# How do you find the square root of an imaginary number of the form a+bi?

Dec 24, 2015

Alternatively, solve without using trigonometry to find that the square roots of $a + b i$ are:

$\pm \left(\left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) + \left(\frac{b}{\left\mid b \right\mid} \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i\right)$

#### Explanation:

Suppose $a + b i = {\left(c + \mathrm{di}\right)}^{2}$

How do we solve for $c$ and $d$?

${\left(c + \mathrm{di}\right)}^{2} = {c}^{2} + 2 c \mathrm{di} + {d}^{2} {i}^{2} = \left({c}^{2} - {d}^{2}\right) + \left(2 c d\right) i$

So we want to solve:

${c}^{2} - {d}^{2} = a$

$2 c d = b$

From the second of these, we find:

$d = \frac{b}{2 c}$

So:

${d}^{2} = {b}^{2} / \left(4 {c}^{2}\right)$

Substituting this in the first equation we get:

${c}^{2} - {b}^{2} / \left(4 {c}^{2}\right) = a$

Multiply through by $4 {c}^{2}$ to get:

$4 {\left({c}^{2}\right)}^{2} - {b}^{2} = 4 a {c}^{2}$

Subtract $4 a {c}^{2}$ from both sides to get:

$4 {\left({c}^{2}\right)}^{2} - 4 a \left({c}^{2}\right) - {b}^{2} = 0$

From the quadratic formula, we find:

${c}^{2} = \frac{4 a \pm \sqrt{{\left(4 a\right)}^{2} + 16 {b}^{2}}}{8} = \frac{a \pm \sqrt{{a}^{2} + {b}^{2}}}{2}$

For $c$ to be Real valued we require ${c}^{2} \ge 0$, hence we need to choose the root with the $+$ sign...

${c}^{2} = \frac{a + \sqrt{{a}^{2} + {b}^{2}}}{2}$

Hence:

$c = \pm \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}$

Then:

$d = \pm \sqrt{{c}^{2} - a}$

$= \pm \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2} - a}$

$= \pm \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}$

The remaining question is: What signs do we need to choose?

Since $2 c d = b$:

If $b > 0$ then $c$ and $d$ must have the same signs.

If $b < 0$ then $c$ and $d$ must have opposite signs.

If $b = 0$ then $d = 0$, so we don't have to worry.

If $b \ne 0$ then we can use $\frac{b}{\left\mid b \right\mid}$ as a multiplier to match the signs as we require to find that the square roots of $a + b i$ are:

$\pm \left(\left(\sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} + a}{2}}\right) + \left(\frac{b}{\left\mid b \right\mid} \sqrt{\frac{\sqrt{{a}^{2} + {b}^{2}} - a}{2}}\right) i\right)$

Footnote

The question asked what is the square root of $a + b i$.

For positive Real numbers $x$, the principal square root of $x$ is the positive one, which is the one we mean when we write $\sqrt{x}$. This is also the square root that people commonly mean when they say "the square root of $2$" and suchlike, neglecting the fact that $2$ has two square roots.

For negative Real numbers $x$, then by convention and definition, $\sqrt{x} = i \sqrt{- x}$, That is, the principal square root is the one with a positive coefficient of $i$.

It becomes more complicated when we deal with the square roots of Complex numbers in general.

Consider:

$\sqrt{- \frac{1}{2} - \frac{\sqrt{3}}{2} i} = \pm \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right)$

Which sign do you prefer?