Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm studying for a complex analysis exam, and I'm stuck on this problem from an old exam:

Let $g$ be a holomorphic function on $|z|<R,R>1$, with $|g(z)|\leq 1$ for all $|z|\leq 1$.

(a) Show that for all $t\in C$ with $|t|<1$, the equation $$z=tg(z)$$ has a unique solution $z=s(t)$ in the disc $|z|<1$.

(b) Show that $t\mapsto s(t)$ is a holomorphic function on the disc $|t|<1$. (Hint: find an integral formula for $s$.)

share|improve this question

1 Answer 1

up vote 10 down vote accepted

In (a), you want to show that $f_t(z) = z - tg(z)$ has a unique zero in the unit disk. Since $|tg(z)| < 1$ for $|z| \leqq 1$, this function is ripe for an application of Rouché's theorem. For (b), you should be able to modify the argument principle in order to pick out the zero of $f_t$, instead of merely counting it. More explicitly, look at the function $$ P(t) = \int_{|z| = 1} \frac{zf_t'(z)}{f_t(z)} dz. $$

share|improve this answer
    
Dylan, when it asks for a unique solution, does that mean there is only one fixed-point within the unit disk for each $t$? Should one even worry about that part? –  mathmath8128 Jul 5 '11 at 5:49
    
I think Rouché's theorem will tell you that there is only one zero of $f$ in the disc. Counting multiplicity, even. I'll edit this. –  Dylan Moreland Jul 5 '11 at 5:55
    
I think I would like you to say more about part b). Thanks. –  Cantor's Paradise Jul 7 '11 at 18:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.