Finding all the values of $\sqrt[3]{7-4i}$ I'm reading about De Moivre's Formula and the Roots of Unity, and one of the exercises is to find all the different values of 
$$
\sqrt[3]{7-4i}
$$
I know that you can find the $n$th root of 1 with $x^{n-1}+x^{n-2}+\dots+x+1=0$, but I don't know how to find the $n$th root of an arbitrary complex number. I tried sketching it geometrically on an Argand diagram by plotting some kind of triangle, but I'm not sure it was correct and didn't help find the roots anyway. 
Is there a strategy for doing this (preferably in line with the content of De Moivre formula and roots of unity)?
 A: Hint: We have $$\sqrt[3]{7-4i}=(7-4i)^{1/3}$$
$$r=\sqrt{7^2+(-4)^2}=\sqrt{65}$$ $$\tan \theta=\left|\frac{-4}{7}\right|=\frac{4}{7}$$ Since, the point $(7, -4)$ lies in the fourth quadrant hence we have   $$\sin \theta=\frac{\frac{4}{7}}{\sqrt{1+\left(\frac{4}{7}\right)^2}}=\frac{4}{\sqrt{65}}$$ $$\cos \theta=\frac{1}{\sqrt{1+\left(\frac{4}{7}\right)^2}}=\frac{7}{\sqrt{65}}$$ Now, we have $$(7-4i)^{1/3}=r^{1/3}(\cos(-\theta)+i\sin(-\theta))^{1/3}$$
$$=\sqrt[3]{65}\left(\cos \theta -i\sin \theta\right)^{1/3}$$ $$=\sqrt[3]{65}\left(\cos\frac{(2k\pi+\theta)}{3} -i\sin\frac{(2k\pi+\theta)}{3}\right)$$  Now, substitute $k=0, 1, 2$
Can you proceed further? 
A: $$ 7-4i=\sqrt{7^2+4^2} e^{i\theta}\\tan \theta =\frac{-4}{7} \rightarrow \theta \sim -29.7^0$$ now consider :$$a=7-4i=\\a=\sqrt{65} e^{i\theta}=\sqrt{65} e^{-29.7^0i}=\sqrt{65} e^{(-29.7^0+2k\pi )i}\\$$ and now 
$$\sqrt[3]{a}= a^{\frac{1}{3}}=(\sqrt{65} e^{(-29.7^0+2k\pi )i})^{\frac{1}{3}}\\(\sqrt{65})^{\frac{1}{3}} e^{\frac{(-29.7^0+2k\pi )}{3}i}$$ 
so put $$ n=0,1,2$ to have the roots
A: $$\sqrt[3]{7-4i}=$$
$$\left|\sqrt[3]{7-4i}\right|e^{\arg\left(\sqrt[3]{7-4i}\right)i}=$$
$$\left|7-4i\right|^{\frac{1}{3}}e^{\arg\left(\sqrt[3]{7-4i}\right)i}=$$
$$\sqrt{\Re(7-4i)^2+\Im(7-4i)^2}^{\frac{1}{3}}e^{\arg\left(\sqrt[3]{7-4i}\right)i}=$$
$$\sqrt{7^2+4^2}^{\frac{1}{3}}e^{\arg\left(\sqrt[3]{7-4i}\right)i}=$$
$$\sqrt{65}^{\frac{1}{3}}e^{\arg\left(\sqrt[3]{7-4i}\right)i}=$$
$$65^{\frac{1}{2}\cdot\frac{1}{3}}e^{\arg\left(\sqrt[3]{7-4i}\right)i}=$$
$$65^{\frac{1}{6}}e^{\arg\left(\sqrt[3]{7-4i}\right)i}=$$
$$\sqrt[6]{65}e^{\arg\left(\sqrt[3]{7-4i}\right)i}=$$
$$\sqrt[6]{65}e^{-\frac{1}{3}\tan^{-1}\left(\frac{4}{7}\right)i}$$
