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my answer would be $i$ because I used the definition $i=\sqrt-1$ and replaced the $i$ with that

I'm not sure if I got it right though

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    $\begingroup$ Do you mean $(\sqrt{i})^i$ or $\sqrt{i^i}$? $\endgroup$
    – Wojowu
    Mar 8 '19 at 15:22
  • $\begingroup$ @Wojowu in either case, the result is the same. $\endgroup$
    – Decaf-Math
    Mar 8 '19 at 15:26
  • $\begingroup$ Try considering taking logarithm on both sides of $y=(\sqrt{i})^i$. $\endgroup$ Mar 8 '19 at 15:27
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    $\begingroup$ You can see $i^i$. $\endgroup$ Mar 8 '19 at 15:27
  • $\begingroup$ You can see square roots of complex numbers here, in particular $\sqrt{i^i}$. $\endgroup$ Mar 8 '19 at 15:28
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$$\sqrt{i}^i=i^{i/2}=\exp \left(\left(\dfrac{\pi i}{2}+2\pi i n\right)\cdot \dfrac{i}{2}\right)=\exp \dfrac{-\pi}{4}\cdot\exp-\pi n, \ n\in \mathbb{Z}$$

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  • $\begingroup$ Complex exponentiation doesn't work like real exponentiation. There are infinitely many values of $i^i$ (and two square roots of each.) $\endgroup$
    – saulspatz
    Mar 8 '19 at 15:36
  • $\begingroup$ This uses one of the values of $\log i$. There are many more, which will result in a different value $\endgroup$ Mar 8 '19 at 15:37
  • $\begingroup$ @RossMillikan Would adding $2\pi n, \ n\in \mathbb{Z}$ in the expression for $i$ generalize? $\endgroup$ Mar 8 '19 at 15:38
  • $\begingroup$ Yes, that's how to get them all. $\endgroup$
    – saulspatz
    Mar 8 '19 at 15:38
  • $\begingroup$ Thanks @saulspatz I'll edit my answer right away. $\endgroup$ Mar 8 '19 at 15:39

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