Solutions for Math 311 Assignment #1
(1) Show that
(a) Re(iz) = − Im(z);
(b) Im(iz) = Re(z).
Proof. Let z = x + yi with x = Re(z) and y = Im(z). Then
Re(iz) = Re(−y + xi) = −y = − Im(z) and Im(iz) = Im(−y +
xi) = x = Re(z).
(2) Verify the associative law for multiplication of complex numbers. That is, show that
(z1 z2 )z3 = z1 (z2 z3 )
for all z1 , z2 , z3 ∈ C.
Proof. Let zk = xk + iyk for k = 1, 2, 3. Then
(z1 z2 )z3 = ((x1 + y1 i)(x2 + y2 i))(x3 + y3 i)
= ((x1 x2 − y1 y2 ) + i(x2 y1 + x1 y2 ))(x3 + y3 i)
= (x1 x2 x3 − x3 y1 y2 − x2 y1 y3 − x1 y2 y3 )
+ i(x2 x3 y1 + x1 x3 y2 + x1 x2 y3 − y1 y2 y3 )
and
z1 (z2 z3 ) = (x1 + y1 i)((x2 + y2 i))(x3 + y3 i))
= (x1 + y1 i)((x2 x3 − y2 y3 ) + i(x2 y3 + x3 y2 ))
= (x1 x2 x3 − x3 y1 y2 − x2 y1 y3 − x1 y2 y3 )
+ i(x2 x3 y1 + x1 x3 y2 + x1 x2 y3 − y1 y2 y3 )
Therefore,
(z1 z2 )z3 = z1 (z2 z3 )
(3) Compute
1+i
(a)
;
1−i
(b) (1 − i)4 .
Answer. (a) i (b) −4
(4) In each case, sketch the set of points on the complex plane
determined by:
(a) |z − i| = 2;
1
2
(b) 2 < |z + 3 + i| < 3;
(c) |z − 1| = |z − 2i|;
(d) z 2 + z 2 = 2;
(e) | Im(z + i)| > 3.
Answer. (a) The circle x2 + (y − 1)2 = 4.
(b) The annulus 4 ≤ (x + 3)2 + (y + 1)2 < 9.
(c) The line 2x + 4y + 3 = 0.
(d) The hyperbola x2 − y 2 = 1.
(e) The union of two half planes {y > 4} and {y < −2}.
(5) Show that
Im(z1 − z2 )
|z1 | + |z2 |
≤
|z3 + z4 |
||z3 | − |z4 ||
for all complex numbers z1 , z2 , z3 , z4 satisfying |z3 | 6= |z4 |. Also
give the sufficient and necessary condition under which the
equality holds.
Proof. Since
| Im(z1 − z2 )| ≤ |z1 − z2 | ≤ |z1 + z2 |
and
|z3 + z4 | ≥ |z3 | − |z4 |
we have
Im(z1 − z2 )
| Im(z1 − z2 )|
|z1 | + |z2 |
|z1 | + |z2 |
≤
≤
≤
.
|z3 + z4 |
|z3 + z4 |
||z3 | − |z4 ||
||z3 | − |z4 ||
If the equality holds, we must have Re(z1 −z2 ) = 0, Im(z1 −z2 ) ≥
0 and |z1 − z2 | = |z1 | + |z2 |. It follows that x1 = x2 = 0, y1 ≥ y2
and y1 y2 ≤ 0, where we let zk = xk + iyk for k = 1, 2, 3, 4.
If z1 6= 0 or z2 6= 0, we also have |z3 + z4 | = |z3 | − |z4 |, i.e.,
z3 = λz4 or z4 = λz3 for some λ < 0.
In conclusion, the inequality holds if x1 = x2 = 0, y1 ≥ y2 ,
y1 y2 ≤ 0 and
• z1 = z2 = 0 or
• z3 = λz4 or
• z4 = λz3 for some λ < 0.
(6) Use
(a)
(b)
(c)
the properties of conjugates and moduli to show that
z + 3i = z − 3i;
iz = −iz; √
√
|(2z + 5)( 2 − i)| = 3|2z + 5|.
