# Square Root of Complex Number in Cartesian Form

## Theorem

Let $z \in \C$ be a complex number.

Let $z = x + i y$ where $x, y \in \R$ are real numbers.

Let $z$ not be wholly real, that is, such that $y \ne 0$.

Then the square root of $z$ is given by:

$z^{1/2} = \pm \paren {a + i b}$

where:

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

## Proof

Let $a + i b \in z^{1/2}$.

Then:

 $\ds \paren {a + i b}^2$ $=$ $\ds x + i y$ Definition 4 of Square Root of Complex Number $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds a^2 + 2 i a b - b^2$ $=$ $\ds x + i y$ Square of Sum and $i^2 = -1$

Equating imaginary parts in $(1)$:

 $\ds 2 a b$ $=$ $\ds y$ $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds b$ $=$ $\ds \frac y {2 a}$ rearranging

Equating real parts in $(1)$:

 $\ds a^2 - b^2$ $=$ $\ds x$ $\ds \leadsto \ \$ $\ds a^2 - \paren {\frac y {2 a} }^2$ $=$ $\ds x$ substituting for $b$ from $(2)$ $\ds \leadsto \ \$ $\ds 4 a^4 - 4 a^2 x - y^2$ $=$ $\ds 0$ multiplying by $4 a^2$ and rearranging $\ds \leadsto \ \$ $\ds a^2$ $=$ $\ds \frac {4 x \pm \sqrt {16 x^2 + 16 y^2} } 8$ Quadratic Formula $\ds \leadsto \ \$ $\ds a^2$ $=$ $\ds \frac {x \pm \sqrt {x^2 + y^2} } 2$ dividing top and bottom by $4$ $\text {(3)}: \quad$ $\ds \leadsto \ \$ $\ds a$ $=$ $\ds \pm \sqrt {\frac {x + \sqrt {x^2 + y^2} } 2}$ taking the square root

Note that in $(3)$, only the positive square root of the discriminant $x^2 + y^2$ is used.

This is because the negative square root of $x^2 + y^2$ would yield $\dfrac {x - \sqrt {x^2 + y^2} } 2 < 0$.

As $a \in \R$, it is necessary that $\dfrac {x + \sqrt {x^2 + y^2} } 2 > 0$.

Hence $\sqrt {x^2 + y^2} > 0$.

Then:

 $\ds b$ $=$ $\ds \frac y {2 a}$ from $(2)$ $\ds$ $=$ $\ds \frac y {2 \paren {\pm \sqrt {\dfrac {x + \sqrt {x^2 + y^2} } 2} } }$ $\ds \leadsto \ \$ $\ds b^2$ $=$ $\ds \frac {y^2} {2 \paren {x + \sqrt {x^2 + y^2} } }$ $\ds \leadsto \ \$ $\ds b^2$ $=$ $\ds \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x + \sqrt {x^2 + y^2} } \paren {x - \sqrt {x^2 + y^2} } }$ multiplying top and bottom by $x - \sqrt {x^2 + y^2}$ $\ds \leadsto \ \$ $\ds b^2$ $=$ $\ds \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x^2 - \paren {x^2 + y^2} } }$ Difference of Two Squares $\ds \leadsto \ \$ $\ds b^2$ $=$ $\ds \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {- 2 y^2}$ $\ds \leadsto \ \$ $\ds b^2$ $=$ $\ds \frac {-x + \sqrt {x^2 + y^2} } 2$

But from $(2)$ we have:

$b = \dfrac y {2 a}$

and so having picked either the positive square root or negative square root of either $a^2$ or $b^2$, the root of the other is forced.

So:

if $y > 0$, then $a$ and $b$ are both of the same sign.

Thus:

$b = 1 \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$

if $y < 0$, then $a$ and $b$ are of opposite sign.

Thus:

$b = \paren {-1} \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$

Hence:

$b = \dfrac y {\cmod y} \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$

$\blacksquare$

## Examples

### Example: $i$

$\sqrt i = \pm \left({\dfrac {\sqrt 2} 2 + \dfrac {\sqrt 2} 2 i}\right)$

### Example: $2 + 2 i$

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

### Example: $3 + 4 i$

$\sqrt {3 + 4 i} = \pm \left({2 + i}\right)$

### Example: $-8 + 6 i$

$\sqrt {-8 + 6 i} = \pm \paren {1 + 3 i}$

### Example: $5 - 12 i$

$\sqrt {5 - 12 i} = \pm \paren {3 - 2 i}$