How to correctly find the angle phi in de moivre's formula?

Question

I am in high school, and we started learning De Moivre's formula. I had some problems with my homework concerning rooting of z. So far, this is what I know about the formula:

$\sqrt[n]{z}= \sqrt[n]{r}\left (\cos \left(\dfrac{\phi+2k\pi}{n} \right)+i\sin \left(\dfrac{\phi+2k\pi}{n} \right) \right)$

where $z = a + bi$ and $r = \sqrt{a^2 + b^2}$

Now here's the problem. What is $\phi$ equal to? My professor told us that $\phi = \arctan \left(\dfrac ba \right)$ but this seems incorrect to me. In my homework I had a simple equation of $x^2 + 1 = 0$ where $x = \sqrt{-1}$ and if you try solving $\phi$ with the above formula, you will get it to be 0, whereas the correct answer is $\phi = \pi$

So is $\phi = \arctan \left(\dfrac ba \right)$ a wrong way to find $\phi$, or am I mistaking somewhere?

Answer 1

If you write a complex number $a+bi$ in polar form $r(\cos\phi+i\sin\phi),$ then you must have $a=r\cos\phi$ and $b=r\sin\phi.$ Dividing these equations gives $$\frac ba=\tan\phi,$$ where $r\ne 0.$ So yes, you have that $$\phi=\arctan\left(\frac ba\right).$$

Answer 2

In fact $,\phi=\arctan \frac {-1}0$ as $b<0,a=0$ $\phi $ will lie in the Third Quadrant

For more example, let $z=-1-i,$

$\arctan \frac{-1}{-1}\ne \arctan \frac11$ as $b<0,a<0,\phi $ will lie in the Third Quadrant

The details can be found here

There is a range of values $n$ that can assume, that is $0,1,2\cdots, n-1$ (explained here) so as to give $n$ roots of $x^n=r$

Here $\displaystyle x^2=-1=\cos(2n\pi+\pi)+i\sin(2n\pi+\pi)$

$$\implies x=\cos\frac{(2n+1)\pi}2+i\sin\frac{(2n+1)\pi}2=i\sin\frac{(2n+1)\pi}2$$ where $n=0,1$