Appendix A: Complex Numbers
1.
2.
3.
4. Write z = a + bi and 
.
.
a.im (iz) = im (− b + ai) = a = re z.
c.
.
.
e.
5. a. unit circle (c) line y = x (e) 0 and the positive real axis
7. 
9. a. If z = a + bi and 
, then 
, so 
, then , so
Take positive square roots.
10.
11.
12.
13. a.
= (cos 2θ − sin 2θ) + i(2 cos θ sin θ)
= (cos 2θ − sin 2θ) + i(2 cos θ sin θ)
14. a.
c.
15. a.z = r(cos θ + sin θ) so 
. But 
and 
, so re
. If z−1 = seiϕ, then 1 = zz−1 = rsei(θ+ϕ). Thus rs = 1, so 
, and θ + ϕ = 0 so (one choice for ϕ is) ϕ = − θ. Hence 
.
. But and , so re. If z−1 = seiϕ, then 1 = zz−1 = rsei(θ+ϕ). Thus rs = 1, so , and θ + ϕ = 0 so (one choice for ϕ is) ϕ = − θ. Hence .
16. a. Let 
so that the kth root of unity is 
by DeMoivre's theorem. Now 
by the Hint so,
so that the kth root of unity is by DeMoivre's theorem. Now by the Hint so,
because 
Since 
this implies that 
as required.
Since this implies that as required.
17. a. Have zi = eiθi for angles θi. The angles between them all equal 
(because they are equally spaced). Let θ1 = α as in the diagram. Then
(because they are equally spaced). Let θ1 = α as in the diagram. Then
If we write z = eiβ, then z5 = 1. Now use the hint:
19. Let 
, 
. If 
is a root of f(x), then f(z) = 0. Hence, 
because 
for each i (being real).
, . If is a root of f(x), then f(z) = 0. Hence, because for each i (being real).
21. Let z = reiθ. (a) If t > 0, tz = treiθ has the same angle θ as z. Hence tz is on the line through 0 and z, on the same side of 0 as z. (b) If t = − s, s > 0, then tz = srei(θ+π), and so is on the line through 0 and z, on the other side of 0 from z.
23.
a. If 
, then 
. If b1 ≠ b, then 
is rational, an impossibility. Hence, b1 = b whence a = a1.
, then . If b1 ≠ b, then is rational, an impossibility. Hence, b1 = b whence a = a1.
c.
, so:
, so:
e. Using (c) and (d) above: 
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.