A function that satisfies $|f(z)-\bar z|<0.9$ is not analytic in the unit circle.

Question

I've came accros this excersize:
Suppose that $D=\{z:|z| \le 1\}\subset \mathbb C$ and $$f:D\rightarrow\mathbb C$$ suppose that for every $z\in D$ such that $|z|<1$ $$|f(z)-\bar z|<0.9$$ where $\bar z$ is the complex conjugate of $z$. Prove that $f$ cannot be analytic in $D$.
I started with assuming that $f$ is indeed analytic in order to get a contradiction. My first attempt was to try and get some similarities between $f$ and $g(z)=\bar z$ since they are relatively close to each other, and $g$ is not analytic. This idea quickly failed. Also I tried to integrate $f$ around $D$, or see if it possible that $f$ satisfies the Cauchy–Riemann equations, which also did not get me any further.

Answer 1

Hint: Say $\gamma$ is the unit circle. It would follow that $$\left|\frac1{2\pi i}\int_\gamma(f(z)-\overline z)\,dz\right|\le\frac9{10}.$$But you can easily figure out exactly what that integral is...

Answer 2

For any $\epsilon \in (0,\frac{1}{10})$, we have $$\begin{align} \left|\frac{1}{2\pi i}\int_{|z|=1-\epsilon} f(z) dz\right| \ge & \left|\frac{1}{2\pi i}\int_{|z|=1-\epsilon} \bar{z} dz \right| - \left|\frac{1}{2\pi i}\int_{|z|=1-\epsilon} ( f(z) - \bar{z} ) dz \right|\\ \ge & (1-\epsilon)^2 - \frac{1}{2\pi}\int_{|z|=1-\epsilon}|f(z) - \bar{z}| |dz|\\ \ge & (1-\epsilon)^2 - \frac{9}{10} (1-\epsilon) = (\frac{1}{10}-\epsilon)(1-\epsilon) > 0 \end{align} $$ This implies $\displaystyle\;\int_{|z|=1-\epsilon} f(z) dz \ne 0$. As a result, $f(z)$ cannot be holomorphic over $|z| < 1 - \epsilon$ or it will contradict with Cauchy integral theorem.