Square Root of Complex Number¶

The square root of a real number $a\ge 0$ is a number that gives $a$ when multiplied with itself. For example, $\sqrt{4}=2$ because $2^2=4$. However, we also have $(-2{)}^2=4$, so it seems like $\sqrt{4}=-2$. To avoid getting $2=-2$, we require that $\sqrt{a}$ is not negative. Then all numbers that give $a$ when multiplied with themselves are $\pm \sqrt{a}$, which is why $\pm$ appears in front of square root so often.

Let's generalize this to complex numbers.

With angles and lengths¶

We can define the square root of any complex number $z$ by requiring $(\sqrt{z}{)}^2=z$, just like with real numbers. Because of how multiplication adds angles and multiplies lengths, this means that $$\begin{array}{rl}{\left|\sqrt{z}\right|}^2& =\left|z\right|,\\ 2(\text{angle of }\sqrt{z})& =\text{angle of }z.\end{array}$$ Solving for the length and angle of $\sqrt{z}$, we get $$\begin{array}{rl}\left|\sqrt{z}\right|& =\sqrt{\left|z\right|},\\ \text{angle of }\sqrt{z}& =\frac{\text{angle of }z}{2}.\end{array}$$ As usual, if $z=0$, the angle equation is not needed because the length formula gives $\sqrt{0}=0$. Any nonzero complex number $z\ne 0$ has infinitely many angles, differing from each other by $2\pi$, which can be written as $$\text{all angles of }z=\arg (z)+k2\pi ,$$ where $k$ is any integer. Then we have $$\text{angle of }\sqrt{z}=\frac{\arg (z)+k2\pi }{2}=\frac{\arg (z)}{2}+k\pi .$$ If $k$ is even, then $+k\pi$ adds some number of full turns to the angle and can be ignored. If $k$ is odd, then $+k\pi$ adds some number of full turns, and also adds a half-turn $\pi$ which means flipping to the opposite side of $0$. So, we get two possible values of $\sqrt{z}$, with equal lengths but on opposite sides of $0$. In the below picture, the blue dots are the possible values of $\sqrt{z}$.

The only thing left to do is to decide which of these should be called $\sqrt{z}$. Because of the flipping, the other one will then be $-\sqrt{z}$, and $\pm \sqrt{z}$ will appear commonly just like with real numbers.

Note that now we can put any complex number inside square root, including negative real numbers.

Examples:

• The length of $\pm \sqrt{-1}$ is $\sqrt{\left|-1\right|}=1$. Because the angle of $-1$ is 180 degrees, one of $\pm \sqrt{-1}$ has angle 90 degrees. So, one of $\pm \sqrt{-1}$ is $i$ and the other is $-i$. Because $i$ and $-i$ are both on the imaginary axis, with $i$ above $-i$, we have $\sqrt{-1}=i$.

  

• The length of $\pm \sqrt{4}$ is $2$. Because the angle of $4$ is $0$, one of $\pm \sqrt{4}$ has angle $0/2=0$, i.e. one of $\pm \sqrt{4}$ is a positive real number. The other one will then be a negative real number. So, $\pm \sqrt{4}$ are $2$ and $-2$. Because $2$ is on the right side of the imaginary axis, $\sqrt{4}=2$. This works with any positive number instead of $4$, so the square root we have defined is compatible with the square root of nonnegative real numbers.

The below picture shows how distinguisihing $\sqrt{z}$ and $-\sqrt{z}$ works. Blue complex numbers can be written as $\sqrt{z}$, with some complex number $z$, and red numbers are the corresponding $-\sqrt{z}$ numbers.

Using the convention $-\pi <\arg (z)\le \pi$, the distinguishing of $\pm \sqrt{z}$ becomes $$\arg (\sqrt{z})\in (-\frac{\pi }{2},\frac{\pi }{2}]\text{and}\arg (-\sqrt{z})\notin (-\frac{\pi }{2},\frac{\pi }{2}].$$ Now we get $$-\pi <2\arg (\sqrt{z})\le \pi .$$ Because angles get added when multiplying complex numbers, $2\arg (\sqrt{z})$ is an angle of $(\sqrt{z}{)}^2=z$. Another angle of $z$ is $\arg (z)$, for which we also have $-\pi <\arg (z)\le \pi$. Because only one angle of $z$ is between $-\pi$ and $\pi$, we must have $\arg (z)=2\arg (\sqrt{z})$.

