I am solving complex variable and I have solve to one problem that $\sqrt{i}^{\sqrt{i}}$. If I had $i^i$ then it become $e^{i \log i}$ and: $$i=\cos(\pi/2)+i\sin(\pi/2)\implies i=e^{\pi/2}$$ so $\log i=2ni\pi+\log e^{\pi/2}$ it become $i(4n+1)\pi/2$ which shows $e^{i(i(4n+1)\pi/2}= e^{-(4n+1)\pi/2}$ therefore $e^{i \log i}=e^{-(4n+1)\pi/2}$.

I tried stack exchange for the first time and this is my first question please help me out. I tried my best to explain.

question: show that $\sqrt{i}^{\sqrt{i}}$= $e^{-\pi/4\sqrt{2}}(\cos π/4(√2)+i \sin π/4(√2) )$

  • $\begingroup$ @matt:-how you edit it so correctly.i am new to SE i am curious to know.e^ilogi correctly.where we get pie symbol. $\endgroup$ – Ankur Saxena Jan 10 '13 at 11:15
  • $\begingroup$ If you click edit, you can see how I got the formulae to display correctly. It's using syntax called LaTeX, which you may already be familiar with. There is a guide here: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – mdp Jan 10 '13 at 11:19
  • $\begingroup$ @Matt:-thank sir $\endgroup$ – Ankur Saxena Jan 10 '13 at 11:21
  • 1
    $\begingroup$ This expression is not well defined. There is no unique definition of $x^y$ for $x,y\in\Bbb C$, nor for that matter of $\sqrt i$. You can force a unique value (among infinitely many possibilities) by choosing a branch of the logarithm, but there is no reason that value so obtained is any better than any other. If you get this expression as giving the solution of a genuine problem (which is unlikely), it would be doubtful that the value computed would solve the problem. $\endgroup$ – Marc van Leeuwen Jan 10 '13 at 17:24

$$\sqrt{i}^{\sqrt{i}} = e^{i \frac{\pi}{4} e^{i \frac{\pi}{4}}} $$

$$ = e^{i \frac{\pi}{4} \left [ \cos{\left ( \frac{\pi}{4} \right )} + i \sin{\left ( \frac{\pi}{4} \right )} \right ] } $$

$$ = e^{i \frac{\pi}{4} \left ( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right )} $$

$$ = e^{-\frac{\sqrt{2} \pi}{8}} \left [ \cos{\frac{\sqrt{2} \pi}{8}} + i \sin{\frac{\sqrt{2} \pi}{8}} \right ] $$

  • $\begingroup$ thank sir but i show correct answer it is much closer than that correct one. $\endgroup$ – Ankur Saxena Jan 10 '13 at 11:43
  • $\begingroup$ This answer is the solution shown above. I just rationalized the denominator, i.e. $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. $\endgroup$ – Ron Gordon Jan 10 '13 at 11:50
  • $\begingroup$ i got it but what you have done in 1st and 2nd step sir.please explain me i trying it & i have to solve this question. $\endgroup$ – Ankur Saxena Jan 10 '13 at 11:54
  • $\begingroup$ Not sure how it could be more clear. Sorry. $\endgroup$ – Ron Gordon Jan 10 '13 at 11:57
  • $\begingroup$ okey thank alot sir....... $\endgroup$ – Ankur Saxena Jan 10 '13 at 11:59

Try substituting e.g. $\sqrt i = \pm \left( \dfrac{1 + i}{\sqrt 2} \right)$.

  • $\begingroup$ This answer was very simple, but very spot on. All that you needed to know was $2i = (1 + i)^2$. $\endgroup$ – George N. Missailidis Aug 2 '17 at 21:29

Surely, you can find $a+ib$ such that for a given $t$ we have $t=\sqrt{a+ib}$. So we don't have any difficulties to solve $t=\sqrt{i}$ for a given $t$. Now take $x=\sqrt{i}^{\sqrt{i}}$ so $$x^2=i^{\sqrt{i}}=i^t$$ where $t=a+ib$ for some $a,b$. Therefore $$x^2=i^{\sqrt{i}}=i^t=i^{a+ib}=i^a\cdot (i^i)^b$$ and you already noted you could evaluate $i^i$, so find $x$.

  • $\begingroup$ Nice strategy to use. +1 $\endgroup$ – Namaste Feb 19 '13 at 0:20

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