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Find the principal value of $(Log(1-i\sqrt3))^i$.

Now $r=2$ and $\theta = tan^{-1}(-\sqrt3) = -\pi / 3$ so we have $log(1 + i\sqrt3) = ln2 + i(\frac{-\pi}{3} + 2\pi n)$. So PV of $(Log(1-i\sqrt3))^i$ is $e^{i(ln2 + i(\frac{-\pi}{3} + 2\pi n))} = e^{(\frac{\pi}{3} - 2\pi n) + iln2}$ which gives us $e^{\frac{\pi}{3} + iln2}$?

Is that right?

Also I am trying to find the modulus and PV of the argument of the complex number $cot(iLog2)$ but am not sure how to proceed?

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I think your answer's correct, but why not going directly by the very definition and by the main logarithmic branch?:

$$\text{Log}\,(1-\sqrt3\,i):=\log|1-\sqrt3\,i|+i\arg(1-\sqrt3\,i)=$$

$$=\log2+i\arctan\frac{-\sqrt3}1=\log 2-\frac\pi3i\implies$$

$$\text{Log}^i\,(1-\sqrt3\,i)=e^{i\,\text{Log}\,(1-\sqrt3\,i)}=e^{\frac\pi3+i\log2}$$

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    $\begingroup$ You made a mistake there, $\text{Log}^i\,(1-\sqrt3\,i) \neq e^{i\,\text{Log}\,(1-\sqrt3\,i)}$, but rather $\text{Log}^i\,(1-\sqrt3\,i) = e^{i\,\text{Log} \text{Log}\,(1-\sqrt3\,i)}$. $\endgroup$ – Ondřej Čertík Sep 28 '15 at 3:38

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