Find the 4 roots of the equation: $z^{4}+4=0$

Question

Find the 4 roots of the equation: $z^{4}+4=0$

Note: I can't use the form: $e^{i\theta}$

My attempt:

Note we have: $z=(-4)^{\frac{1}{4}}$, consider $w=-4$, then $w^{1/4}=z$ this implies $w=z^4$.

Consider the polar form of $z$ and $w$:
$|w|=r=4$,
$Arg(w)=\pi=\theta$

Then, $$w=4\cos\pi+i\sin\pi$$ $$z=p\cos\phi+i\sin\phi$$

This implies: $w=z^4$ iff $4=p^4$ and $4\phi=\pi+2k\pi$ iff $p=4^{1/4}$ and $\phi=\frac{\pi+2k\pi}{4}$ with $k=0,1,2,3$

Then the roots are:

$$z_k=4^{1/4}(\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}})$$ with $k=0,1,2,3$

is correct this?

Answer 1

You solution looks OK; the writing can be improved slightly though. I will follow your argument and rewrite it as follows.

The equation is equivalent to $z^4=-4$.

Now let $z=r(\cos \theta+i\sin\theta)$ and write $$ -4=4(\cos \pi +i\sin \pi)\tag{1} $$ By the De Moivre's formula, $$ z^4=r^4(\cos (4\theta)+i\sin (4\theta))\tag{2} $$ Comparing (1) and (2), we have $$ r=\sqrt[4]{4},\quad 4\theta=(2k+1)\pi,\quad k\in\mathbb{Z}. $$

Thus the four roots are $$ z_k=4^{1/4}(\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}})$$ with $k=0,1,2,3$.

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[Added:] The solution can be simplified further:

• Observe that $4^{1/4}=2^{2/4}=\sqrt{2}$.

• The expression $\cos{\frac{\pi+2k\pi}{4}}+i\sin{\frac{\pi+2k\pi}{4}}$ can be found exactly by hand and thus can be used to simply the solution. For instance, $$ z_0=\sqrt{2}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=1+i. $$ I will leave the cases $k=1,2,3$ to you.

Answer 2

It is $$(z^2)^2+(2^2)^2=0$$ or $$a^2+b^2=0$$ and this is $$(a-bi)(a+bi)=0$$ where $$a=z^2,b=2^2$$

Answer 3

$z^2=\pm 2i$ and then the roots are $\pm1\pm i$.

Answer 4

Consider the factorization of $z^4+4$: $$z^4+4=z^4\color{blue}{+4z^2}+4\color{blue}{-4z^2}=(z^2+2)^2-(2z)^2=(z^2+2z+2)(z^2-2z-2)$$ Then we get $z^2+2z+2=0$ or $z^2-2z+2=0$, then finish it by using quadratic formula.