3
Proof. (a)
z + 3i = z + 3i = z − 3i.
(b)
iz = iz = −iz
(c)
√
√
|(2z + 5)( 2 − i)| = |2z + 5|| 2 − i|
√
√
= 3|2z + 5| = 3|2z + 5|
Solutions for Math 311 Assignment #2
(1) Find the principal argument Arg(z) when
i
(a) z =
;
1√+ i
(b) z = ( 3 + i)2010 .
Solution. (a)
i
1 i
Arg
+
= Arg
= Arg
1+i
2 2
!
√
2 πi/4
e
= π/4
2
(b)

√
Arg(( 3 + i)2010 ) = Arg 22010
√
3 i
+
2
2
!2010 

= Arg((eπi/6 )2010 ) = Arg(e335πi ) = Arg(eπi ) = π
(2) Show that if Re(z1 ) > 0 and Re(z2 ) > 0, then
Arg(z1 z2 ) = Arg(z1 ) + Arg(z2 ).
Proof. We always have
Arg(z1 z2 ) = Arg(z1 ) + Arg(z2 ) + 2kπ
for some integer k. Since Re(z1 ) > 0 and Re(z2 ) > 0,
−π/2 < Arg(z1 ) < π/2 and − π/2 < Arg(z2 ) < π/2
and hence
−π < Arg(z1 ) + Arg(z2 ) < π.
And since
−π < Arg(z1 z2 ) ≤ π,
4
−2π < Arg(z1 z2 ) − (Arg(z1 ) + Arg(z2 )) < 2π
Therefore, −2π < 2kπ < 2π and −1 < k < 1. Hence k = 0. (3) Use the identity
n
X
k=0
to derive
n
X
sin(kθ) =
k=1
zk =
1 − z n+1
1−z
cos(θ/2) − cos((2n + 1)θ/2)
2 sin(θ/2)
Proof. Let z = eiθ . Then
n
n
n
X
X
X
1 − ei(n+1)θ
ikθ
cos(kθ) + i
sin(kθ) =
e =
1 − eiθ
k=0
k=1
k=0
1 − ei(n+1)θ
1 − ei(n+1)θ
=
(1 − cos θ) − i sin θ
2 sin(θ/2)(sin(θ/2) − i cos(θ/2))
1 − ei(n+1)θ
i(e−θi/2 − e(2n+1)iθ/2 )
=
=
2 sin(θ/2)(−i)eθi/2
2 sin(θ/2)
(sin((2n + 1)θ/2) + sin(θ/2)) + i(cos(θ/2) − cos((2n + 1)θ/2)
=
2 sin(θ/2)
=
Therefore,
n
X
k=1
sin(kθ) =
cos(θ/2) − cos((2n + 1)θ/2
.
2 sin(θ/2)
(4) Find all the square roots of (a) 1 + i and (b) 1 +
them in rectangular coordinates.
√
3i. Express
Solution. (a)
q
√
√
√
√
4
4
1+i=
2eπi/4 = 2ekπi+πi/8 = ± 2eπi/8
p
√
By
half-angle
formula,
cos(π/8)
=
4
+
2/2 and sin(π/8) =
p
√
4 − 2/2. Therefore,
√
q
q
4
√
√
√
2
1+i=±
4+ 2+i 4− 2
2
5
(b)
q
√
√ √
√
√
1 + 3i = 2 eπi/3 = 2ekπi+πi/6 = ± 2eπi/6 = ±
√
√ !
6
2
+i
2
2
(5) Find the four zeros of the polynomial z 4 + 4 and use these to
factor z 4 + 4 into quadratic factors with real coefficients.
