Solve for $z \in \mathbb{R}$ : $z^6 = -64$

Question

I can only think of the solutions 2i and -2i, but there should be more solutions. I am very new to complex numbers and equations and was wondering if anyone could help with the following question:

Solve for $z \in \mathbb{R}$ : $z^6 = -64$

Answer 1

If $z\in \Bbb R$ then $z^6 \ge 0$ as it is the square of $z^3$. So no solutions.

Answer 2

Hint:

Solve using the exponential form of $z$: if $z=r\mathrm e^{i\theta}$, the equation becomes $$z^6=r^6\mathrm e^{6i\theta}=-64=2^6\mathrm e^{i\pi}$$ so that \begin{cases} r^6=2^6 \quad\text{(and }r>0),\\ 6\mkern 1mu\theta\equiv \pi\mod 2\pi. \end{cases} Can you end the computations?

Answer 3

$-64$ can be written as $-64=64(\cos\pi+i\sin\pi)$.

Set $z=r(\cos\theta+i\sin\theta)$

$$z^6=r^6\left(\cos(6\theta)+i\sin(6\theta)\right)$$ To solve $z^6= -64$ we must solve $$r^6\left(\cos(6\theta)+i\sin(6\theta)\right)=64(\cos\pi+i\sin\pi)$$ Which gives us $r^6=64\to r=2$ and $6\theta=\pi+2k\pi$ for $k=0,1,2,3,4,5$

This means $\theta_k=\frac{\pi}{6}+k\frac{\pi}{3}$ for $k=0,1,2,3,4,5$

Substitute the six values of $k$ and get the six roots of the equation.

Answer 4

I would like you to go through this and this.

Proceeding as the above solutions, you mentioned that you got the solution $2i$. The value of $n$ here is $6$. So, we get a hexagon as shown

As you can see that none of the complex numbers which satisfy this equation lie on the real number line, so your equation has no real solutions.

Answer 5

$z^6+64=0 \iff (z^3)^2-(8i)^2=0 \iff (z^3-8i)(z^3+8i)=0 \iff z^3-8i=0$ or $z^3+8i=0$

For example $z^3-8i=0 \iff \Big( \frac{z}{-2i} \Big) ^3=1 \iff w^3=1$ with $w= \frac{z}{-2i}$. The other case is analogous.

Can you take it from here?