Imaginary number challenge(2)

$\large |(\sqrt[2-i]{1+i})^{3+i}|$

$=\large |(1+i)^{\frac{3+i}{2-i}}|$

$=\large |(1+i)^{\frac{3+i}{2-i} \times \frac{2+i}{2+i}}|$

$=\large |(1+i)^{\frac{5+5i}{5}}|$

$=\large |(1+i)^{1+i}|$

$=\large |(\sqrt{2}e^{\frac{\pi}{4}i})^{1+i}|$

$=\large |(\sqrt{2}e^{\frac{\pi}{4}i})((\sqrt{2})^{i}e^{-\frac{\pi}{4}})|$

$=\large |(\sqrt{2}e^{-\frac{\pi}{4}})((\sqrt{2})^{i}e^{\frac{\pi}{4}i})|$

$=\large |(\sqrt{2}e^{-\frac{\pi}{4}})(e^{(\ln{\sqrt{2}}+\frac{\pi}{4})i})|$

$=\large (\sqrt{2}e^{-\frac{\pi}{4}})|(e^{(\ln{\sqrt{2}}+\frac{\pi}{4})i})|$

$=\large (\sqrt{2}e^{-\frac{\pi}{4}})(1)$

$=\large (\sqrt{2}e^{-\frac{\pi}{4}})$

$\Rightarrow 63a^{3}b= 63(2)^{3}(4) = 2016$

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$e^{\frac{\pi}{4}} = \cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})$

$e^{\frac{\pi}{4}} = \frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$

$(\sqrt{2})e^{\frac{\pi}{4}} = (\sqrt{2})\frac{\sqrt{2}}{2}+(\sqrt{2})i\frac{\sqrt{2}}{2}$

$\sqrt{2}e^{\frac{\pi}{4}} = 1+i$

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$|e^{ix}| = \sqrt{\cos(x)^{2} + \sin(x)^{2}}$

$|e^{ix}| =1$

$\Rightarrow |(e^{(\ln{\sqrt{2}}+\frac{\pi}{4})i})| = 1$