Analytic functions close to $\bar{z}$ Is there an analytic function $f\colon\Bbb{C}\longrightarrow \Bbb{C}$ such that for any $z$ on the unit circle $\lvert f(z) - \overline{z}\rvert < 1 $?
 A: Let me give an alternative proof. Suppose $f$ is the desired function. Then for $z\bar{z} = 1$:
$$ |z f(z) - 1| = |z| |f(z) - \bar{z}| < 1 $$
so $g(z) = zf(z) - 1$ is an analytic function satisfying $|g(z)| < 1$ whenever $|z| = 1$. By the maximum modulus principle 
$$ 1 =  |0 \cdot f(0) - 1| = |g(0)| < 1$$
which gives a contradiction. 
A: No, there is no such function, because if there were, we would have
$$1 = \left\lvert\frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \overline{z}\, dz\right\rvert = \left\lvert\frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \bigl(\overline{z} - f(z)\bigr)\, dz \right\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \lvert \overline{z} - f(z)\rvert \, dt < 1$$
by Cauchy's integral theorem and the standard estimate.
We find $\frac{1}{2\pi i}\int_{\lvert z\rvert = 1} \overline{z}\,dz = \frac{1}{2\pi i}\int_0^{2\pi} e^{-it} ie^{it}\,dt = 1$ by direct evaluation, or by observing that $\overline{z} = 1/z$ on the unit circle. Cauchy's integral theorem asserts that $\int_{\lvert z\rvert = 1} f(z)\, dz = 0$, from which we get the second equality.

A more geometric argument:
The condition $\lvert \overline{z} - f(z)\rvert < 1$ on the unit circle implies that, as mappings of the unit circle to $\mathbb{C}^\ast$, $f$ is homotopic to $\overline{z}$, in particular, both have the same winding number around $0$, namely $-1$. But the winding number of $f$ (restricted to the unit circle) around $0$ is
$$\frac{1}{2\pi i} \int_{\lvert z\rvert = 1} \frac{f'(z)}{f(z)}\, dz$$
which is the number of zeros of $f$ in the unit disk (counted with multiplicity): a non-negative integer.
