Let $f:D_2(0) -> \mathbb{C}$ be holomorphic and suppose that $|f(z)| \leq |z-3|^2$ for all $z \in \mathbb{C}$ with $|z|=1$. Use estimation theorem to prove that $|f(0)| \leq 10$
Now I know that the estimation theorem is $|\int_{\gamma}f| \leq \int_{a}^{b} |f(\gamma(t)) \gamma '(t)|dt$
Following through the answer in the book:
$$|\int_{\gamma} \frac{f(z)}{z}dz| \leq \int_{0}^{2\pi}|\frac{f(\gamma(t))}{\gamma(t)}||\gamma'(t)|dt \leq \int_{0}^{2\pi}|\gamma(t)-3|^2$$
How does this equal the end point?
$$\int_{0}^{2\pi}|\gamma(t)-3|^2=\int_{0}^{2\pi}(\cos(t)-3)^2+\sin(t)^2dt$$
Again what identity has been used here? $$\int_{0}^{2\pi}(\cos(t)-3)^2+\sin(t)^2dt=\int_{0}^{2\pi}(\cos^2(t)+\sin^2(t)-6\cos(t)+9)dt = \int_{0}^{2\pi}(10-6\cos(t))dt = [10t+6\sin(t)]_{0}^{2\pi} = 20\pi$$
Implying $|f(0)|=|\frac{1}{2i\pi}\int_{\gamma}\frac{f(z)}{z}dz| \leq 10$
I can follow the last, its just the two points above I would appreciate being explained
Although you don't mention the path used, in the calculation it looks as if $\gamma(t)=e^{it}=\cos t + i \sin t$, so (using $\sin^2 t + \cos^2 t =1$) we obtain $|\gamma(t)|=|\gamma'(t)|=1$ from which the inequality in the first point should be clear and for the second point we have $|\gamma(t)-3|^2 =|\cos t -3 + i \sin t |^2 = (\cos t -3 )^2+\sin^2 t$.