Equations and nth roots in the complex field

Problem

Solve in $\mathbb{C}$:

(a) $z^{2}+4z+13=0$, and plot the solutions on the Gauss plane.

(b) Find all cube roots of $8$ and verify that the corresponding vectors form equal angles with one another.

(c) Find all solutions of $z^{4}+16=0$ in algebraic form.

Solution

(a) The discriminant is $\Delta=4^2-4\cdot13=16-52=-36$, so $\sqrt\Delta=6i$ and $z=\frac{-4\pm6i}{2}=-2\pm3i.$ The two solutions correspond to the points $(-2,3)$ and $(-2,-3)$ on the Gauss plane, symmetric about the real axis.

(b) Writing $8=8e^{i0}$, the cube roots are $2e^{i2k\pi/3}$ for $k=0,1,2$: $z_0=2,\qquad z_1=2e^{i2\pi/3}=-1+\sqrt3\,i,\qquad z_2=2e^{i4\pi/3}=-1-\sqrt3\,i.$ All three vectors have modulus $2$, and their arguments differ by $2\pi/3=120°$: they are the vertices of an equilateral triangle inscribed in the circle of radius $2$.

(c) We have $z^4=-16=16e^{i\pi}$, so the four solutions are $2e^{i(\pi/4+k\pi/2)}$ for $k=0,1,2,3$: $\sqrt2+\sqrt2\,i,\quad -\sqrt2+\sqrt2\,i,\quad -\sqrt2-\sqrt2\,i,\quad \sqrt2-\sqrt2\,i.$

$\boxed{z=-2\pm3i;\qquad \sqrt[3]{8}=\{2,\ -1+\sqrt3 i,\ -1-\sqrt3 i\};\qquad \sqrt[4]{-16}=\{\pm\sqrt2\pm\sqrt2\,i\}}$