Broken rules¶

The rules $$\sqrt{zw}=\sqrt{z}\sqrt{w}\text{and}\sqrt{\frac{z}{w}}=\frac{\sqrt{z}}{\sqrt{w}}$$ do not work for all complex numbers $z$ and $w$. To show that the first rule doesn't work, we can choose $z=w=-1$. Then we get $$\sqrt{-1}\sqrt{-1}=(\sqrt{-1}{)}^2=-1$$ and $$\sqrt{(-1)(-1)}=\sqrt{1}=1.$$ For the second rule, we can choose $z=1$ and $w=-1$. We get $$\sqrt{\frac{1}{-1}}=\sqrt{-1}=i$$ and $$\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{i}=\frac{-i}{i(-i)}=\frac{-i}{-(-1)}=-i.$$ From these examples, it seems like the rules might work with additional $\pm$ signs, as in $$\sqrt{zw}=\pm \sqrt{z}\sqrt{w}\text{and}\sqrt{\frac{z}{w}}=\pm \frac{\sqrt{z}}{\sqrt{w}}.$$ This is indeed true. To see why, notice that $$(\sqrt{z}\sqrt{w}{)}^2=(\sqrt{z}{)}^2(\sqrt{w}{)}^2=zw.$$ The only two numbers that give $zw$ when squared are $\sqrt{zw}$ and $-\sqrt{zw}$, and because $\sqrt{z}\sqrt{w}$ is one of them, we must have $$\sqrt{z}\sqrt{w}=\pm \sqrt{zw},$$ or written slightly differently, $$\sqrt{zw}=\pm \sqrt{z}\sqrt{w}.$$ We can similarly derive the rule for division.

For any positive real number $p$, we have $$\text{angle of }zp=\text{angle of }z,$$ and because the angle of square root depends only on the angle of the complex number inside, we get $$\text{angle of }\sqrt{zp}=\text{angle of }\sqrt{z}.$$ On the other hand, because $\sqrt{p}$ is also positive, we also have $$\text{angle of }\sqrt{z}\sqrt{p}=\text{angle of }\sqrt{z}.$$ By combining these results, we get $$\text{angle of }\sqrt{zp}=\text{angle of }\sqrt{z}\sqrt{p},$$ so the $\pm$ in $\sqrt{zp}=\pm \sqrt{z}\sqrt{p}$ must be $+$, and it actually works without $\pm$. That also works for $p=0$, because then both sides are zero.

Using this, we can for example calculate $$\sqrt{-5}=\sqrt{(-1)5}=\sqrt{-1}\sqrt{5}=i\sqrt{5}.$$ We also get $$\sqrt{\frac{z}{p}}=\sqrt{z\frac{1}{p}}=\sqrt{z}\sqrt{\frac{1}{p}}=\sqrt{z}\frac{\sqrt{1}}{\sqrt{p}}=\frac{\sqrt{z}}{\sqrt{p}}$$ for any positive $p$, where we used $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$ with positive real numbers $a=1$ and $b=p$. We obviously can't allow $p=0$.

