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What is $(1-i)^{1+i}$ ?

I did:

$(1-i)^{1+i} = \left(e^{ln(1-i)}\right)^{1+i} = e^{(1+i)\cdot ln(1-i)} = e^{(1+i)\cdot [ ln\sqrt{2} + i\cdot(-\pi/4) + 2ki\pi ]}$ $ = e^{(ln\sqrt{2} + i\cdot(-\pi/4) + 2ki\pi)} \cdot e^{i\cdot(ln\sqrt{2} - \pi/4 + 2ki\pi)}$

This is it?

Daniel Robert-Nicoud
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Vazrael
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2 Answers2

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HINT :

$$1-i = \sqrt{2}e^{-i\pi/4}$$ Then the calculations depend under which form your want it but here's another hint for later : $$\sqrt{2}^{1+i} = \sqrt{2}\cdot2^{i/2} = \sqrt{2}\cdot e^{i\log(2)/2}$$

user88595
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When you work with logarithms in the complex plane you have to be very careful. The logarithm cannot be defined in a good way on the whole complex plane (even if you remove the $0$). You have to chose a branch of log, that means choosing a particular definition of logarithm. I would take the principal branch (i.e. the one with imaginary part lying in $(-\pi,\pi)$ and which is undefined on $\mathbb{R}_{\le 0}$) and do it as follows:

$$(1-i) = \sqrt{2}\ e^{-i\frac{\pi}{4}}$$ (you can see why this is true e.g. identifying $\mathbb{C}$ with $\mathbb{R}^2$ and drawing the position of $(1-i)$ as a "vector" in this space). Then you get: $$(1-i)^{1+i}=\left(\sqrt{2}\ e^{-i\frac{\pi}{4}}\right)^{1+i} = 2^{\frac{1+i}{2}}e^{-i\frac{\pi}{4}(1+i)}$$ This gives you: $$2^{\frac{1+i}{2}} = e^{\frac{i}{2}\ln(2)}\sqrt{2}$$ $$e^{-i\frac{\pi}{4}(1+i)} = e^{-i\frac{\pi}{4}}e^{\frac{\pi}{4}}$$ and thus: $$(1-i)^{1+i}=e^{-i\frac{\pi}{4}}e^{\frac{\pi}{4}}e^{\frac{i}{2}\ln(2)}\sqrt{2}=$$ $$(1-i)e^{\frac{\pi}{4}}\left(\cos\left(\frac{\ln(2)}{2}\right)+i\sin\left(\frac{\ln(2)}{2}\right)\right)$$

For more details on branches of log, try to give a look at this page.

Daniel Robert-Nicoud
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  • Well, you had to work with the complex logarithm anyway, didn't you? – DonAntonio Dec 08 '13 at 12:53
  • @DonAntonio Not really, I reduced the problem to the logarithm of a real number. I believe you know that trying to keep the number of logarithms of ("strictly") complex numbers to a minimum tends to avoid the confusion coming from having to choose a branch of log (in particular if you are not used to working like that). – Daniel Robert-Nicoud Dec 08 '13 at 12:58
  • The very second equality in your second line of mathematics (the one with $;(1-i)^{1+i};$ not only uses the complex logarithm but it also does so assuming a very particular branch for it, apparently the main one. I understand the problems with the complex logarithm, yet someone trying to answer a question like the OP's has to deal with it sooner or later, so I think it's better to do it sooner and warn the student about some of the problems here than dodge it. – DonAntonio Dec 08 '13 at 13:02
  • @DonAntonio Ok, you win :) How should I edit my answer to make the point crear(er)? – Daniel Robert-Nicoud Dec 08 '13 at 13:11
  • It's not, of course, a matter of winning-losing, but only to remark to the OP some problems/aspects that may arise in his question. If you want to add something to your question you could probably mention that, trying to avoid as much the problematic you mention in your first line (and perhaps explaining a little why you think that about working with the comp. log.), you' try this or that approach, etc. Remarking that what the OP did is fine can also help, imo. – DonAntonio Dec 08 '13 at 13:14
  • @DonAntonio Better? – Daniel Robert-Nicoud Dec 08 '13 at 13:42
  • Indeed so, @Daniel +1 – DonAntonio Dec 08 '13 at 13:44
  • Thank you very much; I just didn't get why "$(1−i)=\sqrt{2} e^{−i \pi/4}$" this is true – Vazrael Dec 08 '13 at 21:39
  • @K.L. You probably have seen (else you will see it shortly) that $e^{i\phi} = \cos(\phi)+i\sin(\phi)$. Apply this to $\sqrt{2}\ e^{i\pi/4}$. – Daniel Robert-Nicoud Dec 08 '13 at 23:06