## Square Root of i

Do you remember the imaginary number $i$, which stands for $\sqrt{-1}$ ?  We couldn't describe $\sqrt{-1}$ with a real number, since the square of a positive number is positive, and the square of a negative number is positive.  There is no real number whose square is negative.  The square root of a real number is not always a real number.  It turns out that $\sqrt{-1}$ is a rather curious number, which you can read about in Imaginary Numbers.

But have you ever thought about $\sqrt{i}$ ?  Won't we need a $j$, or some other invention to describe it?  Actually, no.  It turns out that $\sqrt{i}$ is another complex number.  We won't need a $j$.

When we first encountered the number $i$, we also learned about complex numbers, or numbers of the form $a+bi$.  It turns out that square roots of complex numbers are always other complex numbers.

Consider   $2+3i$   for the moment.  We can square that number:

\begin{equation*} (2+3i)^2 = (2+3i)(2+3i) = 4+6i+6i+9i^2 = 4+12i-9 = -5+12i \end{equation*}

Therefore, the square root of   $-5+12i$   is   $2+3i$.   So now we have demonstrated one case where the square root of a complex number is another complex number.

### An algebraic derivation

So let's assume that there is a number   $a+bi$   which represents $\sqrt{i}$.  Since

\begin{equation*} (a+bi)^2 = a^2+2abi+b^2 i^2 = (a^2-b^2)+(2ab)i \end{equation*}

and since this result should equal the number $i$, we obtain the following system of equations:

 \left\{ \begin{align} a^2-b^2 &=0 \\ 2ab &=1 \end{align} \right.

When the first equation is solved for the variable $a$, we learn that the variables $a$ and $b$ are either equal or opposites.  From the second equation, we know that the product is a positive one-half, so the variables $a$ and $b$ must be equal.  (If they were opposite, the product would have been negative.)  Therefore, the second equation becomes   $2a^2=1$,   whose solution is   $a = \pm \dfrac{\sqrt{2}}{2}$.

Therefore, the number $i$ has two square roots (just like positive numbers do).  They are   $\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} i$   and   $-\dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2} i$.   (You can check them both.  They both work!)

### Epilogue

Once you become accustomed to the idea that $i$ is just another number, then it becomes easy to accept the idea that we could use the number $i$, or any other imaginary number, in (almost) any operation that we could use a real number.  You can read more about finding Square Roots of Imaginary Numbers, Using Algebra.  Or read more about finding all manner of roots using the relationship between Imaginary Numbers and Trigonometry.