# Prove that there exists analytic $f$ such that $f(z) = 1/\bar{z}$ on the boundary

I'm doing some self-study in complex analysis, and came to the following question:

Let $D(a,1) \subset \mathbb{C}$ be the disk of radius $1$ with center at $a \in \mathbb{C}$, and let $\partial D(a,1)$ be the boundary of $D(a,1)$. Prove that $|a| <1$ if and only if there exists a function $f$ analytic on $D(a,1)$ and continuous up to $\partial D(a,1)$ such that $f(z) = 1/\bar{z}$ for $z \in \partial D(a,1)$.

I don't know what area of the theory to try to apply here. I know that $|a| <1$ iff $\; 0 \in D(a,1)$, and I know that the function $1/\bar{z}$ is nowhere analytic. However, I can't see how to get to the desired conclusion.

Thanks.

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Note that $\bar z=\bar a + \overline{z-a}$, and on the boundary $z-a$ has unit length. Therefore $\overline{z-a}=\frac{1}{z-a}$ on the boundary (but only there). This should help you construct $f$ -- that is, the "only if" direction. For "if", one approach would be to combine the same construction with something like Cauchy's theorem. – Henning Makholm Jul 7 '12 at 1:08

If you know about harmonic functions, then you can use the maximum principle on the real and imaginary parts of $f(z) - {1 / \bar{z}}$ to show that if $a > 1$, then if such a $f(z)$ existed then $f(z)$ would have to be ${1 / \bar{z}}$ for all $z$.
If $a < 1$, I'd use an explicit construction like Henning Makholm suggests.