10 1. The Complex Numbers
Proof. By Theorem 1.1.8 we know that
| Re(zn) Re(w)| |zn w|,
| Im(zn) Im(w)| |zn w| and
|zn w| | Re(zn) Re(w)| + | Im(zn) Im(w)|
for each n. The first two of these inequalities, together with Remark 1.2.2, im-
ply that if zn w, then Re(zn) Re(w) and Im(zn) Im(w). The third
inequality, together with the fact that the sum of two sequences converging to
zero also converges to zero, implies the converse that is, if Re(zn) Re(w) and
Im(zn) Im(w), then zn w.
Example 1.2.5. Show that the sequence
{2−n
+ in/(n + 1)} converges to i.
Solution: This follows from the previous theorem and the fact that
2−n
0
and n/(n + 1) 1.
Example 1.2.6. Show that if {zn} and {wn} are two convergent sequences of
complex numbers with zn z and wn w, then zn +wn z+w and znwn zw.
Solution: If zn = xn + iyn, z = x + iy, wn = un + ivn and w = u + iv, then
zn + wn = xn + un + i(yn + vn),
z + w = x + u + i(y + v),
znwn = xnun ynvn + (xnvn + ynun)i
and
zw = xu yv + (xv + yu)i.
We know that xn x, yn y, un u and vn v. We also know the rules
about limits of products, sums and differences of sequences of real numbers. These
rules imply xn + un x + u, yn + vn y + v, xnun ynvn xu yv and
xnvn + ynun xv + yu. Since the real and imaginary parts of zn + wn and znwn
converge to the real and imaginary parts, respectively, of z+w and zw, the previous
theorem implies that zn + wn z + w and znwn zw.
Series of Complex Numbers. A series of complex numbers is a formal sum of
the form

k=0
zk = z0 + z1 + z2 + · · · + zk + · · · ,
with zk C. The series converges if its sequence of partial sums {sn} converges,
where
(1.2.1) sn =
n
k=0
zk.
In this case, if s = limn→∞ sn, then we say s is the sum of the series and write
s =

k=0
zk.
Just as with real series, if a series
∑∞
k=0
zk converges, then its terms must tend
to zero. Thus, if {zk} fails to have limit 0, then the series diverges. This test for
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