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My progression:

$(1-i)^{i+1} = e^{(i+1) * \ln(1-i)}$. I get stuck after this point.

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How do you define $\ln(1-i)$? – Seirios Sep 25 '13 at 9:08
    
Can you evaluate $\ln (1-i)$ ? – Stefan Sep 25 '13 at 9:08
    
$ln(1-i) = ln(\sqrt{2}e^{i\arctan(-1/1)}) = ln(\sqrt{2}) + i7\pi/4?$ – David Sep 25 '13 at 9:10
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$\ldots+2\pi ki(i+1)$, $k \in \mathbb{Z}$. – njguliyev Sep 25 '13 at 9:21
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Yes. ${}{}{}{}$ – njguliyev Sep 25 '13 at 9:28
up vote 4 down vote accepted

The expression $z^w$ where $z, w \in \mathbb{C}$ is not uniquely determined. In fact, we define $$z^w = e^{w\log z}$$ where $\log z$ is any logarithm of $z$. There are infinitely many choices of $\log z$, and for most values of $z$ and $w$ there will be infinitely many possible values for $z^w$.

To get something unique, you will have to specify a particular branch of the complex logarithm, but when you do so. $z^w$ won't be defined for all $z$ (or at the very least $z^w$ won't be continuous in $z$, depending on what your conventions with branches are).

In your particular case $\log(1-i) = \ln \sqrt 2 - \dfrac{i\pi}4 + 2\pi i k$ for some arbitrary integer $k$, and

\begin{align} (1-i)^{1+i} &= e^{(1+i)\log(1-i)} \\ &= e^{(1+i)(\ln \sqrt 2 - \frac{i\pi}4 + 2\pi i k)} \\ &= e^{ \ln \sqrt 2 + \frac\pi4-2\pi k + i(\ln\sqrt 2 - i\frac{\pi}4 + 2\pi k)} \\ &= \sqrt 2 e^{\frac\pi4-2\pi k\pi}\cdot e^{i(\ln\sqrt2-\frac\pi4)} \\ &= \sqrt 2 e^{\frac\pi4-2\pi k\pi}\cdot \big( \cos (\ln\sqrt2-\frac\pi4) + i \sin(\ln\sqrt2-\frac\pi4) \big) \\ \end{align}

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Dutifully upvoted. – Did Sep 25 '13 at 16:48

$$\ln\left((1-i)^{(1+i)}\right)=(1+i)\ln(1-i)=(1+i)\left[\frac{1}{2}\ln(2)-\frac{1}{4}i\pi\right]$$ So: $$\ln[(1-i)^{(1+i)}]=F=\frac{1}{2}\ln(2)(1+i)+\frac{1}{4}\pi(1-i)$$ and: $$(1-i)^{(1+i)}=\exp(F)$$ and then: $$(1-i)^{(1+i)}=(1+i)\left(\sin\left(\frac{1}{2}\ln(2)\right)-i\cos\left(\frac{1}{2}\ln(2)\right)\right)\exp\left(\frac{1}{4}\pi\right)$$

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How did you go from $(1-i)^{1+i}$ to $(1-i)\ln(1+i)$? – David Sep 25 '13 at 9:27
    
@String: I agree. I correct it right now – Riccardo.Alestra Sep 25 '13 at 9:58
    
@David: Sorry. It was a mistake. I just corrected it – Riccardo.Alestra Sep 25 '13 at 10:04
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This is still incorrect. – mrf Sep 25 '13 at 10:55
    
@mrf: corrected it – Riccardo.Alestra Sep 25 '13 at 11:25

$n \in {\mathbb Z}$.

\begin{align} \left(1 - {\rm i}\right)^{1 + {\rm i}} &= \left[\sqrt{2\,}\,{\rm e}^{{\rm i}\left(-\pi/4 + 2n\pi\right)}\right] ^{1 + {\rm i}} = \left(\sqrt{2\,}\,\right)^{1 + {\rm i}} {\rm e}^{{\rm i}\left(-\pi/4 + 2n\pi\right) - \left(-\pi/4 + 2n\pi\right)} \\[3mm]&= \sqrt{2\,}\,{\rm e}^{\pi/4 - 2n\pi}\, \left(\sqrt{2\,}\,\right)^{\rm i} {\rm e}^{{\rm i}\left(-\pi/4 + 2n\pi\right)} = \left(1 - {\rm i}\right){\rm e}^{\pi/4 - 2n\pi}\, 2^{{\rm i}/2} \\[3mm]&= {\rm e}^{\pi/4 - 2n\pi}\, {\rm e}^{{\rm i}\ln\left(2\right)/2}\left(1 - {\rm i}\right) = {\rm e}^{\pi/4 - 2n\pi}\, \left[% \cos\left({1 \over 2}\ln\left(2\right)\right) + {\rm i}\sin\left({1 \over 2}\ln\left(2\right)\right) \right]\left(1 - {\rm i}\right) \\ ----&------------------------------------ \end{align}

\begin{align} &n \in {\mathbb Z}\,, \\[5mm] &\color{#ff0000}{\left(1 - {\rm i}\right)^{1 + {\rm i}}} \\&= \color{#ff0000}{{\rm e}^{\pi/4 - 2n\pi}\times \left\{% \left[ \cos\left({1 \over 2}\ln\left(2\right)\right) + \sin\left({1 \over 2}\ln\left(2\right)\right)\right] + {\rm i}\left[% -\cos\left({1 \over 2}\ln\left(2\right)\right) + \sin\left({1 \over 2}\ln\left(2\right)\right)\right] \right\}} \end{align}

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