44 CHAPTER 1 • COMPLEX NUMBERS- EXAMPLE 1.26 Show that the right half-plane H = {z: Re (z) > O} is a
domain.
Solution First we show that H is connected. Let zo and z 1 be any two
points in H. We claim the obvious, that the straight-line segment C given by
E<J.uation (1-48) lies entirely within H. To prove this claim, we let z (t*) =
z 0 + (z 1 - zo) t., for some t* E [O, 1], be an arbitrary point on C. We must show
that Re(z(t•)) > 0. Now,
Re(z(t*)) = Re(zo + (z1 - zo)t•)
= Re(zo(l-t*)) +Re(z1t•)
= (1 -t*) Re (zo) + t*Re(z1). (1-53)If t* = 0, the last expression becomes Re (zo), which is greater than zero because
zo E H. Likewise, if t• = 1, then E<J.uation (1-53) becomes Re (zi), which also is
positive. Finally, if 0 < t* < 1, then each term in E<J.uation (1-53) is positive, so
in this case we also have Re (z (t')) > 0.
To show that H is open, we suppose without loss of generality that theinequality Re(zo) $ Re(z 1 ) holds. We claim that D, (zo) <;;; H, where e =
Re (z 0 ). We leave the proof of this claim as an exercise.A domain, together with some, none, or all its boundary points, is called
a region. For example, the horizontal strip { z : 1 < Im ( z) $ 2} is a region.
A set formed by taking the union of a domain and its boundary is called a
closed region; thus, { z : 1 $ Im (z) $ 2} is a closed region. A set S is said to
be a bounded set if it can be completely contained in some closed disk, that
is, if there exists an R > 0 such that for each z in S we have lzl $ R. The
rectangle given by {z: lxl $ 4 and IYI $ 3} is bounded because it is contained
inside the disk D5(0). A set that cannot be enclosed by any closed disk is called
an unbounded set.
We mentioned earlier that a simple closed curve is positively oriented if its
interior is on the left when the curve is traversed. How do we know, though,
that any given simple closed curve will have an interior and exterior? Theorem
1.6 guarantees that this is indeed the case. It is due in part to the work of the
French mathematician Camille Jordan (1838-1922).