Find product of solutions of $x^6=-64$ 
If the six solutions of $x^6=-64$ are written in the form $a+bi$,where
  $a$ and $b$ are real, then find the product of those solutions with $a>0$.

The answer in my book is given as $4$ but I don't see why.
Can you give me some hint on that?
 A: The points create a regular hexagon centered at $0$ (with a vertex at $\pm2i$). Each root has its absolute value equal to $\sqrt[6]{64}=2$. There are two roots with $a>0$:
$$2(\cos(\pi/6)\pm i\sin(\pi/6))$$
and their product is equal to
$$4(\cos(\pi/6)- i\sin(\pi/6))(\cos(\pi/6)+ i\sin(\pi/6))$$
$$=4(\cos(\pi/6)^2+\sin(\pi/6)^2)=4$$
A: $$x^6=-64\Longleftrightarrow$$
$$x^6=64e^{\pi i}\Longleftrightarrow$$
$$x=\left(64e^{(2\pi k+\pi)i}\right)^{\frac{1}{6}}\Longleftrightarrow$$
$$x=2e^{\frac{1}{6}(2\pi k+\pi)i}\Longleftrightarrow$$
$$x=\begin{cases}2e^{\frac{1}{6}(2\pi\cdot0+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot1+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot2+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot3+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot4+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot5+\pi)i}\end{cases}\Longleftrightarrow$$
$$x=\begin{cases}2e^{\frac{\pi i}{6}}\\
2i\\
2e^{\frac{5\pi i}{6}}\\
2e^{-\frac{5\pi i}{6}}\\
-2i\\
2e^{-\frac{\pi i}{6}}\end{cases}\Longleftrightarrow$$
$$x=\begin{cases}2e^{\pm\frac{\pi i}{6}}\\
\pm 2i\\
2e^{\pm\frac{5\pi i}{6}}
\end{cases}\Longleftrightarrow$$
$$x=\begin{cases}\sqrt{3}\pm1i\\
\pm 2i\\
-\sqrt{3}\pm1i
\end{cases}\Longleftrightarrow$$
$$x=\begin{cases}\sqrt{3}\pm i\\
\pm 2i\\
-\sqrt{3}\pm i
\end{cases}$$
With $k\in\mathbb{Z}$ and $k:0-5$

So only the first one is the one you need, notice that for $a,b\in\mathbb{R}$
$(a+bi)(a+bi)^*=(a+bi)(a-bi)=a^2+b^2$:
$$\left(\sqrt{3}+i\right)\left(\sqrt{3}-i\right)=\left(\sqrt{3}\right)^2+1^2=3+1=4$$
A: I know this is late, but here is a solution for students with intermediate knowledge.
To find the solutions, restate the equation.
$z^6+64=0$
This can be rewirtten as $(z^3+8i)(z^3-8i)=0$
$(z+2i)(z^2-2zi-4)(z-2i)(z^2+2zi-4)$
Using the quadratic formula, $z = i±\sqrt{-1+4} = i±\sqrt{3}$ and $z = -i±\sqrt{-1+4} = -i±\sqrt{3}$
These solutions with a positive $a$ can be restated as $\sqrt{3}-i$ and $\sqrt{3}+i$. To find the product of these, multiply them both.
$(\sqrt{3}-i)(\sqrt{3}+i) = 3+1 = 4$
The product of the solutions with $a>0$ is 4.
