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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.

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@labbhattacharjee Thank you! It was very helpful –  Siyanda Oct 27 '13 at 12:16

2 Answers 2

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.

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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

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 ?

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