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I have an issue (sorry for bad usage of signs) proving

\begin{equation} \int\limits_{1}^{i} \mathrm{e}^{z^{2}} \mathrm{d}z \le \mathrm{e}\sqrt{2} \end{equation}

I'm also supposed to tell why its independent of way. I thought that I might be able to change the contour to a quarter of the unit circle, $|z|=1$, but then my length is gonna be $\pi/2$ ... Thanks in advance for any help!

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  • $\begingroup$ Have you heard of the Estimation Lemma? $\endgroup$ – user363087 Sep 18 '16 at 16:10
  • $\begingroup$ That's exactly what I was trying to do, but I dont know how to implement it... $\endgroup$ – A.Maine Sep 18 '16 at 16:46
  • $\begingroup$ You'll first need to find the length of the line from $z = 1$ to $z = i$, which can be done via Pythagoras. You'll then need to show that the maximum value of $e^{z^2}$ is $e$, if $z$ is a complex number lying on that line. $\endgroup$ – user363087 Sep 18 '16 at 16:49
  • $\begingroup$ This video might provide a nice tutorial for you. $\endgroup$ – Mark Viola Sep 18 '16 at 16:53
  • $\begingroup$ You need absolute values on that inequality. $\endgroup$ – zhw. Sep 18 '16 at 17:22
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\begin{equation} \int\limits_{1}^{i} \mathrm{e}^{z^{2}} \mathrm{d}z \le \mathrm{e}\sqrt{2} \end{equation}

This inequality does not make sense since the integral is a complex number. It should be

$$\left\lvert \int_1^i e^{z^2}\, dz\right\rvert \le e\sqrt{2}$$

To obtain this estimation, note that every point on the line segment from $1$ to $i$ takes the form $1 - t + ti$, $0 \le t \le 1$. Thus $\lvert e^{z^2}\rvert = e^{\operatorname{Re}(z^2)} = e^{(1 - t)^2 - t^2} = e^{1 - 2t} \le e$. The length of the segment from $1$ to $i$ is $\sqrt{2}$. Hence, by the ML-estimate your integral is bounded by $e\sqrt{2}$.

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  • $\begingroup$ Sorry for asking, but the -t^2, is that one necessary, I dont understand why we put (1-t)^2 - t^2 instead of (1-t+it)^2 or only (1-t)^2, sorry for a probably bad question $\endgroup$ – A.Maine Sep 18 '16 at 17:23
  • $\begingroup$ @A.Maine if $z = x + yi$, then $z^2 = (x^2 - y^2) + 2xy i$; in particular, the real part of $z^2$ is $x^2 - y^2$. That's why we have $\operatorname{Re}(z^2) = (1 - t)^2 - t^2$ for $z = 1 - t + ti$. $\endgroup$ – kobe Sep 18 '16 at 17:25
  • $\begingroup$ now I understand, thank you very much for taking time $\endgroup$ – A.Maine Sep 18 '16 at 17:28
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On the line segment from $1$ to $i$, we have the estimate \begin{align} |e^{z^2}|\leq e^{|z|^2} \leq e. \end{align}

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