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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 $s$.)

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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 $$ P(t) = \int_{|z| = 1} \frac{zf_t'(z)}{f_t(z)} dz. $$

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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

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