Is there a number that ${x^2=i}$ where ${i^2=-1}$ Imagine that there is a number ${x^2=i}$ where ${i^2=-1}$ where should it be as it is not in ${C}$ , and how to draw it , and where ?
 A: There is absolutely such an $x$ (two, in fact), and they both do lie in $\mathbb{C}$.
The nice thing about squares in complex numbers is that they work in a very predictable way: they double the argument and square the modulus.
The complex number $i$ has modulus $1$ and argument $\frac{\pi}{2}$. This suggests immediately that the number which has modulus $1$ and argument $\frac{\pi}{4}$ -- namely, $\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ -- is a square root of $i$, for instance.
Note that you should have expected this to have solutions in $\mathbb{C}$, as $\mathbb{C}$ is algebraically closed: that is, all polynomials with coefficients in $\mathbb{C}$ factor completely in $\mathbb{C}$.  Such a number $x$ is a root of $x^2-i=0$.
A: Recall that complex numbers can be written in polar form:
$$a+bi=\sqrt{a^{2}+b^{2}}e^{i\varphi}$$
Where $\varphi=\arg(a+bi)$, in this case we have $\sqrt{a^{2}+b^{2}}=1$ and $\varphi=\arg(i)=\frac{\pi}{2}$. We therefore have:
$$i=e^{\frac{i\pi}{2}}\implies x=\pm(e^{\frac{i\pi}{2}})^{\frac{1}{2}}=\pm e^{\frac{i\pi}{4}}=\pm\frac{1}{\sqrt{2}}(1+ i)$$
A: Since 
$$\Large i=e^{i\frac \pi 2}$$
then
$$\Large x=\pm e^{i\frac \pi 4}\Rightarrow x^2=i$$
A: $(1+i)^{2} = 1 + 2i - 1 = 2i$. Divide by $2$ to get $i$.
A: Lets, instead of $x^2=i$, we write $z^2=i$, where $z=u+iv$, here $u,v\in\mathbb{R}$.
$$z^2=(u+iv)^2=i$$
We know that, $(u+iv)^2=u^2+2ivu-v^2$
Therefore,
$$z^2=(u+iv)^2=u^2+2ivu-v^2=i$$ 
Solving 
$$u^2+2ivu-v^2=i$$
We will have $u^2-v^2$, the "real part", is must be zero, since in the right side of the equation there is no "real part".
And $2uv$, the "complex part" must be $1$, since in the right side of the equation there is $i$.
Therefore, the task is finding those $u,v$, and that we doing by solving the following system of equations:
$$\left\{\begin{matrix}
u^2-v^2=0\\ 
2uv=1
\end{matrix}\right.$$
Try to finish this off. 
A: Try squaring $(1 + i)/\sqrt{2}.$  I'll bet you can easily find another!
