# Find the square roots of the complex number. (Enter your answers as a comma-separated list. Round terms to four decimal places.) ?

## 5+5i

Apr 23, 2017

$\pm \left(\left(\sqrt{\frac{5 \sqrt{2} + 5}{2}}\right) + \left(\sqrt{\frac{5 \sqrt{2} - 5}{2}}\right) i\right)$

$\approx \pm \left(2.4567 + 1.0176 i\right)$

#### Explanation:

General method

Here's a non-trigonometric method for finding square roots of complex numbers...

Suppose we are given any complex number $a + b i$ and want to find its square roots in the form $x + y i$.

We want to solve:

$a + b i = {\left(x + y i\right)}^{2}$

$\textcolor{w h i t e}{a + b i} = \left({x}^{2} - {y}^{2}\right) + 2 x y i$

Equating real and imaginary parts, we have:

$\left\{\begin{matrix}{x}^{2} - {y}^{2} = a \\ 2 x y = b\end{matrix}\right.$

From the second equation, we find:

$y = \frac{b}{2 x}$

Substituting $\frac{b}{2 x}$ for $y$ in the first equation, we have:

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

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

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

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

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

$\textcolor{w h i t e}{0} = {\left(2 {x}^{2}\right)}^{2} - 2 \left(2 {x}^{2}\right) a + {a}^{2} - \left({a}^{2} + {b}^{2}\right)$

$\textcolor{w h i t e}{0} = {\left(2 {x}^{2} - a\right)}^{2} - {\left(\sqrt{{a}^{2} + {b}^{2}}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(2 {x}^{2} - a - \sqrt{{a}^{2} + {b}^{2}}\right) \left(2 {x}^{2} - a + \sqrt{{a}^{2} + {b}^{2}}\right)$

Hence:

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

So:

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

In order for $x$ to be real, we need the $+$ sign and hence:

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

Then:

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

Since $2 x y = b$, the sign we need for $y$ is the same as that for $x$ if $b > 0$ and the opposite of that for $x$ if $b < 0$.

So if $b \ne 0$ then the square roots of $a + b i$ are:

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

$\textcolor{w h i t e}{}$
Example

To find the square roots of $5 + 5 i$, put $a = 5$ and $b = 5$.

Then:

$\sqrt{{a}^{2} + {b}^{2}} = \sqrt{{5}^{2} + {5}^{2}} = 5 \sqrt{2}$

$\frac{b}{\left\mid b \right\mid} = \frac{5}{5} = 1$

So the square roots are:

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

$= \pm \left(\left(\sqrt{\frac{5 \sqrt{2} + 5}{2}}\right) + \left(\sqrt{\frac{5 \sqrt{2} - 5}{2}}\right) i\right)$

$\approx \pm \left(2.45673236 + 1.017611864 i\right)$

$\approx \pm \left(2.4567 + 1.0176 i\right)$