1.1 - 1.4 Homework Solutions
(2.2b) Prove that Im(iz) = Re z.
Write z = x + iy where x, y are real. Then
iz =
=
=
Im(iz) =
=
i(x + iy)
ix − y
−y + ix
x
Re z
(2.4) Solve the equation z 2 − 2z + 2 = 0.
Write z = x + iy where x, y are real. Then
z 2 − 2z + 2 = (x + iy)(x + iy) − 2(x + iy) + 2
= x2 − y 2 + i2xy − 2x − i2y + 2
= (x2 − y 2 − 2x + 2) + i(2xy − 2y)
If z 2 − 2z + 2 = 0 then both x2 − y 2 − 2x + 2 = 0 and 2xy − 2y = 0. We begin with the
equation from the imaginary part.
2xy − 2y
2y(x − 1)
2y = 0
y=0
=
=
or
or
0
0
x−1=0
x=1
If y = 0, then x2 − 2x + 2 = 0 which has no real solutions (the discriminant b2 − 4ac =
(−2)2 − 4(1)(2) = −4). Since x is assumed to be real, no solutions result from y = 0.
If x = 1, then
12 − y 2 − 2(1) + 2 = 0
1 = y2
±1 = y
There are two solutions, z = 1 + i and z = 1 − i.
(3.2) Prove that
(z −1 )−1 = z
when (z 6= 0).
We have learned that for z = x + iy
x
−y
1
= 2
+i 2
2
z
x +y
x + y2
1
Using this same formula again,
³
´
³
´
−y
x
−
x2 +y 2
x2 +y 2
1
= ³
´2 ³
´2 + i ³
´2 ³
´2
1/z
−y
−y
x
x
+
+
x2 +y 2
x2 +y 2
x2 +y 2
x2 +y 2
³
´
³
´
= ³
³
= ³
x
x2 +y 2
x2 +y 2
(x2 +y 2 )2
x
x2 +y 2
1
x2 +y 2
´
´ + i³
³
´ + i³
y
x2 +y 2
x2 +y 2
(x2 +y 2 )2
y
x2 +y 2
1
x2 +y 2
´
´
´
= x + iy
= z
(3.4) Prove that if z1 z2 z3 = 0, then at least one of the three factors is zero.
First note that we can rewrite z1 z2 z3 = 0 as (z1 z2 )z3 = 0.
We have already shown that if z4 z5 = 0, then either z4 = 0 or z5 = 0. Using this fact
with z4 = z1 z2 and z5 = z3 , then either z1 z2 = 0 or z3 = 0. But if z1 z2 = 0, then either
z1 = 0 or z2 = 0.
So either z1 = 0, z2 = 0 or z3 = 0.
(4.A) Prove that |z1 + z2 | ≥ |z1 | − |z2 | for every pair of complex numbers z1 and z2 .
The trick to proving this inequality is to notice that if you move the |z2 | to the other
side, the inequality can be written as
|z1 | ≤ |z1 + z2 | + |z2 |
This is very similar to the Triangle Inequality. In fact, it follows directly from the triangle
inequality and the fact that |z1 | = |(z1 − z2 ) + z2 |. Apply the Triangle Inequality to that
expression and you get
|z1 | = |(z1 − z2 ) + z2 | ≤ |z1 + z2 | + |z2 |
which (once you move the |z2 | back over) is what we needed to prove.
√
(4.4) Verify that 2|z| ≥ |Re z| + |Im z|.
Start by putting the left side in terms of x and y.
√ p
√
2|z| =
2 x2 + y 2
p
=
2x2 + 2y 2
The book has a hint, suggesting we use the fact that (|x| − |y|)2 ≥ 0 because (|x| − |y|) is
real so the square makes it positive. Remember that |Re z| = |x| and |Im z| = |y|. We will
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rearrange the hint inequality to get the inequality we need to prove.
(|x| − |y|)2
|x|2 − 2|x||y| + |y|2
x2 − 2|x||y| + y 2
x2 + y 2
2x2 + 2y 2
2x2 + 2y 2
p
2x2 + 2y 2
√
2|z|
≥
≥
≥
≥
≥
≥
0
0
0 (|x|2 = x2 , |y|2 = y 2 since x, y real)
2|x||y|
x2 + 2|x||y| + y 2
(|x| + |y|)2
≥ |x| + |y|
≥ |Re z| + |Im z|
(4.5) Sketch the set of points determined by the given conditions.
(a) |z − 1 + i| = 1 can be rewritten as |z − (1 − i)| = 1, so this equation represents the
circle with center 1 − i and radius 1.
(b) |z + i| ≤ 3 is the same as |z − (−i)| ≤ 3, which represents the points which are a
distance 3 or less from −i. In other words, the circle centered at −i with radius 3, along
with its inside.
(c) |z − 4i| ≥ 4 represents the points which are a distance 4 or more from 4i. In other
words, the circle centered at 4i with radius 4, along with everything outside that circle.
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