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.
[Math] Proof for complex numbers and square root
complex numbersproof-writing
Related Solutions
Most of the time complex numbers are introduced using $z=a+i b$ along with the basic addition rule and some more complicated multiplication rules. But really, it is much more natural to think of them in polar coordinates when doing multiplication; especially if you keep them in the form $z=r e^{i \phi}$.
You're right to start with the multiplicative identity $1 \cdot z=z$. If we look at the unit circle, $1=1+i(0)$, and look what is happening when we multiply by $z$ we see two things: the radius changes from $1\rightarrow r$, and the angle rotates from $0\rightarrow \phi$.
Therefore, to find inverse we need to find some $z'$ such that $1 \cdot z \cdot z'=1$. This $z'$ operation will also be a scaling and a rotation operation, but in this case, what angle should you rotate by, and what value should you scale by, to return to 1?
Once this concept clicks, you can really see that $z=r e^{i \phi}$ is the natural choice for complex multiplication, and makes it almost as easy as $z=a+ib$ for complex addition.
Hint: irrespective of $|z|$, you will get $\tan(\theta)=\frac{4}{3}$. So, taking the inverse will give you $\theta$ in radians.
Best Answer
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.
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 $\pi$).