Roots of complex numbersEdexcel International A Level Further Maths: Revision notes

To solve $z^n=w$, write $w$ in polar form with the general argument, then use De Moivre's theorem: $w=r\,e^{i(\theta+2k\pi)}\ \Rightarrow\ z=r^{1/n}\,e^{i\left(\frac{\theta+2k\pi}{n}\right)},\quad k=0,1,\dots,n-1.$ There are exactly $n$ distinct roots, because $k=n$ gives the same root as $k=0$. Take the real positive $n$th root of the modulus.

All roots have the same modulus $r^{1/n}$ and their arguments are equally spaced by $\frac{2\pi}{n}$. For $z^3=8$: $|z|=2$, arguments $0,\frac{2\pi}{3},\frac{4\pi}{3}$, so the roots are $2,\ 2e^{\pm2\pi i/3}$. The non-real roots are $2\left(-\frac12\pm\frac{\sqrt3}2i\right)=-1\pm i\sqrt3$. Give final arguments in the range $-\pi<\theta\le\pi$ by subtracting $2\pi$ where needed.

Solve $z^4=-8+8\sqrt3\,i$. $|w|=16$ and $\arg w=\frac{2\pi}{3}$ (second quadrant), so $w=16e^{i(2\pi/3+2k\pi)}$. Then $z=2e^{i\left(\frac\pi6+\frac{k\pi}{2}\right)}$, giving $2e^{i\pi/6},\ 2e^{2\pi i/3},\ 2e^{-5\pi i/6},\ 2e^{-i\pi/3}$, equally spaced by $\frac\pi2$.

The solutions of $z^n=1$ are $z=e^{2\pi ik/n}$. They lie on the unit circle, and the roots of unity sum to zero for $n\ge2$ because $z^n-1$ has no $z^{n-1}$ term. For $n=5$: $1+2\cos\frac{2\pi}{5}+2\cos\frac{4\pi}{5}=0$, so $\cos\frac{2\pi}{5}+\cos\frac{4\pi}{5}=-\frac12$, using $e^{i\alpha}+e^{-i\alpha}=2\cos\alpha$. Non-real roots occur in conjugate pairs when the coefficients are real.

If $(w+2)^3=-27i$, solve $z^3=-27i$ for $z=w+2$, then subtract $2$ from each root: $z=3e^{-i\pi/6},\,3i,\,3e^{-5\pi i/6}$ gives $w=z-2$. For $(w+1)^5=w^5$, divide by $w^5$ to get $\left(\frac{w+1}{w}\right)^5=1$. Then $\frac{w+1}{w}=e^{2\pi ik/5}$, $k=1,2,3,4$ (not $k=0$), and $w=\frac{1}{e^{2\pi ik/5}-1}=-\frac12-\frac i2\cot\frac{k\pi}{5}$, using $e^{i\phi}-1=e^{i\phi/2}\cdot2i\sin\frac\phi2$.