Math 440/508 Quiz 1 Solution, Fall 2017
1. Does there exist a function f:C→Cthat is holomorphic at every point on the unit
circle S1={z∈C:|z|= 1}and not holomorphic anywhere else in the complex plane? If
yes, provide such a function with complete justification. If not, explain why not.
(10 points)
Solution. Yes, such a function exists.
Consider the function f(z)=(|z| − 1)2. Then fis infinitely real-differentiable for all z6= 0.
For such z,∂f
∂¯z= 2(|z| − 1)z
¯z,
which is nonzero unless |z|= 1. We have proved in class that a smooth function fis
holomorphic if and only if it satisfies the Cauchy-Riemann equations, namely ∂f/∂ ¯z= 0.
Therefore we conclude that fis holomorphic at z0∈C\{0}if and only if z0satisfies |z0|= 1.
Further if z0= 0, a direct computation shows that
lim
h→0
f(h)−f(0)
h= lim
h→0
|h|2−2|h|
h=−2 lim
h→0
|h|
h
does not exist, as can be seen by choosing sequences happroaching 0 along the real and
imaginary axis respectively. This proves that fis not holomorphic at zero either, completing
the proof.