Math 412 Group Work #8 Spring 2009
SOLUTIONS
Problem 1. Suppose that {zn}converges. Prove that {|zn|} converges. Is
the converse true?
SOLUTION: Suppose {zn} → ζfor some ζ∈C. Let > 0 be given. Then
there exists n∈Nsuch that whenever n > N, we have |zn−ζ|< . But then
|zn|−|ζ| ≤ |zn−ζ|< and |ζ|−|zn| ≤ |ζ−zn|=|zn−ζ|< . Thus,
||zn|−|ζ|| <
for all n∈N. Hence, {|zn|} → |ζ|.
The converse is not true, as the sequence zn= (−1)nshows.
Problem 2. Evaluate each limit when they exist:
(a): lim
n→∞
n3(1 + 3i)
2n3+ 4n+ 1.
SOLUTION: We have
lim
n→∞
n3(1 + 3i)
2n3+ 4n+ 1 = lim
n→∞
1+3i
2 + 4
n2+1
n3
=1+3i
2.
(b): lim
n→∞ (−1)n+(n+ 1)2i
n2+ 4 .
SOLUTION: Letting zn= (−1)n+(n+1)2i
n2+4 , we see that Re(zn) = xn= (−1)n, which
does not converge. Thus, by Theorem 4.1, {zn}does not converge.
(c): lim
n→∞ h√n+ 1 −√n+ 2ii.
SOLUTION: Note that
lim
n→∞ h√n+ 1 −√ni= lim
n→∞ h√n+ 1 −√ni√n+1+√n
√n+1+√n= lim
n→∞
1
√n+1+√n= 0.
Thus, the real part of the above sequence converges to 0, while the imaginary part is
constant 2i. Thus, by Theorem 4.1, the limit is 2i.
Problem 3. Suppose that {zn}converges. Show that {zn}is bounded in
two ways:
