AP EAMCET · Maths · Complex Number
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
- A \(\frac{2}{e}\)
- B e
- C \(e+1\)
- D \(e^2\)
Answer & Solution
Correct Answer
(B) e
Step-by-step Solution
Detailed explanation
\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…
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