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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
$\ldots+2\pi ki(i+1)$, $k \in \mathbb{Z}$. – njguliyev Sep 25 '13 at 9:21
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

Hint: $i+1=\sqrt 2e^{2n\pi i+i\pi/4}, 1-i=\sqrt 2e^{2m\pi i-i\pi/4}$, and $$ (i+1)\ln(1-i)=(\frac{-i\pi}{4}+2\pi m i)\times \sqrt2ln\sqrt 2e^{2n\pi i+i\pi/4}\\ =(\frac{\pi}{4}-2m\pi)\times \sqrt2ln\sqrt 2e^{2n\pi i-i\pi/4}\\ =(1-i)(\frac{\pi}{4}-2m\pi)\times ln\sqrt 2 $$

Hint: Please take a look at the discussion below and also another answer from mrf. One has to be prudent in defining the logarithm of complex numbers.

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This only gives one of the infinitely many possible values of $(1-i)^{1+i}$. – mrf Sep 25 '13 at 10:54
If this is obvious to you, why posting a misleading answer? – Did Sep 25 '13 at 11:15
Let us be serious for one second: the main difficulty students face when dealing with complex valued logs is the multivalued aspect. To omit it completely from a hint (yes a hint posted to answer an MSE question) is (I am sorry to say so)... misleading. (I am not quite sure that advocating "the imprecision of answer... not critical for the solution" and a "real contribution... in method" in this case proves what you think. The point IS critical, and yes, this IS a question of method.) – Did Sep 25 '13 at 15:15
@TheChaz2.0 I think that I understand what you mean and that I agree. The source of the trouble might be to pretend that there exists a function named $\log$ defined on the complex plane and such that, for every complex numbers $z$ and $z'$, $\log(zz')=\log(z)+\log(z')$. As we know, there is not--which makes the reading of some answers on this MSE page rather... dizzying, to say the least (which was probably your point). – Did Sep 25 '13 at 16:01
@Did: Right! And it's not just in answers here... the Complex book I'm going through (Saff & ??) spits out these equalities with only a single comment early on about how they could make things "more precise" (which they don't, really). Now about the Answer above, I guess my thinking was that if the OP's confusion wouldn't increase/decrease based on this answers. – The Chaz 2.0 Sep 25 '13 at 16:30

$n \in {\mathbb Z}$.

\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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$$\ln\left((1-i)^{(1+i)}\right)=(1+i)\ln(1-i)=(1+i)\left[\frac{1}{2}\ln(2)-\frac{1}{4}i\pi\right]$$ 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
This is still incorrect. – mrf Sep 25 '13 at 10:55
@mrf: corrected it – Riccardo.Alestra Sep 25 '13 at 11:25

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