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I have a equation that should divide a circle in even parts. As I found its called circle-dividing equation. I'v found same information how to solve a equation which has a form like this:

$$z^6 = 1$$

one of the first steps was to transform the equation using MOIVRE to

$$z^6 = 1+0i$$

$\cos(6 \varphi) = 1$ // real number part

$\sin(6 \varphi) = 0$ // imag number part

but my equation is:

$$z^3 = 8i$$

How can I resolve this?

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3 Answers 3

up vote 2 down vote accepted

$$z^3=8i=8e^{\dfrac{i\pi}2}=8e^{\left(2k\pi+\dfrac\pi2\right)i}$$ where $k$ is any integer

$$z=2e^{\dfrac{(4k+1)\pi i}6}$$ where $k\equiv0,1,2\pmod3$ (See this)

Now use Euler Formula

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$$r^3(\cos(3\varphi)+i\sin(3\varphi))=8i = 8\left(\cos\frac\pi2+i\sin\frac\pi2\right)$$

So $$ r^3=8\text{ and }3\varphi=\frac\pi2. $$

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Hint you can rewrite the given equation to $$ \left(\frac z2\right)^3=i $$ and solve for $\frac z2$.
You could also rewrite it to $$ \left(\frac z2 i\right)^3=1 $$ by multiplying both sides with $i^3$.

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