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\(\displaystyle z=(1-i)^{i-1}\)

\(\displaystyle 1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4} \right)+i\sin\left(-\frac{\pi}{4} \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}\)

Hence:

\(\displaystyle z=\left(e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i} \right)^{i-1}=e^{\frac{\pi}{4}-\frac{1}{2}\ln(2)}e^{\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)i}=\frac{e^{\frac{\pi}{4}}}{\sqrt{2}}\left(\cos\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)+i\sin\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right) \right)\)

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\(\displaystyle z=(1-i)^{i-1}\)

\(\displaystyle 1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4} \right)+i\sin\left(-\frac{\pi}{4} \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}\)

Hence:

\(\displaystyle z=\left(e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i} \right)^{i-1}=e^{\frac{\pi}{4}-\frac{1}{2}\ln(2)}e^{\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)i}=\frac{e^{\frac{\pi}{4}}}{\sqrt{2}}\left(\cos\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)+i\sin\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right) \right)\)

Yeah thats right but however i have to include the 2K\(\displaystyle {\pi}\)

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I see this as wholly unnecessary, but write it in if it is required.Yeah thats right but however i have to include the 2K\(\displaystyle {\pi}\)

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You can find more values if you take one turn or several turns..My Professor told m that..I see this as wholly unnecessary, but write it in if it is required.

- Jan 17, 2013

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I think it becomes necessary if we study complex numbers , Of course we always consider the Principle Logarithm if we are working on the real numbers .I see this as wholly unnecessary, but write it in if it is required.

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\(\displaystyle 1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4}+2k\pi \right)+i\sin\left(-\frac{\pi}{4}+2k\pi \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i+2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}e^{2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}\)

This is why I found it unnecessary in this case. Am I missing something?

- Jan 17, 2013

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I think the OP talks about the logarithm since $a^b$ is actually a multivalued function in the sense that b is a complex number .

\(\displaystyle 1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4}+2k\pi \right)+i\sin\left(-\frac{\pi}{4}+2k\pi \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i+2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}e^{2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}\)

This is why I found it unnecessary in this case. Am I missing something?

Consider the simpler example

\(\displaystyle \Large i^i = e^{i\log(i)}=e^{i(\ln(1)+(\frac{\pi}{2}+2k\pi) i )}=e^{-\frac{\pi}{2}-2k\pi }\)

Ofcourse we could do

\(\displaystyle \Large i^i= (e^{i\frac{\pi}{2}})^i=e^{-\frac{\pi}{2}}\)

But this will be not complete since it only gives the solution when \(\displaystyle k=0\).

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\(\displaystyle z=\frac{e^{\frac{\pi}{4}(8k+1)}}{\sqrt{2}}\left( \cos\left(\frac{\pi}{4}(8k+1)+\frac{1}{2}\ln(2) \right)+i\sin\left(\frac{\pi}{4}(8k+1)+\frac{1}{2}\ln(2) \right) \right)\) ?

W|A told me only when $k=0$ is this a solution.

My apologies for any confusion I caused to the OP.

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