So we actually found $\sqrt{i}$, the square root of i. In fact, the number $i$ had two square roots, $\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} i$ and $-\dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2} i$. So can we find square roots of other imaginary numbers? Sure.
Let's try to find the square roots of $3+4i$. Since
\begin{equation*} (a+bi)^2 = a^2+2abi+b^2 i^2 = (a^2-b^2)+(2ab)i \end{equation*}we need to solve the system of equations
$ \left\{ \begin{align} a^2-b^2 &=3 \\ 2ab &=4 \end{align} \right. $ |
We can't solve this system as quickly as we did when we found the square root of $i$, but it still can be done. Solving the second equation for the variable $b$, we get $b=\dfrac2a$. Substituting this quantity into the first equation, we get $a^2-\dfrac{4}{a^2}=3$. Clearing the fractions gives $a^4-3a^2-4=0$. This can be solved by factoring, so $a^2=4$ or $a^2=-1$. But we want real values for the variable $a$, so we discard $-1$. Therefore, $a^2=4$, and the values of $a$ are $2$ and $-2$. Using $b=\dfrac2a$, we find $b=\pm 1$. So the two square roots of $3+4i$ are $2+i$ and $-2-i$.
Let's try to find the square roots of $2+3i$. Once again, since
\begin{equation*} (a+bi)^2 = a^2+2abi+b^2 i^2 = (a^2-b^2)+(2ab)i \end{equation*}we need to solve the system of equations
$ \left\{ \begin{align} a^2-b^2 &=2 \\ 2ab &=3 \end{align} \right. $ |
Solving the second equation for the variable $b$, we get $b=\dfrac{3}{2a}$. Substituting this quantity into the first equation, we get $a^2 - \dfrac{9}{4a^2} = 2$. Clearing the fractions gives $4a^4-8a^2-9=0$. This can be solved using the quadratic formula, and we get $a^2 = 1 \pm \dfrac{\sqrt{13}}{2}$. But we want real values for the variable $a$, so we discard the negative sign. Therefore, we get $a = \pm \sqrt{1+\dfrac{\sqrt{13}}{2}} = \pm \dfrac12 \sqrt{4+2\sqrt{13}}$. (Messy, but true!) Using $b=\dfrac{3}{2a}$, we find $b= \pm \dfrac{3}{\sqrt{4+2\sqrt{13}}}$. So the two square roots of $2+3i$ are $\dfrac12 \sqrt{4+2\sqrt{13}} + \dfrac{3}{\sqrt{4+2\sqrt{13}}} i$ and $ -\dfrac12 \sqrt{4+2\sqrt{13}} - \dfrac{3}{\sqrt{4+2\sqrt{13}}} i$.
A Square Root Calculator is also available. It gives the square roots of complex numbers in radical form, as discussed on this page.
There is another way to find roots, using trigonometry. You can read more about this relationship in Imaginary Numbers and Trigonometry.