Find product of solutions of $x^6=-64$

Question

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?

Answer 1

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

$$4(\cos(\pi/6)- i\sin(\pi/6))(\cos(\pi/6)+ i\sin(\pi/6))$$ $$=4(\cos(\pi/6)^2+\sin(\pi/6)^2)=4$$

Answer 2

$$x^6=-64\Longleftrightarrow$$ $$x^6=64e^{\pi i}\Longleftrightarrow$$ $$x=\left(64e^{(2\pi k+\pi)i}\right)^{\frac{1}{6}}\Longleftrightarrow$$ $$x=2e^{\frac{1}{6}(2\pi k+\pi)i}\Longleftrightarrow$$ $$x=\begin{cases}2e^{\frac{1}{6}(2\pi\cdot0+\pi)i}\\ 2e^{\frac{1}{6}(2\pi\cdot1+\pi)i}\\ 2e^{\frac{1}{6}(2\pi\cdot2+\pi)i}\\ 2e^{\frac{1}{6}(2\pi\cdot3+\pi)i}\\ 2e^{\frac{1}{6}(2\pi\cdot4+\pi)i}\\ 2e^{\frac{1}{6}(2\pi\cdot5+\pi)i}\end{cases}\Longleftrightarrow$$ $$x=\begin{cases}2e^{\frac{\pi i}{6}}\\ 2i\\ 2e^{\frac{5\pi i}{6}}\\ 2e^{-\frac{5\pi i}{6}}\\ -2i\\ 2e^{-\frac{\pi i}{6}}\end{cases}\Longleftrightarrow$$ $$x=\begin{cases}2e^{\pm\frac{\pi i}{6}}\\ \pm 2i\\ 2e^{\pm\frac{5\pi i}{6}} \end{cases}\Longleftrightarrow$$ $$x=\begin{cases}\sqrt{3}\pm1i\\ \pm 2i\\ -\sqrt{3}\pm1i \end{cases}\Longleftrightarrow$$ $$x=\begin{cases}\sqrt{3}\pm i\\ \pm 2i\\ -\sqrt{3}\pm i \end{cases}$$

With $k\in\mathbb{Z}$ and $k:0-5$

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So only the first one is the one you need, notice that for $a,b\in\mathbb{R}$

$(a+bi)(a+bi)^*=(a+bi)(a-bi)=a^2+b^2$:

$$\left(\sqrt{3}+i\right)\left(\sqrt{3}-i\right)=\left(\sqrt{3}\right)^2+1^2=3+1=4$$