# Problems finding the n-th root of a negative number

I am having troubles understanding how to calculate the n-th root of a negative number.

To explain what exactly I am struggling with, let me demonstrate my thought process alongside the calculation of the n-th root of a positive number (in $$\mathbb{C}$$), for example

$$\tilde{z}^4 = 2 \implies \tilde{z} = \sqrt{2}, \text{ where } \tilde{z} \in \mathbb{C}$$

To my understanding, this can be solved by the following formula, among others:

$$\sqrt[n]{z} = \sqrt[n]{r} \times e^{i\frac{\phi + 2k\pi}{n}}, k \in \{ 0, 1, ..., n-1 \}$$

where $$r = |z| = \sqrt{a^2 + b^2}$$ and $$\phi = \arctan \frac{b}{a}$$. Given the equation from above, this gives

$$\sqrt{2} = \sqrt{2} \times e^{i\frac{k\pi}{2}}, k \in \{ 0, 1, ..., n-1 \}$$

since $$a = 2, b = 0 \implies r = \sqrt{2^2 + 0^2} = 2, \phi = \arctan \frac{0}{2} = 0$$. This results in

$$z = \left\{ \sqrt{2}, \sqrt{2} \times e^{\frac{i\pi}{2}}, \sqrt{2} \times e^{-\frac{i\pi}{2}} , -\sqrt{2} \right\}$$ which is correct.

However, this approach seemingly fails for negative numbers $$z$$, as this would produce the same (now obviously wrong) results. I have found that for negative numbers, the result is correct if $$\phi = \pi$$, but either way there is a mistake in my thinking when applying the above formula to negative numbers, and I need help figuring out where exactly I am going wrong.

This is not necessarily related to the main question asked, but what would happen if I were to take the n-th root of a purely imaginary number (or rather, a number $$k \in \mathbb{C}, \text{where } \mathfrak{R}(k) = 0$$), as this would lead to $$\phi = \arctan \frac{b}{0}$$.

Set $$z=re^{i\phi}$$
and obtain $$\sqrt[n]{z}=\sqrt[n]{re^{i\phi}}=\sqrt[n]{r}\cdot e^{\frac{i\phi}{n}}$$ where $$r=|z| \text{ and } \phi=\arctan{\frac{\Im (z)}{\Re (z)}}+2\pi k$$
Now for example if $$z =-2$$, a "negative" number (note that $$\mathbb{C}$$ is not a ordered Field, so negative complex numbers don't really exist), and $$\sqrt[n]{-2}=\sqrt[n]{2e^{i(\pi+2\pi k)}}=\sqrt[n]{2}\cdot e^{\frac{i(\pi+2\pi k)}{n}}, k\in\{0,1,...,n-1\}$$ which obviously exists.
If the real part of $$z$$ is 0, we just set $$\phi=\pm\frac{\pi}{2}$$, depending on whether the imaginary part of $$z$$ is positive or negative.
The standard definition of the arctan function gives an angle in $$(-\pi/2,\pi/2)$$. If you are taking $$n$$-th roots of a non-zero complex number you may find it to start by writing $$a+bi=r(\cos\theta+i\sin\theta) \text{ where }r >0, 0\le \theta<2\pi.$$ The formula $$\theta = \arctan(b/a)$$ works fine if $$a>0,b \ge 0$$. Otherwise, it must be modified. If $$a<0$$, let $$\theta = \pi+ \arctan(b/a)$$ . If $$a>0,b<0$$ let $$\theta = 2\pi+ \arctan(b/a).$$