Solving a complex equation of the form $f(z)=0$

Question

I'm trying to solve the following equation for $z\in \mathbb{C}$: $(z^3-64)(z^3+64)=0$.
I'm not sure I'm doing it right, and need some guidance...

I split this into two:
1. $z^3-64=0$
2. $z^3+64=0$

Now, concentrating on the first equation for now, I have:
$z^3-64=0$
$ \Rightarrow(x+yi)^3-64=0$
$\Rightarrow x^3+3x^2yi+3x(yi)^2+(yi)^3-64=0$
$\Rightarrow x^3+3x^2yi-3xy^2-y^3i-64=0$

Splitting again, I have:
1. $x^3-3xy^2-64=0$ ("real" part)
2. $3x^2yi-yi=0\Rightarrow 3x^2y-y=0$ ("imaginary" part)

And now solving for $x$ and $y$...
Am I doing it right so far?
If not, what's the way to solve it, and in general, what's the way to solve an equation of the form $f(z)=0$ where $z\in\mathbb{C}$?

And last thing, correct me if I'm wrong, but this equation has 12 solutions. So my question is: What if I'm only interested in the solutions for $z$ that are in first quadrant of the Gaussian plane, is there a way to eliminate in advance some of the work that should be done to solving this equation?

In the notation of this answer, set $w=64$ and $z=\rho e^{i\theta}$. — Git Gud, Oct 23 '14 at 11:16

score 3 · Accepted Answer · answered Oct 23 '14 at 11:17

3

You should use polar coordinates instead.

I'll demonstrate on the first part: $$z^3 = 64$$ Let $z = r(\cos v + i \sin v)$ and $64 = 64(\cos (2 \pi n) + i \sin (2 \pi n)$, then we get:

$$r^3(\cos v + i \sin v)^3 = 64(\cos (2 \pi n) + i \sin (2 \pi n)$$ $$r(\cos v + i \sin v)^3 = 4(\cos (2 \pi n) + i \sin (2 \pi n))^{1/3}$$

Using de Moivre's formula:

$$r(\cos v + i \sin v)^3 = 4(\cos (\frac{2 \pi n}{3}) + i \sin (\frac{2 \pi n}{3}))$$

Meaning $r = 4$ and $v = \frac{2 \pi n}{3}$. For $n=0,1,2$ we have three different solutions to $z$. Convert back to the form $a+bi$ and we are done. Do the same with the next part.

answered Oct 23 '14 at 11:17

naslundx

9,720

Great, thanks. It is all coming back to me now...
When do I want to use polar coordinates and when the method I used? – so.very.tired Oct 23 '14 at 11:21
1

For a binomial equation like this, on the form $z^n = w$, the polar version is always simpler. :) – naslundx Oct 23 '14 at 11:25

score 1 · Answer 2 · answered Oct 23 '14 at 11:37

1

Hints;

$$z^3-64=0 \iff z^3-4^3=0 \iff (z-4) (z^2+4 z+16)=0$$

$$z^3+64=0 \iff z^3+4^3=0 \iff (z+4) (z^2-4 z+16)=0$$

Concluding;

$$f(z)=0 \iff (z-4) (z+4) (z^2-4 z+16) (z^2+4 z+16) = 0$$

I think you can tackle this.

answered Oct 23 '14 at 11:37

UserX

4,930

score 0 · Answer 3 · answered Oct 23 '14 at 11:05

0

Answer. $$ z=\pm 2 \quad\text{and}\quad z=\pm 1\pm i \sqrt{3}. $$

answered Oct 23 '14 at 11:05

Yiorgos S. Smyrlis

83,933

1

Thanks, but I'm not interested in the final solution. – so.very.tired Oct 23 '14 at 11:07

Solving a complex equation of the form $f(z)=0$

3 Answers3