Square Roots of Complex Numbers

# Square Roots of Complex Numbers

Consider the set of real numbers, $\mathbb{R}$. If $x \in \mathbb{R}$, then there exists a $w \in \mathbb{R}$ with $x = w^2$ if and only if $x \geq 0$. For example, if $x = 4$, then $w = 2$ (or $w = -2$) satisfies this equation. However, if $x = -2$, then we must have that $w = \sqrt{-2} \not \in \mathbb{R}$ to satisfying this equation. In other words, for negative real numbers $x$ that may not exist another *real* number whose square is equal to $x$.

What's nice about the set of complex numbers is that this problem disappears as we will prove in the following theorem.

Theorem 1: If $z \in \mathbb{C}$ then there exists a $w \in \mathbb{C}$ such that $z = w^2$. |

**Proof:**Let $z = a + bi$ and let $w = c + di$. Now we want to find $c$ and $d$ such that:

\begin{align} \quad z = a + bi = (c + di)^2 = [c^2 - d^2] + 2cdi = w^2 \end{align}

- Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal, and so, we must have that:

\begin{align} \quad a = c^2 - d^2 \end{align}

(3)
\begin{align} \quad b = 2cd \end{align}

- Now notice that $(c^2 + d^2)^2 = c^4 + 2c^2d^2 + d^4$ and $(c^2 - d^2)^2 = c^4 - 2c^2d^2 + d^4$. Therefore:

\begin{align} \quad (c^2 + d^2)^2 &= (c^2 - d^2)^2 + 4c^2d^2 \\ \quad &= (\underbrace{c^2 - d^2}_{a})^2 + (\underbrace{2cd}_{b})^2 \\ \quad &= a^2 + b^2 \end{align}

- Therefore, since $c^2 + d^2 \geq 0$ we have that:

\begin{align} \quad c^2 + (c^2 - a) &= \sqrt{a^2 + b^2} \\ \quad 2c^2 &= \sqrt{a^2 + b^2} + a \\ \quad c^2 &= \frac{\sqrt{a^2 + b^2} + a}{2} \\ \quad c &= \pm \sqrt{ \frac{ \sqrt{a^2 + b^2} + a}{2}} \end{align}

- And similarly:

\begin{align} \quad c^2 + d^2 &= \sqrt{a^2 + b^2} \\ \quad (a + d^2) + d^2 &= \sqrt{a^2 + b^2} \\ \quad 2d^2 &= \sqrt{a^2 + b^2} - a \\ \quad d^2 &= \frac{\sqrt{a^2 + b^2} - a}{2} \\ \quad d &= \pm \sqrt{ \frac{ \sqrt{a^2 + b^2} - a}{2}} \end{align}

- Let $\alpha$ and $\beta$ be defined as:

\begin{align} \quad \alpha = \sqrt{ \frac{ \sqrt{a^2 + b^2} + a}{2}} \quad \mathrm{and} \quad \beta = \sqrt{ \frac{ \sqrt{a^2 + b^2} - a}{2}} \end{align}

- Notice that $\alpha, \beta \in \mathbb{R}$, so such a $w \in \mathbb{C}$ exists with $z = w^2$ and:

\begin{align} \quad \sqrt{z} = w = \pm (\alpha \pm \beta i) \quad \blacksquare \end{align}

- Notice that since $b = 2cd$ so either $b > 0$ or $b < 0$ (in the case where $b = 0$, $z = a$, so $w = \sqrt{a} \in \mathbb{C}$ is the squareroot of $x = a + 0i$ and is rather uninteresting.

**Case 1:**If $b > 0$ then this implies that $c, d > 0$ or $c, d < 0$, and so $w = \alpha + \beta i$ or $w = -\alpha - \beta i$.

**Case 2:**If $b < 0$ then this implie that $c > 0$ and $d < 0$, in which case $w = \alpha - \beta i$, or, $c < 0$ and $d > 0$ in which case $w = -\alpha + \beta i$. $\blacksquare$

Corollary 1: Let $z = a + bi \in \mathbb{C}$. Then:a) The square roots $\pm \sqrt{z}$ are real numbers if and only if $z$ is a real number and $z > 0$.b) The square roots $\pm \sqrt{z}$ are imaginary numbers if and only if $z$ is a real number and $z < 0$.c) The square roots $\pm \sqrt{z}$ are equal if and only if $z = 0$. |

*We will only prove part (a). The other parts are similar.*

**Proof of a):**$\Rightarrow$ Suppose that $\pm \sqrt{z} \in \mathbb{R}$. Then from Theorem 1, $\beta = 0$, $\sqrt{z} = \alpha$, $\sqrt{z} = - \alpha$, and $\alpha \in \mathbb{R}$ so $\pm \sqrt{z} \in \mathbb{R}$.

- $\Leftarrow$ Let $z \in \mathbb{R}$ and suppose that $z > 0$. Then $\pm \sqrt{z}$ are obviously well-defined real numbers. $\blacksquare$