The maximum value of the modulus of $e^{z^2}$ on the set $\{z \in C / 0 \leq \operatorname{Re}(z) \leq 1,0…
The maximum value of the modulus of $e^{z^2}$ on the set $\{z \in C / 0 \leq \operatorname{Re}(z) \leq 1,0 \leq \operatorname{Im}(z) \leq 1\}$ is
- $\frac{2}{e}$
- e
- $e+1$
- $e^2$
Solution
$\begin{aligned} & \text { Let } z=x+i y 0 \leq x \leq 1,0 \leq y \leq 1 \\ & \Rightarrow \quad e^{z^2}=e^{x^2-y^2+2 i x y} \\ & \Rightarrow \quad e^{z^2}=e^{x^2-y^2} \cdot e^{i(2 x y)} \\ & \Rightarrow \quad\left|e^{z^2}\right|=e^{x^2-y^2} \\ & {\left[\left|e^{i(2 x y)}\right|=1\right]} \\ & 0 \leq x^2 \leq 1,0 \leq y^2 \leq 1 \\ & \Rightarrow \quad 0 \leq x^2 \leq 1 \\ & -1 \leq-y^2 \leq 0 \\ & \Rightarrow \quad-1 \leq x^2-y^2 \leq 1 \\ & \Rightarrow \text { maximum of }\left|e^{z^2}\right|=e \\ & \end{aligned}$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
Practice more Complex Number questions on Aicharya