Is $\int\limits_0^R e^{-(1+i)r^2}dr$ the same as $\int\limits_0^R e^{-({1\over2}+{i\over2})r^2}dr$

Question

In a corrected exercise there is this equality

$\Gamma_R^3$ is the line $\{re^{i\pi/8}~: ~r\in[0,R]\}$

$-\sqrt2(re^{i\pi/8})^2=-\sqrt2(r^2e^{i\pi/4})=-\sqrt2r^2({\sqrt2\over2}+i{\sqrt2\over2})$ so I don't get how this is right. Mybe I'm missing something...

Thanks

Answer 1

Continuing from your calculation,$$-\sqrt2(re^{i\pi/8})^2=-\sqrt2(r^2e^{i\pi/4})=-\sqrt2r^2({\sqrt2\over2}+i{\sqrt2\over2})=-r^2(1+i)\sqrt2\cdot\frac{\sqrt2}2=-r^2(1+i)$$since $\sqrt2\cdot\frac{\sqrt2}2=1$.