588 ANSWERS- T he right half-plane given by Re(z) > 0 is mapped onto the region in the
right half-plane satisfying u^2 - v^2 > 0 and lies to the right of u^2 - v^2 = 0.
This is the region between the Enes u = v and u = -v in t he right half of
the w planfl. A similar analysis can be applied to the case where b = O.Section 2.3. Limits and Continuity: page 77la. -3+ 5i.
le. -4i.
le. 1 - ~i.- lim (e"'cosy+ix^2 y) = lim (e"cosy+ix^2 y). The result now fol-
z-+•o (:z;, 11)-(xo, yo)
lows by Theorem 2.1 since the real and imaginary parts of the last expression
have limits that imply t he desired conclusion. You should show the details
for this, of course.- 12 -
Sa. lim J::L = lim ~ = lim z = 0.
z-o z z - 0 z z~o
7a. i.
7c. 1.- No. To see why, approach 0 along t he real and imaginary axes, respectively.
lla. If z -+ - 1 along the upper semicircle r = 1, 0 < 0 ~ 1r, then lim f ( z) =
z-- 1
IJ-+1r lim (cos(~) + isin (U] =cos(~) + isin (~) = i.- The real part is continuous since lim xell = lim xeY = xoeYo. A
-+•o (x,y)~(xo,yo}
similar argument shows the imaginary part is continuous. The function f is
then continuous by Theorem 2.1.
- No. The limit d oes not exist. Show why.
17. Rewrite f as in problem 9, and mimic the argument for part a with an
arbitrary negative real number taking t he role of -1.- Let €: > 0 be given. Since Jim f (z) = 0, there is some number o such that
z-zo
f (z) E D fr (0) whenever z E D~ (zo). Show t his implies that if z E D6 (O),
then If (z) g (z) -OI =If (z)l lg {z)I < €,so that f (z) g (z) E D, (0).
2la. We have remarked that example (2.16) shows that the function h(z) = z is
continuous for all z. Since f is continuous for all z, we can apply Theorem
(2.4) to the function f o h to conclude that g (z) = f (h (z)) = f (z) is
continuous for all z.