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The question says to find all the values of $(1+i)^{(1-i)}$

I have trouble figuring out firstly, exactly what values are being looked for. I can toy around with the equation a bit to try to make it look "acceptable" (i.e $ax + byi$ format) but get stuck along the way. So I need help with:

a) figuring out what values are needed. i.e. what does the question $mean$ and some brief background or diagram that explains, in a practical sense, what I'm supposed to be looking for.

b) the algebra that can lead me to a reasonable solution.

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MY ATTEMPT:

$$ (1+i)^{(1-i)} = (1+i)^{(1-i)}.\frac{(1+i)^{(1+i)}}{(1+i)^{(1+i)}} = \frac{(1+i)^2}{(1+i)^{(1+i)}} = *$$

*is where I get stuck.

Siyanda
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3 Answers3

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$\forall n\in \mathbb{Z}$, \begin{align*} e^{2n\pi i} &= 1 \\ 1+i &= \exp \left[ \frac{\ln 2}{2}+i\pi \left( 2n+\frac{1}{4} \right) \right] \\ (1+i)^{1-i} &= \exp \left[ \frac{1-i}{2} \ln 2+i(1-i) \left( 2n+\frac{1}{4} \right) \pi \right] \\ &= \exp \left[ \frac{\ln 2}{2}+\left( 2n+\frac{1}{4} \right) \pi+ i\left( -\frac{\ln 2}{2}+2n\pi+\frac{\pi}{4} \right) \right] \\ &= \sqrt{2} \, \exp \left[ \left( 2n+\frac{1}{4} \right) \pi+ i\left( \frac{\pi}{4}-\frac{\ln 2}{2} \right) \right] \\ &= \sqrt{2} \, e^{\left( 2n+\frac{1}{4} \right) \pi} \left[ \cos \left( \frac{\pi}{4}-\frac{\ln 2}{2} \right)+ i\sin \left( \frac{\pi}{4}-\frac{\ln 2}{2} \right) \right] \end{align*}

Ng Chung Tak
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In general, $a^b=e^{b\ln a}$. Now in complex analysis, the logarithm function can be considered multi-valued, so if $A$ is one possible value for $\ln a$, the other possibilities are $A+2\pi i n$ with an integer $n$. Now substitute that into your formula.

Harald Hanche-Olsen
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  • Hi, thanks for the tip! Using it I ended up with the following answer: $$ 2(1+i) e^{(\frac{\pi}{4} + 2k \pi)+(\frac{\pi}{4} - 2k{\pi})i}$$ – Siyanda Oct 26 '13 at 12:09
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Continue developing the exponential using Moivre formula and expand again to isolate the real and imaginary parts. You will end with a nice formula in the form of (a + I b). Are you able to continue with this ?

Claude Leibovici
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