The maximum value of the modulus of e^{z^2} o | vedclass

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(A) Let $z = x + iy$, where $0 \leq x \leq 1$ and $0 \leq y \leq 1$. Then $z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$. Thus, $|e^{z^2}| = |e^{x^2 - y^2} \cd

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$e$

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(A) Let $z = x + iy$, where $0 \leq x \leq 1$ and $0 \leq y \leq 1$. Then $z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$. Thus, $|e^{z^2}| = |e^{x^2 - y^2} \cdot e^{i(2xy)}| = e^{x^2 - y^2} \cdot |e^{i(2xy)}|$. Since $|e^{i(2xy)}| = 1$, we have $|e^{z^2}| = e^{x^2 - y^2}$. To maximize $e^{x^2 - y^2}$, we need to maximize the exponent $f(x, y) = x^2 - y^2$ subject to $0 \leq x \leq 1$ and $0 \leq y \leq 1$. The maximum value of $x^2 - y^2$ occurs when $x$ is maximum $(x=1)$ and $y$ is minimum $(y=0)$. Thus, the maximum value is $e^{1^2 - 0^2} = e^1 = e$.

The maximum value of the modulus of $e^{z^2}$ on the set $\{z \in \mathbb{C} : 0 \leq \operatorname{Re}(z) \leq 1, 0 \leq \operatorname{Im}(z) \leq 1\}$ is

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