Without angles and lengths¶

We find a formula for $\sqrt{a+bi}$ that doesn't require calculating the angle of $a+bi$. Let $x+yi=\sqrt{a+bi}$. Then we have $$a+bi=(x+yi{)}^2=x^2+2xyi+i^2{y}^2=x^2-y^2+2xyi.$$ By comparing real and imaginary parts, we get $$\{\begin{array}{rl}{x}^2-y^2& =a\\ 2xy& =b.\end{array}$$ We already know how to calculate the square root of a real number (both negative and nonnegative), so we can assume $b\ne 0$. From the second equation, we see that then $x$ and $y$ are nonzero too. So, we can divide by $x$ and we get $y=\frac{b}{2x}$ from the second equation. Let's plug that into the first equation: $$\begin{array}{rl}{x}^2-{\left(\frac{b}{2x}\right)}^2& =a\\ x^2-\frac{b^2}{4x^2}& =a\\ (x^2{)}^2-\frac{b^2}{4}& =a{x}^2\\ (x^2{)}^2-a{x}^2-\frac{b^2}{4}& =0\end{array}$$ By applying the quadratic formula (TODO) with $x^2$ instead of $x$, we get $$\begin{array}{rl}{x}^2& =\frac{-(-a)\pm \sqrt{(-a{)}^2-4(-\frac{b^2}{4})}}{2}\\ & =\frac{a\pm \sqrt{a^2+b^2}}{2}=\frac{a\pm r}{2},\end{array}$$ where $r=\sqrt{a^2+b^2}$ is the length of $a+bi$. Because $b$ is a nonzero real number, we have $b^2>0$ and so $$r=\sqrt{a^2+b^2}>\sqrt{a^2}=\left|a\right|.$$ This means that $a-r$ is negative, regardless of whether $a$ is positive, negative or zero. Because $x^2$ is not negative, we must choose $+$ and we get $$x^2=\frac{r+a}{2}.$$ Because the point corresponding to the square root of a complex number cannot be on the left side of the imaginary axis, we must have $x\ge 0$. So, we get $$x=\sqrt{\frac{r+a}{2}}.$$ We actually have $x>0$ because $2xy\ne 0$. Now $y=\frac{b}{2x}$ has the same sign as $b$; it's positive if $b$ is positive, and negative if $b$ is negative. To actually find $y$, we can use $x^2-y^2=a$, and we get $$y^2=x^2-a=\frac{r+a}{2}-a=\frac{r+a-2a}{2}=\frac{r-a}{2}.$$ So, we have $$y=\pm \sqrt{\frac{r-a}{2}},$$ where $\pm$ is $+$ when $b$ is positive and $-$ when $b$ is negative.

If $b=0$, then we have $r=\left|a\right|$ and the right side of the formula becomes $$\sqrt{\frac{\left|a\right|+a}{2}}\pm i\sqrt{\frac{\left|a\right|-a}{2}}.$$ If $a$ is positive or zero, this is $\sqrt{a}\pm 0$, and if $a$ is negative, this is $0\pm i\sqrt{\left|a\right|}$. So, if we choose $+$ whenever $b=0$, this result also works in that case.

Example: ${\sin }^2(\theta )$ and ${\cos }^2(\theta )$¶

Let $\theta$ be any angle, and let $z$ be a complex number on the origin-centered unit circle with angle $\theta$. Like in the derivation of de Moivre's formula, $z^2$ is also on the unit circle with angle $2\theta$, and we have $$\begin{array}{rl}z& =\cos (\theta )+i\sin (\theta ),\\ z^2& =\cos (2\theta )+i\sin (2\theta ).\end{array}$$

Because $z=\pm \sqrt{z^2}$, we have $$\cos (\theta )+i\sin (\theta )=\pm \sqrt{\cos (2\theta )+i\sin (2\theta )}.$$ Now we can use the formula for $\sqrt{a+bi}$ with $a=\cos (2\theta )$ and $b=\sin (2\theta )$. Because $z^2$ is on the unit circle, its length is $r=\sqrt{a^2+b^2}=1$, and we get $$\cos (\theta )+i\sin (\theta )=\pm (\sqrt{\frac{1+\cos (2\theta )}{2}}\pm i\sqrt{\frac{1-\cos (2\theta )}{2}}).$$ By comparing real and imaginary parts, we get $$\begin{array}{rl}\cos (\theta )& =\pm \sqrt{\frac{1+\cos (2\theta )}{2}}\\ \sin (\theta )& =\pm \sqrt{\frac{1-\cos (2\theta )}{2}},\end{array}$$ where we combined the two $\pm$ signs into another $\pm$ sign for the imaginary part of the right side. Finally, to get rid of all $\pm$ signs and square roots, we can square both sides to get $$\begin{array}{rl}{\cos }^2(\theta )& =\frac{1+\cos (2\theta )}{2}\\ {\sin }^2(\theta )& =\frac{1-\cos (2\theta )}{2}.\end{array}$$