ANSWERS 59300 00- I:; (a+ ib) (xn + iyn) = I:: [(axn - by,.)+ i(bxn + ayn)]. By Theorem 4.4,
n = l n=l
this expression equals (au - bv) + i (bu+ av) = (a+ib)(u+iv). Explain why
in detail.
Duplicate the part of the theorem that shows lim Xn = u, but replace Xn
tl~OO
with Yn and u with v.
Following the hint , for e > 0 there exist numbers N. and M, such that
n > N. implies lzn -(d < ~' and n > M. implies lz.. -(21 < ~· Let
L. =Max {N,, M,}. Then n > L, implies 1(1 -(21=1(1 - Zn+ z,. - (2 1 ~
1 (1 - z..I + lz.. -(2 1 < e.
Let e > 0 and suppose Jim Zn= 0. This means there exists N, such that
n -oo
n > N. implies Zn E D, (0), that is, lzn -OI < e. But t hen llz,.1- OI =
!Zn -OI < e, so a lso we have lz..I E D. (0). Therefore, Jim lznl = O. The
n-oo
other direction is similar. Show the details.
Section 4.2. Julia and Mandelbrot Se t s: page 139
la. If z = r (cos B + isinB) f 0 , show N (z) = 4 (r - :) cosB + i4 (r + :) sinB.
The result follows from this-explain!le. If zo f 0 is real, then obviously z 1 = N (z 1 ) = 4 ( zo -z' 0 ) is real. Assume
z,. is real for some n > 1. T hen Zn+! = N (zn) = 4 (Zn -.: ) is also real,
provided Zn f O.3. For f ( z) = az + b, if our initial guess is zo, then z 1 = zo -az~+b = -~. But
this is the solution to the equation f ( z) = 0, so our iteration stops either
here or with zo if by chance we had set z 0 = - ~.- The Julia set for f-2 (z) = z^2 - 2 is connected by T heorem 4.9 because the
orbit of 0 under f - 2 are {- 2, 2, 2, 2, ... }, which is a bounded set.
7. Suppose c E M, and let {z.k} be the orbits of 0 under fc · By definition of
M, there is some real number N such that lzkl < N for all k. Let {w1<} be
t he orbit of 0 under Jc.. Show by induction that Wk = zk for all k. Once you
have that, it is straightforward to conclude that the set {wk} is bounded.
9. There are many examples. The number -2 is in the Mandelbrot set, but
its negative, 2, is not. Whether you use this example or not, justify your
assertion!