Let $f(z)=2ize^{(i+1)z^2+2i}$. Find $\max_{z\in \bar{B}(0,2)}|f(z)|$.
My attempt:
Let $z=re^{i\theta}, 0\le r\le 2, \theta\in[0,2\pi[.$ Then $f(z)=2ire^{i\theta}e^{(i+1)r^2e^{2i\theta}+2i }$ and thus, $$ |f(z)|=2re^{r^2\cos(2\theta)}.$$
I'm not sure how to continue. I think that I have to find $(r,\theta)$ that maximizes the value of $|f(z)|$. Now, $\cos(2\theta)$ with $\theta\in[0,2\pi[$ reaches a maximum for $\theta = \pi/2$. This means that $z=ri, 0\le r\le 2$ and $|f(z)|=2re^{r^2}$. And this is maximized for $r=2$. In conclusion, $\max_{z\in\bar{B}(0,2)}|f(z)|=4e^4$.
Is this a correct approach? Thanks.
Assuming $\overline B(0,2) = \{z \in \mathbb{C} : |z| \le 2\} $, thanks to the maximum modulus principle we can say the maximum of $|f(z)|$ is reached for $z=2e^{i\theta}$ for $\theta \in [0,2\pi)$.
Evaluating $|f(z)|$ for $z=2e^{i\theta}$ we get $|f(2e^{i\theta})| = 4e^{4(\cos(2\theta) - \sin(2\theta))}$ which is maximized when $\cos(2\theta) - \sin(2\theta)$ reaches his max: $$\max(\cos(2\theta) - \sin(2\theta)) = \sqrt{2} \quad \text{for} \quad \theta = \pi n + \arctan(1-\sqrt2) \approx_{n=1} 2.74 \in [0,2\pi)$$
In conclusion: $$\max|f(z)| = 4e^{4\sqrt2}$$
Edit:
\begin{split} |f(z)| & = |2ir|\cdot |e^{i\theta}| \cdot |e^{ir^2e^{2i\theta}}| \cdot |e^{r^2e^{2i\theta}}| \cdot |e^{2i}| \\ & = 2r \cdot |e^{ir^2[\cos(2\theta) +i\sin(2\theta)]}| \cdot |e^{r^2[\cos(2\theta) +i\sin(2\theta)]}| \\ & = 2r \cdot |e^{ir^2\cos(2\theta)}| \cdot |e^{-r^2\sin(2\theta)}| \cdot |e^{r^2cos(2\theta)}| + \cdot |e^{ir^2\sin(2\theta)]}| \\ & =2r \cdot |e^{-r^2\sin(2\theta) +r^2 \cos(2\theta)}| \end{split}