Physicist: The beauty of complex numbers (numbers that involve 
) is that the answer to this question is a surprisingly resounding: nopers.
The one thing that needs to be known about is that, by definition, 
. Other than that it behaves like any other number or variable. It turns out that the square root is 
. You can check this the same way that you can check that 2 is the square root of 4: you square it.
![\begin{array}{ll}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^2\\[2mm]=\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\\[2mm]=\frac{1}{\sqrt{2}}\left(1+i\right)\frac{1}{\sqrt{2}}\left(1+i\right)\\[2mm]=\frac{1}{2}\left(1+i\right)\left(1+i\right)\\[2mm]=\frac{1}{2}\left(1+i+i+i^2\right)\\[2mm]=\frac{1}{2}\left(1+i+i-1\right)\\[2mm]=\frac{1}{2}\left(2i\right)\\=i\end{array}](https://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7D%5Cleft%28%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%2B%5Cfrac%7Bi%7D%7B%5Csqrt%7B2%7D%7D%5Cright%29%5E2%5C%5C%5B2mm%5D%3D%5Cleft%28%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%2B%5Cfrac%7Bi%7D%7B%5Csqrt%7B2%7D%7D%5Cright%29%5Cleft%28%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%2B%5Cfrac%7Bi%7D%7B%5Csqrt%7B2%7D%7D%5Cright%29%5C%5C%5B2mm%5D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5Cleft%281%2Bi%5Cright%29%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5Cleft%281%2Bi%5Cright%29%5C%5C%5B2mm%5D%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%281%2Bi%5Cright%29%5Cleft%281%2Bi%5Cright%29%5C%5C%5B2mm%5D%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%281%2Bi%2Bi%2Bi%5E2%5Cright%29%5C%5C%5B2mm%5D%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%281%2Bi%2Bi-1%5Cright%29%5C%5C%5B2mm%5D%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%282i%5Cright%29%5C%5C%3Di%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\begin{array}{ll}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^2\\[2mm]=\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\\[2mm]=\frac{1}{\sqrt{2}}\left(1+i\right)\frac{1}{\sqrt{2}}\left(1+i\right)\\[2mm]=\frac{1}{2}\left(1+i\right)\left(1+i\right)\\[2mm]=\frac{1}{2}\left(1+i+i+i^2\right)\\[2mm]=\frac{1}{2}\left(1+i+i-1\right)\\[2mm]=\frac{1}{2}\left(2i\right)\\=i\end{array}")
And like any other square root, the negative, 
, is also a solution. So, does have a square root, and it’s not even that hard to find it. No new “super-imaginary” numbers need to be invented.
This isn’t a coincidence. The complex numbers are “algebraically closed“, which means that no matter how weird a polynomial is, it’s roots are always complex numbers. The square roots of , for example, are the solutions of the polynomial 
. So, any cube root, any Nth root, any power, any combination, any whatever of any complex number: still a complex number.
That hasn’t stopped mathematicians from inventing new and terrible number systems. They just didn’t need to in this case.
17 Responses to Q: “i” had to be made up to solve the square root of negative one. But doesn’t something new need to be made up for the square root of i?