I have an issue when wanting to convert $z=-1$ to the polar form.
Indeed, assuming that $z$ is of the form $z=a+ib$, I start by finding $z=(-1)+(0)i$, thus $a=-1$ and $b=0$.
I know that the polar form of $z$ is $z=re^{i\theta}$, where $r=\sqrt{a^2+b^2}$, thus $r=\sqrt{(-1)^2}=1$.
Then, I need to find $\theta$. I know that $\theta=\operatorname{arg}(z)=\arctan(\frac ba)=\arctan(\frac {0}{-1})=0$.
Therefore, my final answer would be $z=1\times e^{i0}=1$, which is incorrect, as WolframAlpha tells me the correct answer is $z=e^{i\pi}=-1$. I understand that I messed up when finding $\theta=0$ instead of $\theta=\pi$.
Any help is appreciated.
Thank you
Your attempt is good, but your formula for the argument $\theta=\operatorname{arg}(z)=\arctan\left(\frac{b}{a}\right)=\arctan\left(\frac {0}{-1}\right)=0$ only works in special cases (when $a > 0$). Actually the formulas are: $$ \begin{align*} z &= a + b \cdot \mathrm{i}\\ z &= r \cdot \operatorname{cis}(\theta) = r \cdot e^{\theta \cdot \mathrm{i}}\\ \\ r &= |z| = \sqrt{a^{2} + b^{2}}\\ \theta &= \arg(z) = \operatorname{arctan2}\left( b, ~a \right)\\ \\&\text{with}\\\\ \operatorname{arctan2}\left( b, ~a \right) &= \begin{cases} \arctan\left({\frac {b}{a}}\right) \qquad\quad~~~ \text{for } a > 0\\ \arctan\left({\frac {b}{a}}\right) + \pi \qquad \text{for } a < 0 \quad b > 0\\ \pi \qquad\qquad\qquad\quad~~~ \text{for } a < 0 \quad b = 0\\ \arctan\left({\frac {b}{a}}\right) - \pi \qquad \text{for } a < 0 \quad b < 0\\ \frac{\pi}{2} \qquad\qquad\qquad\quad~~ \text{for } a = 0 \quad b > 0\\ -\frac{\pi}{2} \qquad\qquad\qquad~~~ \text{for } a = 0 \quad b < 0\\ \end{cases} + 2 \cdot k \cdot \pi \end{align*} $$
If you use that you'll get: $$ \begin{align*} z &= -1\\ z &= -1 + 0 \cdot \mathrm{i}\\ z &= \sqrt{\left( -1 \right)^{2} + 0^{2}} \cdot e^{\arctan2\left( 0, ~-1 \right) \cdot \mathrm{i}}\\ z &= 1 \cdot e^{(\pi + 2 \cdot k \cdot \pi) \cdot \mathrm{i}}\\ z &= e^{\left( \pi + 2 \cdot k \cdot \pi \right) \cdot \mathrm{i}}\\ z &= e^{\pi \cdot \mathrm{i}} \tag{if k = 0}\\ \end{align*} $$
The formula $e^{\pi \cdot \mathrm{i}} = -1$ is also known as Euler's formula, named after Leonhard Euler.
However, if you don't want to calculate the angle $\theta$, you can also consider the complex number $z = a + b \cdot \mathrm{i} = x + y \cdot \mathrm{i}$ as a vector (versor) with $z ~\widehat{=} \left[\begin{matrix} a\\ b \end{matrix}\right] = \left[\begin{matrix} x\\ y \end{matrix}\right]$ and draw this in a coordinate system. Measure the angle from the positive a-axis / x-axis to the vector (versor) with a protractor. If the angle you are measuring is in degrees, to get the result in radians you can use $\theta = \text{angle}^{\circ} = \frac{\text{angle}}{180} \cdot \pi$.