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For what values of $\alpha$ does $1^{\alpha}$ does $1^{\alpha} = 1$. What are the possible values of $1^{\alpha}$? What are the values of $1^{\frac{1}{2}}$? (Hint: use the definition of $z^{\alpha}$.)

Attempt: Recall the definition of a complex $\alpha$ constant where $z \neq 0$, then $z^{\alpha} = e^{\alpha \log z}$.

Then, the possible values for $1^{\alpha}$ using the definition are: $1^{\alpha} = e^{\alpha \log 1} = e^{\alpha [\log 1 + i\arg 1]} = e^{\alpha i 2 k\pi } $.

And when $\alpha = \frac{1}{2}$ we have $ 1^ {\frac {1}{2}} = e^{\frac{1}{2} i 2 k\pi } = e^{ik\pi} $.

I don't know how to continue.

I dont know for what values of $\alpha$ does $1^{\alpha}$ does $1^{\alpha} = 1$. Can someone please help me? I would really appreciate it. Thank you in advance.

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  • $\begingroup$ How are you defining $\log z$? Is it a single-valued function or multi-valued? If single-valued with $\log 1=0$ then it is true for all values $\alpha.$ If multivalued, then are you asking when all values of $1^{\alpha}$ are $1$? Because if multivalued, then $1$ is always one of the values of $1^{\alpha}$. $\endgroup$ – Thomas Andrews Apr 1 '15 at 0:29
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Hint: You want $e^{2\pi k \alpha i} = 1$. What values of $z$ give you $e^{iz} = 1$? You'll want to distinguish between the cases $k = 0$ and $k \ne 0$.

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  • $\begingroup$ when $z = \pi$ we have $e^{i\pi} = 1$ . And when $k = 0$ we have $e^{2\pi k\alpha i} = e^0 = 1$ When k is not equal to zero, then when $k = 2\pi$ or any multiple of $2\pi$ it will be equal to 1? $\endgroup$ – Mahidevran Apr 1 '15 at 0:27
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    $\begingroup$ @Mahidevran $e^{i\pi}=-1$, not $1$. $\endgroup$ – AlexR Apr 1 '15 at 0:29
  • $\begingroup$ When $z$ is an integer multiple of $2\pi$ you get $1$. So you want $k\alpha$ to be ... Here $k$ is an integer, it's not going to be $2\pi$. $\endgroup$ – Robert Israel Apr 1 '15 at 2:49
  • $\begingroup$ k has to be the integer 1? $\endgroup$ – Mahidevran Apr 1 '15 at 3:02
  • $\begingroup$ No, it doesn't. Any integer will work if $1$ does. $\endgroup$ – Robert Israel Apr 1 '15 at 6:51

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