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You have a good approach, but I think you have made an algebraic slip somewhere. To avoid a complex argument for the natural log function, I wrote:
\(\displaystyle z=(1-i)^{i-1}\)
\(\displaystyle 1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4} \right)+i\sin\left(-\frac{\pi}{4} \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}\)
Hence:
\(\displaystyle z=\left(e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i} \right)^{i-1}=e^{\frac{\pi}{4}-\frac{1}{2}\ln(2)}e^{\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)i}=\frac{e^{\frac{\pi}{4}}}{\sqrt{2}}\left(\cos\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right)+i\sin\left(\frac{\pi}{4}+\frac{1}{2}\ln(2) \right) \right)\)
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\(\displaystyle {\pi}\)
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.
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.
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 .\(\displaystyle z=(1-i)^{i-1}\)
\(\displaystyle 1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4}+2k\pi \right)+i\sin\left(-\frac{\pi}{4}+2k\pi \right) \right)=\sqrt{2}e^{-\frac{\pi}{4}i+2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}e^{2k\pi i}=e^{\frac{1}{2}\ln(2)-\frac{\pi}{4}i}\)
This is why I found it unnecessary in this case. Am I missing something?