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