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While trying to find the fourier transform of $\Large \frac{1}{1 + x^4} $, using the definition and the residue theorem has required me to evaluate nasty looking expressions like $$\large \rm e^{-ike^{i\frac{\pi}{4}}} .$$

Mathematica tells me this is the same as $$ \rm e^{\frac{k}{\sqrt{2}}} \cos(\frac{k}{\sqrt{2}}) - i \rm e^{\frac{k}{\sqrt{2}}} \sin(\frac{k}{\sqrt{2}}) $$

Now these two expressions aren't obviously equivalent to me, and my question is, how can I get from the first expression to the second expression?

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Note that

$e^{\frac{i\pi}{4}}=cos(\frac{\pi}{4})+isin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}(1+i)$ so

$e^{-ike^{\frac{i\pi}{4}}}=e^{-ik\frac{\sqrt{2}}{2}(1+i)}=e^{\frac{k}{\sqrt{2}}-\frac{ik}{\sqrt{2}}}=e^{\frac{k}{\sqrt{2}}}(\cos(\frac{k}{\sqrt{2}})-isin(\frac{k}{\sqrt{2}}))$

which gives the form you want.

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It's just Euler's formula $$ e^{\sigma + i\varphi} = e^\sigma\cos\varphi + ie^\sigma\sin\varphi $$ applied twice. First to get $$ e^{i\frac{\pi}{4}} = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \text{,} $$ and then again to get $$ \begin{align} e^{-ike^{i\frac{\pi}{4}}} &= e^{-ik\left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right)} = e^{\frac{k}{\sqrt{2}} - i\frac{k}{\sqrt{2}}} = e^{\frac{k}{\sqrt{2}}}\cos\left({-\frac{k}{\sqrt{2}}}\right) + ie^{\frac{k}{\sqrt{2}}}\sin\left({-\frac{k}{\sqrt{2}}}\right) \\ &= e^{\frac{k}{\sqrt{2}}}\cos\left({\frac{k}{\sqrt{2}}}\right) - ie^{\frac{k}{\sqrt{2}}}\sin\left({\frac{k}{\sqrt{2}}}\right) \text{.} \end{align} $$

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