Recall from the chapter on prerequisite material the basic properties of complex numbers,
complex absolute value and so forth. See Definition 1.12.3 and the following material
after this definition. If you have a sequence of complex numbers
{zk}
where
z_{k} = x_{k} + iy_{k} then to say that
|zk|
is bounded is to say that
∘ -------
x2k + yk2
is bounded. In
other words, the ordered pairs
(xk,yk)
are in a bounded subset of ℝ^{2}. Also to say that
lim_{k→∞}
|zk − z|
= 0 is the definition of what you mean by lim_{k→∞}z_{k} = z and it is the
same as saying that lim_{k→∞}
(xk,yk)
=
(x,y)
where z = x + iy. Thus, if you have a
bounded sequence of complex numbers
{zk}
, you must have
{xk}
and
{yk}
both be a
bounded sequence in ℝ and so
{xk}
is contained in some interval
[a,b]
and
{yk}
is
contained in some interval
[c,d]
. Thus such a bounded sequence must have a
subsequence, still denoted as z_{k} such that x_{k}→ x ∈
[a,b]
and y_{k}→ y ∈
[c,d]
. This yields
the following simple observation sometimes called the Weierstrass Bolzano theorem.
Theorem 15.13.1Let
{zk}
be a sequence of complex numbers such that
|zk|
is a bounded sequence of real numbers. Then there exists a subsequence
{znk}
and a complex number z such that lim_{k→∞}z_{nk} = z. Sets of the formK ≡
{z ∈ ℂ : |z| ≤ r}
are sequentially compact.
Proof: It only remains to verify the last assertion. Letting
{zk}
⊆ K, the above
discussion shows that there exists z and a subsequence
{znk}
such that z_{nk}→ z. It only
remains to verify that z ∈ K. However, this is clear from the triangle inequality.
Indeed,
|z| ≤ |z − zk|+ |zk| ≤ |z − zk|+ r
Hence,
|z| ≤ lk→im∞ |z − znk|+ r = r.■
Lemma 15.13.2Every polynomial p
(z)
having complex coefficients iscontinuous. That is, if z_{k}→ z, then p
(zk)
→ p
(z)
.
Proof:Recall that if z_{k} = x_{k} + iy_{k},z = x + iy, the convergence of z_{k} to z is
equivalent to convergence of x_{k} to x and convergence of y_{k} to y. Also,
n n ∑n ( n ) k n−k k
z = (x+ iy) = k (i) x y
j=1
Thus breaking into real and imaginary parts,
p(z) = Re p(z)+ iIm p(z)
and each of Rep
(z)
and Imp
(z)
are polynomials in x and y as defined in Definition
15.7.2. Therefore, these are each continuous functions of