Square Roots of Imaginary Numbers, Using Algebra

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.

First example

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   $-2-i$.

Second example

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   $-\dfrac12 \sqrt{4+2\sqrt{13}} - \dfrac{3}{\sqrt{4+2\sqrt{13}}} i$.

Calculator

A Square Root Calculator is also available.  It gives the square roots of complex numbers in radical form, as discussed on this page.