Find all the functions f(z) analytic in D: |z|<1 which assumes only
pure imaginary values in D

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Request for Question Clarification by dannidin-ga on 10 Dec 2002 01:22 PST

fmunshi,

The only functions f(z) analytic in D which assume only pure imaginary
values in D are the constant functions: f(z)=i*t, where t is a real
number.

One way to prove this is to use the answer to your question "Open
mapping property of analytic functions"
https://answers.google.com/answers/main?cmd=threadview&id=122121

There, it is shown that if Re f(z) is constant in the domain, then f
itself must be constant. Here we have functions where Re f(z) is
identically 0. So f must be a constant, and since it is pure imaginary
it must an imaginary constant i*(real number).

Is this acceptable as an answer? I did not post this as an answer yet
because

a) you still did not rate livioflores's answer of your other question,
so you may not accept his solution of that question (seems correct to
me, though).

b) perhaps you want a solution which is more self-contained.

Tell me if this is o.k. and I will repost this as an answer. If
something is missing tell me and I will still try to answer the
question.

Regards,
dannidin

Hi fmunshi,

Here is a solution using only the 
Fundamental Theorem of Calculus
and the Cauchy-Riemann Equations.

Let f(z) = u(z) + i v(z) .

Notice u(z) = 0 for all z in D.

This implies 

(du/dx)(z) = 0 and (du/dy)(z) = 0  for all z in D .

Applying the Cauchy-Riemann Equations gives

(dv/dy)(z) = 0 and (dv/dx)(z) = 0  for all z in D .

Now we want to show v is constant.

For x in D and on the real line, we have 
by the Fundamental Theorem of Calculus

v(x) = v(0) + Int((dv/dx)(t), t= 0..x)
     = v(0)
since (dv/dx)(z) = 0 for all z in D.

So v is constant on the real line and equals v(0).

Now for z = x + iy in D, we have by the 
Fundamental Theorem of Calculus

v{z} = v{x} + Int((dv/dy)(x + it), t= 0..y)
     = v(x)
Since (dv/dy)(z) = 0 for all z in D.

But since x is on the real line, v(z) = v(x) = v(0) 
for all z in D, i.e. v is constant.

Thus f must be constant since f(z) = u(z) + iv(z) = iv(0).

Finally since f(z) = iv(0) is analytic for any real number 
v(0), the functions that you are looking for
are those which are just a pure imaginary constant.