Proof for complex numbers and square root

Question

Use the polar form of complex numbers to show that every complex number $z\neq0$ has two square roots.

I know the polar form is $z=r(\cos(\alpha)+i \sin(\alpha))$. I'm just not sure how to do this one.

Answer 1

De Moivre's formula is $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for any integer $n$. Therefore, if $w = s(\cos\theta + i\sin\theta)$ is a non-zero complex number such that $w^2 = z$, then

$$s^2(\cos(2\theta) + i\sin(2\theta)) = r(\cos\alpha + i\sin\alpha).$$

Therefore $s^2 = r$, so $s = \pm\sqrt{r}$, but $s > 0$, so $s = \sqrt{r}$.

Furthermore, $\cos(2\theta) = \cos\alpha$, so $2\theta = \alpha + 2k\pi$ where $k \in \mathbb{Z}$, or $2\theta = -\alpha + 2l\pi$ where $l \in \mathbb{Z}$.

Likewise, $\sin(2\theta) = \sin\alpha$, so $2\theta = \alpha + 2m\pi$ where $m \in \mathbb{Z}$, or $2\theta = \pi - \alpha + 2n\pi$ where $n \in \mathbb{Z}$.

The only possibility for $2\theta$ that will give both $\cos(2\theta) = \cos\alpha$ and $\sin(2\theta) = \sin\alpha$ is $2\theta = \alpha + 2k\pi$ for some $k \in \mathbb{Z}$. Therefore $\theta = \frac{\alpha}{2} + k\pi$ where there are no restrictions on $k$, other than being an integer. So the complete set of solutions to $w^2 = z$ is

$$\left\{\sqrt{r}\left(\cos\left(\frac{\alpha}{2}+k\pi\right) + i\sin\left(\frac{\alpha}{2} + k\pi\right)\right) \mid k \in \mathbb{Z}\right\}.$$

However, as $\cos$ and $\sin$ are $2\pi$-periodic, there are many different values of $k$ which gives the same complex number. If $k$ is even, then $k = 2t$ for some $t \in \mathbb{Z}$, so

\begin{align*} \sqrt{r}\left(\cos\left(\frac{\alpha}{2}+k\pi\right) + i\sin\left(\frac{\alpha}{2} + k\pi\right)\right) &= \sqrt{r}\left(\cos\left(\frac{\alpha}{2}+2t\pi\right) + i\sin\left(\frac{\alpha}{2} + 2t\pi\right)\right)\\ &= \sqrt{r}\left(\cos\left(\frac{\alpha}{2}\right) + i\sin\left(\frac{\alpha}{2}\right)\right). \end{align*}

If $k$ is odd, then $k = 2t+1$ for some $t \in \mathbb{Z}$, so

\begin{align*} \sqrt{r}\left(\cos\left(\frac{\alpha}{2}+k\pi\right) + i\sin\left(\frac{\alpha}{2} + k\pi\right)\right) &= \sqrt{r}\left(\cos\left(\frac{\alpha}{2}+(2t+1)\pi\right) + i\sin\left(\frac{\alpha}{2} + (2t+1)\pi\right)\right)\\ &= \sqrt{r}\left(\cos\left(\frac{\alpha}{2}+\pi+2t\pi\right) + i\sin\left(\frac{\alpha}{2} + \pi 2t\pi\right)\right)\\ &= \sqrt{r}\left(\cos\left(\frac{\alpha}{2} + \pi\right) + i\sin\left(\frac{\alpha}{2}+\pi\right)\right). \end{align*}

Using the fact that $\cos(\beta + \pi) = -\cos\beta$ and $\sin(\beta + \pi) = -\sin\beta$ we can simplify this further:

$$\sqrt{r}\left(\cos\left(\frac{\alpha}{2} + \pi\right) + i\sin\left(\frac{\alpha}{2}+\pi\right)\right) = -\sqrt{r}\left(\cos\left(\frac{\alpha}{2}\right) + i\sin\left(\frac{\alpha}{2}\right)\right).$$

Therefore, the two solutions to $w^2 = z$ are $\pm\sqrt{r}\left(\cos\left(\frac{\alpha}{2}\right) + i\sin\left(\frac{\alpha}{2}\right)\right)$; note that these are distinct, so there are actually two different solutions.

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In general, if you were trying to solve $w^n = z$ with $z \neq 0$, you could use the same method as above. When it came to determine which $k$ gave different complex numbers, you'd consider $k$ modulo $n$, i.e. write $k = nt + j$ where $j = 0, 1, \dots, n - 1$. You'd then find $n$ distinct solutions corresponding to the different values of $j$. Usually, you would have to stop here, as there won't be a corresponding trigonometric formula to simplify the complex numbers (as there was above for the phase shift $\pi$).

Answer 2

Given $z\in \mathbb{C}\setminus\{0\}$, let $u \in \mathbb{C}$ such that: $$ u^2=z. $$ If one sets $$ z=r(\cos\alpha+i\sin\alpha), $$ then $$ u^2=z \iff u=\pm\sqrt{r}(\cos\frac{\alpha}{2}+i\sin\frac{\alpha}{2}). $$

Answer 3

Your question under "Mercy" 's answer makes me wonder if you know how to multiply complex numbers in polar form. Without that, there's not much point in mentioning polar form, as far as I know. Here it is:

$$ \Big( r(\cos\alpha+i\sin\alpha)\Big)\Big(s(\cos\beta+i\sin\beta)\Big) = rs(\cos(\alpha+\beta)+i\sin(\alpha+\beta)). $$

You add the angles and multiply the absolute values. So $$ \Big( s(\cos\beta+i\sin\beta) \Big)^2 = s^2(\cos(2\beta)+i\sin(2\beta)). $$ This is equal to $r(\cos\alpha+i\sin\alpha)$ if and only if $s^2=r$ and $2\beta=\alpha$.

So $\beta$ is half of $\alpha$. But in a sense two different angles are half of $\alpha$: For example, half of $30^\circ$ is either $15^\circ$ or $195^\circ$, which is the same as $-165^\circ$. They're $180^\circ$ apart from each other.

So you can use either of those two half-angles and get a square root. Since they're $180^\circ$ away from each other, one square root is minus the other.