 0) horizontally through the points (0 4 ) on the imaginary axis, we have
4 = 4
0
=
4 4 + 4 0+ 4 =0
w
z
z
z
z
x
i
z
z
x
z
y
z
x
x
x
x
i
y
x
i
y
;
iy
;
x;
w
x
z
z
x
x
x
i
y
x
:
i
;
;
w
y
x
z
x
i
y
i
:
y
Since the limit is not unique, thus, f (z ) does not exist at any point.
0
(2.2) (b) f (z ) = Im(z )
Proof.
4
4 = Im ( + 4 ) Im ( ) = + 4
=
4
4
4 +4 4 +4
where 4 = 4 + 4 and = + .
As 4 approaches (0 0) horizontally through the points (4 0) on the real axis, we have
0
4 = 4
=
4 4 + 4 4 + 0 =0
As 4 approaches (0 0) horizontally through the points (0 4 ) on the imaginary axis, we have
4 = 4
4 =1
=
4 4 + 4 0+ 4
z
w
z
z
z
z
x
i
z
z
y
z
x
y
x
y
i
y
y
x
i
y
;
iy
;
x;
y
w
z
z
y
x
i
y
x
;
;
w
y
z
x
i
:
i
y
y
y
i
y
i
:
Since the limit is not unique, thus, f (z ) does not exist at any point.
0
Page 2 of 3
Complex Analysis I
HOMEWORK 5 { SOLUTION
Fall, 2009
3. (#9 on Page 63) Let f denote the function whose values are
8 2
>
< z
( )=>
f z
(1) Show that if z = 0, then
4z , or 4x4y, plane.
Proof.
Case 1.
z
4
4
4
4
w
z
z
: 0
when z 6= 0,
when z = 0.
= 1 at each nonzero point on the real and imaginary axes in the
= 0:
w
z
=
( + 4z )
f z
4
( )
f z
z
=
z
+ 4 z = (z + 4 z )
2
4
z
0
+ 4z
=
4z ( z + 4z )
z
2
where 4z = 4x + i4y and z = x + iy .
(2) Show that
4
4
w
z
= 1 at each nonzero point (4x; 4x) on the line 4y = 4x in that plane.
(3) Conclude form these observations that f (0) does not exist. Note that to obtain this result, it
is not sucient to consider only horizontal and vertical approaches to the origin in the 4z plane.
(Compare with Example 2, Sec. 19.)
0
Page 3 of 3