MATH 300, ANSWER KEY TO PROBLEM SET 1
1.1.10.
2 2 2
2+i
2+i
−1 − 8i
6 − 17i
36 − 172 − 2 · 6 · 17i
=
=
·
=
=
6i − (1 − 2i)
−1 + 8i −1 − 8i
65
652
=
−253
204
−
i
4225
4225
1.1.20
4
(a) iz = 4 − zi ⇒ 2iz = 4 ⇒ z = 2i
= −2i
(b) z = (1 − 5i) (1 − z) = 1 − 5i − (1 − 5i) z ⇒ (1 + (1 − 5i)) z = 1 − 5i ⇒ z =
= 27−5i
29
(c) z1 = 0, if z 6= 0 ⇒ (2 − i) + 8z = 0 ⇒ z2 = 2−i
8
√
√
(d) z 2 = −16 ⇒ z = ± −16 = ±i 16 = ±4i
1.1.30. Suppose the complex number system has such a subset P. If i belongs to
P then by (iii) i2 = −1 belongs to P and again by (iii) (−1) · i belongs to P This
contradicts condition (i), since not both i and −i belong to P.
So i cannot belong to P. The condition (i) tells us that −i must belong to P.
But then, arguing as above, we see that (−i)(−i)(−i) = i belongs to P which is
what we have just excluded. This contradiction shows that P cannot exist.
1.1.32. First compute
u = (a + b)(c + d) = ac + ad + bc + bd,
v = ac, and w = bd .
This requires only three multiplications. Now the real part ac − bd and the imaginary part bc + ad of (a + ib)(c + id) can be computed using only addition and
subtraction: ac − bd = v − w and bc + ad = u − v − w.
1.2.8. |z − 1|2 = (z − 1)(z − 1) = (z − 1)(z − 1) = |z − 1|2 ⇒ |z − 1| = |z − 1|.
1.2.14. |z1 z2 |2 = (z1 z2 )(z1 z2 ) = z1 z2 z1 z2 = |z1 |2 |z2 |2 ⇒ |z1 z2 | = |z1 ||z2 |.
2
2
2
1.2.16.
h
iLet zh= x + iy, 1 = |z|i = x + y , z 6= 1.
1
1
1−x
1−x
Re 1−z
=Re 1−x−iy
· 1−x+iy
1−x+iy = (1−x)2 +y 2 = 1−2x+1 =
1−x
2(1−x)
=
1
2
1.3.4. First consider the case where k is non-negative. Start with 1 = |z 0 | =
|z|0 (for k = 0) and use the formula |z1 z2 | = |z1 ||z2 | (which we proved in class)
recursively (i.e., by induction on k): if we know that |z k−1 | = |z|k−1 then
|z k | = |z · z k−1 | = |z||z k−1 | = |z||z|k−1 | = |z|k .
For negative k use the identity | zz21 | =
|z1 |
|z2 |
proved in class, with z1 = 1 and
z2 = z |k| .
y
x
1.3.12. z = x + iy = r(cos φ + i sin φ), cos φ = |z|
, sin φ = |z|
, φ = arg(z).
√
1
5π
√
(a) r = 6 2, cos φ = sin φ = − 2 ⇒ φ = 4 + 2πk, k ∈ Z
(b) r = π, cos φ = −1, sin φ = 0 ⇒ φ = π + 2πk, k ∈ Z
(c) r = 10, cos φ = 0, sin φ = 1 ⇒ φ = π2 + 2πk, k ∈ Z
√
(d) r = 2, cos φ = 23 , sin φ = − 21 ⇒ φ = − π6 + 2πk, k ∈ Z.
The principal values of the argument (taking values in the interval (−π, π]) are
given as follows:
(a) Arg(−6 − 6i) = − 3π
4 ,
(b) Arg(−π) = π,
(c) Arg(10i) = π2 ,
√
(d) Arg( 3 − i) = − π6 .