Solving complex eqautions

Question

I have an equation:

$\frac{8\sqrt{3}}{z^4+8}=-i$

I have no idea where to begin solving this! My book doesn't seem to give any hints on equations like this. How do I solve it and what can I search for on information on this? Roots of complex numbers...? Complex equations? I'm just not sure.

Yes. I'm still unsure what is being asked. I have't solved complex equations before. — Paze, Jan 25 '14 at 16:43
What you're asked is to find the extension of the set $\left{z\in \mathbb C\colon \dfrac{8\sqrt{3}}{z^4+8}=-i\right}$. Do you understand this' — Git Gud, Jan 25 '14 at 16:48
No. I don't understand what you mean by 'extension' of the set. — Paze, Jan 25 '14 at 16:50
See the extension wikipedia entry. As an example, the extension of the set ${x\in \mathbb R\colon x^2=1}$ is $-1,1$. — Git Gud, Jan 25 '14 at 16:52

score 2 · Answer 1 · answered Jan 25 '14 at 16:51

2

$8\sqrt{3}=-i(z^4+8) \to 8\sqrt{3}i=(z^4+8) \to z^4 =8\sqrt{3}i-8$

Now you should $8\sqrt{3}i-8$ represent this in polar cordinates $r*e^{i \alpha}$

answered Jan 25 '14 at 16:51

Eli Elizirov

1,820

How does $8\sqrt{3}=-i(z^4+8)$ become $8\sqrt{3}i=(z^4+8)$ ? – Paze Jan 25 '14 at 17:02
@PazeMultiply by $i$ on both sides. – Git Gud Jan 25 '14 at 17:03
Right, the number i behaves differently than others of course.. – Paze Jan 25 '14 at 17:08
Not differently at all. – Lubin Jan 25 '14 at 17:20
No, but you know what I mean. With any other variable, if you multiply by that variable, the equation wouldn't end up like that. – Paze Jan 25 '14 at 17:25
So I've got:
$z^4=16 \left( \frac{1}{2}- \frac {\sqrt{3} }{2} \cdot i \right)$

But I'm unsure how I continue!
– Paze Jan 25 '14 at 18:16
You need to think about z as $z=re^{i\alpha}$ – Eli Elizirov Jan 25 '14 at 21:00
How do you solve $z^3=1$ ? is the same – Eli Elizirov Jan 25 '14 at 21:00

Solving complex eqautions

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