Find $a,b, c,d$ such that $z^{4} + 16 = (z-a)(z-b)(z-c)(z-d) $.

Question

My original idea was choosing $2, -2, 2i, -2i$ but that gives $$z^{4} - 16.$$ Is there a similar way to solve this?

Answer 1

Substitute $z^2:=x$ to get $$x^2+16=(x-4i)(x+4i)$$ Thus $z_{1,2}=\pm2\sqrt{i}$ and $z_{3,4}=\pm2\sqrt{-i}$. What remains is to compute the square root of $i$ and $-i$, respectively. As $i=e^{i\frac\pi2}$ and $-i=e^{i\frac{3\pi}2}$ we get that $$\sqrt{i}=e^{i\frac\pi4},~\sqrt{-i}=e^{i\frac{3\pi}4}$$ Now use Euler's Formula to obtain the solutions which are given by

$$\therefore~z_1=\sqrt2(1+i),~z_2=\sqrt2(-1+i),~z_3=\sqrt2(-1-i),~z_4=\sqrt2(1-i)$$

Answer 2

We have $z^4+16=(z^2+4)^2-8z^2=(z^2+4-2\sqrt{2}z)(z^2+4+2\sqrt{2}z)=[(z-\sqrt{2})^2+2][(z+\sqrt{2})^2+2]=[(z-\sqrt{2})^2-(\sqrt{2}i)^2][(z+\sqrt{2})^2-(\sqrt{2}i)^2]=(z-2-\sqrt{2}i)(z-2+\sqrt{2}i)(z+2-\sqrt{2}i)(z+2+\sqrt{2}i)$. $\implies (a;b;c;d)=(2+\sqrt{2}i;2-\sqrt{2}i;-2+\sqrt{2}i;-2-\sqrt{2}i)$ and permutations.

Answer 3

Sure. There are four roots: $2e^{\frac{\pi i}4},2e^{\frac{3\pi i}4},2e^{\frac{5\pi i}4},2e^{\frac{7\pi i}4}$.

These are gotten by taking any one root, and multiplying by $\rho,\rho^2$ and $\rho^3$ where $\rho$ is a primitive $4$-th root of unity, to get the other three.

You can use Euler's formula to get one (or all) of the roots: since $e^{\pi i}=-1$, we get that, for instance, $z=2e^{\frac{\pi i}4}\implies z^4=-16$.