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Determine all complex values of $(ie^{\pi})^{i}$ and indicate the principal value.

My attempt: $(ie^{\pi})^{i}=e^{i\log(ie^{\pi})}=e^{i[\ln|ie^{\pi}|+i\arg(ie^{\pi})]}$

Note that $\ln|ie^{\pi}|=\ln e^{\pi}=\pi$ and $\arg(ie^{\pi})=\frac{\pi}{2}+2k\pi$ for all integer $k \in \mathbb{Z}$. Hence, $(ie^{\pi})^{i}=e^{i\pi-\frac{\pi}{2}-2k\pi}=-e^{-\frac{\pi}{2}-2k\pi}$

Principal value is $-\frac{\pi}{2}$.

A few parts I'm not quite sure. Like can the principal part be $\frac{\pi}{2}$?

Also can someone help me to check whether I do it correctly or not?

EDIT: From the definition of principal value, I obtain $e^{i\operatorname{Log}(ie^{\pi})}=e^{i\pi-\frac{\pi}{2}}=-e^{-\frac{\pi}{2}}$

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We can write \begin{align} i&=0+1i\\ &=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\\ &=e^{i\frac{\pi}{2}}\tag{1} \end{align}
From equation $(1)$, \begin{align} (ie^\pi)^i&=(e^\frac{i\pi}{2}e^\pi)^i\\ &=(e^\frac{i^2\pi}{2})e^{i\pi}\\ &=e^\frac{-\pi}{2}(-1)\\ &=\frac{-1}{e^\frac{\pi}{2}} \end{align}

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  • $\begingroup$ is the second equality in second line valid? We can expand the complex exponent like real exponent? $\endgroup$ – Idonknow Nov 5 '13 at 17:53

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