602 ANSWERS
- We know that the complex cosine is an entire function that is not a constant.
By Liouville's theorem, it is not bounded.
Sa. IJ<^4 ) (1)1~^41 ~~^0 >. (Explain.)
5b. if <^4 > (O)I ~
41
~!^0 >. (Explain.)7a. If If (z)I ~ m for all z in D, where m > 0, tben the function J is analytic
in D. Apply the maximum modulus theorem to the function J to get your
result.9. Let f (z) = u (z) +iv (z), where v is a harmonic conjugate of u, so that
f is analytic in D. The function F (z) =exp (f (z)) is also analytic in D,
so that IFI does not take on a maximum in D by the maximum modulus
theorem. But IF(z)I = exp(u(z)) for all z (show why). This leads to the
conclusion since u is a real-valued function, and the real-valued function exp
is an increasing function. Explain this last part in detail.- (By contraposition) If f does not have a zero, then J is analytic in D1 (0), so
its maximum occurs on the boundary. Since f is constant on the boundary,
we conclude that both the maximum and the minimum of f are the same,
which means f is constant.
Sectio n 7. 1. Uniform Convergence: page 25 5la. By definition, f (-1) = 1 (^1 1 ) = 4. It appears from the graph that the
value of the upper function is approximately 1 (certainly larger than 4), sothe graph of S,.. must be above the graph off.
n - 1l e. From the graph, we approximate Sn (1) = 5. As Sn (x) = L: xk , we deduce
k=Othat n = 5. Explain.
3a. We see that l&z"I ~ & for z E D 1 (0). By t.he Weierstrass M -test, the
00
series L: &zk converges uniformly on D1 (0) = {z: lzl ~ l}, because t he
k=l 00series L: & converges.
k=l- The crucial step in the theorem is the statement, "Moreover, for all z E
Dr (a) it is clear that jck (z -a)"j = lckl lz - al" ~ lckl r"." If we allowed
00
r = 1, we would not be able to claim that L: lcklr" converges. Explain.
k=O
7a. Let us say that {Jn} and {g,..} converge uniformly on T to f and g, re-
spectively. Let € > 0 be given. The uniform convergence of Un} means