Solution. The 4th roots of −4 are
√ √
√
√
kπi πi
4
4
πi
−4 = 2 e = 2 exp
+
2
4
for k = 0, 1, 2, 3. So
√
√
√
√
z 4 + 4 = (z − 2eπi/4 )(z − 2e3πi/4 )(z − 2e5πi/4 )(z − 2e7πi/4 )
√
√
√
√
= (z − 2eπi/4 )(z − 2e−πi/4 )(z − 2e3πi/4 )(z − 2e−3πi/4 )
√
√
= (z 2 − 2(eπi/4 + e−πi/4 )z + 2)(z 2 − 2(e3πi/4 + e−3πi/4 )z + 2)
= (z 2 − 2z + 2)(z 2 + 2z + 2)
(6) Sketch the following sets and determine which are domains,
which are closed and which are bounded:
(a) |z − 2 + i| ≤ 1;
(b) |2z + 3| > 4;
(c) Im(z) > 1;
(d) |z − 4| ≥ |z|;
(e) Im(z) = 1;
(f) 0 ≤ arg(z) ≤ π/4(z 6= 0).
Justify your answer.
Solution. (a) It is not a domain since it is not open; choose a
point z0 satisfying |z0 −2+i| = 1 and the disk |z −z0 | < r is not
contained in the set for all r > 0. It is closed since {|z − 2 + i| >
1} is open and it is bounded since |z| ≤ |2 − i| + 1 < 4 for all z
satisfying |z − 2 + i| ≤ 1.
(b) It is a domain since it is open and connected. It is not
closed since {|2z + 3| ≤ 4} is not open and it is not bounded
since one can find a sequence zn satisfying |2zn + 3| > 4 such
that limn→∞ |zn | = ∞. For example, let zn = n.
(c) It is a domain since it is open and connected. It is not
closed since {Im(z) ≤ 1} is not open and it is not bounded
since one can find a sequence zn satisfying Im(z) > 1 such that
limn→∞ |zn | = ∞. For example, let zn = 2ni.
6
(d) It is not a domain since it is not open; choose a point z0
satisfying |z − 4| = |z| and the disk |z − z0 | < r is not contained
in the set for all r > 0. It is closed since {|z − 4| < |z|} is open.
It is not bounded since one can find a sequence zn satisfying
|z − 4| ≥ |z| such that limn→∞ |zn | = ∞. For example, let
zn = ni.
(e) It is not a domain since it is not open; choose a point z0
satisfying Im(z) = 1 and the disk |z − z0 | < r is not contained
in the set for all r > 0. It is closed since {Im(z) 6= 1} is open.
It is not bounded since one can find a sequence zn satisfying
Im(z) = 1 such that limn→∞ |zn | = ∞. For example, let zn =
n + i.
(f) It is not a domain since it is not open; choose a point z0
satisfying arg(z) = 0 and the disk |z − z0 | < r is not contained
in the set for all r > 0. It is not closed since 0 does not lie in
the set and the disk |z| < r meets the set for all r > 0. It is not
bounded since one can find a sequence zn satisfying arg(z) = 0
such that limn→∞ |zn | = ∞. For example, let zn = n.
(7) Write the following functions f (z) in the forms f (z) = u(x, y)+
iv(x, y) under Cartesian coordinates and f (z) = u(r, θ)+iv(r, θ)
under polar coordinates:
(a) f (z) = z 2 + z + 1;
1+z
(b) f (z) =
.
1−z
Solution. (a)
f (z) = (x + yi)2 + (x + yi) + 1 = (x2 − y 2 + x + 1) + i(2xy + y)
= r2 e2iθ + reiθ + 1
= (r2 cos(2θ) + r cos θ + 1) + i(r2 sin(2θ) + r sin θ)
(b)
(1 + z)(1 − z)
(1 − z)(1 − z)
1 + 2iy − (x2 + y 2 )
=
(1 − x)2 + y 2
1 − (x2 + y 2 )
2y
=
+
i
(1 − x)2 + y 2
(1 − x)2 + y 2
1 − r2
2r sin θ
=
+i
2
1 + r − 2r cos θ
1 + r2 − 2r cos θ
f (z) =
7
(8) Sketch the region onto which the sector {r ≤ 1, 0 ≤ θ ≤ π/4}
is mapped by the transformation (a) w = z 3 and (b) w = z 4 .
Solution. (a) The image is {|w| ≤ 1, 0 ≤ arg(w) ≤ 3π/4}.
(b) The image is {|w| ≤ 1, 0 ≤ arg(w) ≤ π}.