Let f be an analytic function in the disk ... Prove that

Question

Let $f$ be a function analytic in the disk $D={z:|z-1-i| \leq 4}$ that satisfies $|f(z)| \leq 1$ for all $z$ in $D$.

Prove that $|f^{(3)} (1-i)| \leq \frac{3}{4}$.

Hint: apply Cauchy's Integral Formula to a circle centered at $1-i$

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I know that if a function is analytic then the complex derivative exists at every point. I also know that the disk is a circle centered at $1-i $ with a radius of $4$. I would like help starting and completing this problem.

Answer 1

Cauchy's integral formula is

$$f^{(n)}(z_0)=\frac{n!}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^{n+1}dz}$$

Now, for this example, $n=3$ and $z_0=1-i$. If we take the contour to be parameterized by $z=1-i+\rho e^{i\phi}$, $dz=i\rho e^{i\phi} d\phi$, and $0\le \phi \le 2\pi$, then $f$ is analytic for values of $\rho \le 2$. This is because $f$ is analytic when $|z-(1+i)|\le4$ and the circle $|z-(1-i)|= 2$ is contained in the region of analyticity. Now, we can write for $\rho \le 2$

$$\begin{align} f^{(3)}(1-i)&=\frac{3!}{2\pi i} \int_0^{2\pi} \frac{f(1-i++\rho e^{i\phi})}{\rho^{4}e^{i4\phi}} i\rho e^{i\phi} d\phi\\\\ &=\frac{3}{\pi \rho^3} \int_0^{2\pi} f(1-i++\rho e^{i\phi}) e^{-i3\phi} d\phi\\\\ \end{align}$$

Taking magnitudes of both sides we find

$$\begin{align} |f^{(3)}(1-i)|&=\left|\frac{3}{\pi \rho^3} \int_0^{2\pi} f(1-i++\rho e^{i\phi})e^{-i3\phi} d\phi\right|\\\\ &\le\frac{3}{\pi \rho^3} \int_0^{2\pi} |f(1-i++\rho e^{i\phi})| d\phi\\\\ &\le \frac{3}{\pi \rho^3} 2\pi \max |f(1-i++\rho e^{i\phi})|\\\\ &\le \frac{6}{\rho^3}\\\\ \end{align}$$

Since this is true for all $\rho \le 2$, then the supremum is of this latter bound is $6/2^3$.

Thus

$$|f^{(3)}(1-i)|\le \frac34$$

as was to be shown!

Answer 2

You should be aware that Cauchy's integral formula states $$f^{(n)}(a) =\frac{n!}{2\pi i}\oint_\gamma\frac{f(z)}{(z-a)^{n+1}}\,\mathrm dz$$ where $\gamma$ is a circle lying in $D$. The distance between the point of interest $a=1-i$ and the center $1+i$ of the disk $D$ is $2$, hence we can let $\gamma$ have radius $r$ slightly less than $2$ (and consider the limit as $r\to 2$). We find with unit speed parametrization of $\gamma$ $$| f'''(a)|\le \frac{3}{\pi}\int_{t=0}^{2\pi r}\left|\frac{f(\gamma(t))}{r^4}\right|\,\mathrm dt\le \frac3\pi\frac{2\pi r}{r^4}=\frac 6{r^3}.$$ In the limit $r\to 2$ this gives us $|f'''(a)|\le \frac{6}{2^3}=\frac34$.