What is the square root of "$i$"?

Where $i$ is the square root of negative one. And is there a generalization of the $n$th root of $i$? Also how would this look graphically on the real number axis? Thanks

Solution 1:

If $\;w=a+bi\in\Bbb C\;$, then $\;z=x+yi\;$ fulfills $\;z^2=w\;$ iff

$$z^2=x^2-y^2+2xyi =a+bi\iff \begin{cases}x^2-y^2=a\\{}\\2xy=b\end{cases}$$

The above easy system has at least one solution and at most two different ones, and you'll have to be careful with the signs (pay attention in particular to the sign of $\;b\;$ .

Now, for your question only take $\;a=0\,,\,\,b=1\;$

Solution 2:

To answer your question directly, one choice of $n$th root is as follows:

A square root of $i$ is given by $$ z = \frac 1{\sqrt {2}} (1+i) $$ Graphically, this is a $45^\circ$ ray of length $1$ from the origin of the complex plane.

An $n$th root of $i$ is given by $$ z = \cos(\pi/(2n)) + i\sin(\pi/(2n)) $$ Graphically, this is a $(90/n)^\circ$ ray of length $1$ from the origin of the complex plane.

Solution 3:

If you learned the trigonometric form of complex numbers, the problem is very easy to solve that way.

Just solve $$[R(\cos(\theta)+i\cos(\theta))]^2=i$$

Otherwise, let $z=a+bi$. Then $$z^2=i \Leftrightarrow (a+bi)^2=i \Leftrightarrow a^2-b^2+2abi=i $$

This reduces to $$a^2=b^2$$ $$2ab=1$$

Which is easy to solve.