Show that $\sqrt[i]{i}$ is a real number How can one show that $\sqrt[i]{i}$ is a real number; $\sqrt[i]{i}\in\mathbb{R}$. 
I know for the number to be real given that a calculator produces a value of $4.810477...$ . 
I thought it might be smart to relate the problem to the complex plane, given that the contains the imaginary unit $i$. 
But I do not where to start and how to show this algebraically. Does someone know how to solve this problem?  
 A: By the definition of the complex logarithm, main branch,
$$\log(i)=\log|i|+i\angle i=i\frac{\pi}2.$$
Then $e^{\log(i)/i}=e^{\pi/2}=4.8104773809653516554730356667038\cdots$
With other branches, you will get
$$e^{\pi/2+2k\pi}.$$
For instance 
$$0.00898329102112942788966495190793\cdots$$
A: Let us go through this step by step:


*

*write the number in its radical-form, $\sqrt[b]{a}=a^{\frac{1}{b}}$:


$$\sqrt[i]{i}=i^{\frac{1}{i}}$$


*write $\frac{1}{i}=\frac{i}{i}\cdot\frac{1}{i}=\frac{i}{i^{2}}=\frac{i}{-1}=-i$:


$$\sqrt[i]{i}=i^{-i}$$


*make use of the logarithm properties, $a=e^{\ln(a)}$ and $\ln(a^{b})=b\cdot \ln(a)$:


$$\sqrt[i]{i}=e^{\ln(i^{-i})}=e^{-i\cdot\ln(i)}$$


*complex numbers can be written in Cartesian notation and polar notation, $x+i\cdot y=r\cdot e^{i\cdot \phi}$. 


*

*Let $i$ be a point on the complex plane. It has a real component of $x=0$ and an imaginary component of $y=1$. 

*This means that the point $i$ is at a distance of $r=\sqrt{|x|^{2}+|y|^{2}}=\sqrt{|0|^{2}+|1|^{2}}=1$ from the origin of the plane.

*The point $i$ makes an angle of $\phi=\frac{\pi}{2}$ rad with the real axis. Ofcourse, you could account for multiple whole rotations about the origin of the plane but let us just consider this principle angle.

*It follows that $i=0+i\cdot 1=1\cdot e^{i\cdot \frac{\pi}{2}}$:



$$\sqrt[i]{i}=e^{-i\cdot\ln(e^{i\cdot \frac{\pi}{2}})}=e^{-i^2\cdot\frac{\pi}{2}}=e^{\frac{\pi}{2}}$$
Recognize that
$$\sqrt[i]{i}=e^{\frac{\pi}{2}}\in\mathbb{R}$$
A: It equals $e^{i(\dfrac{\pi}{2}+2k\pi)(\dfrac{1}{i})}=e^{\dfrac{\pi}{2}+2k\pi}$ which is real. (for $k\in N$)
