A creative approach to a common complex numbers problem

Problem: 

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 regular way to do this would be to write the equation as $x^6 = 64e^{i\pi}$, and then get the two solutions with $a>0$ as $2e^{i\pi/6}$ and $2e^{-i\pi/6}$. The answer is the product of these two, which is $4$.

Creative solution!

Note that for any complex number $a+bi$ that satisfies the given equation, either $a<0$, $a=0$, or $a>0$.

If $a=0$, then we have$$(bi)^6 = -64$$$$\implies b^6 (-1) = -64 \implies b = 2, -2,$$which gives us two solutions for $x$, namely $2i$ and $-2i$.

Now, we consider $a>0$ and $a<0$. For the key insight, note that for every solution $z = a+bi$ with $a>0$, there exists the solution $-z = -a-bi$ with $a<0$.

Also, by Vieta's, we know that the product of all of the roots of the equation is $64$.

So, we have$$(2i)(-2i)(z_1)(-z_1)(z_2)(-z_2) = 64$$$$\implies 4(z_1 \cdot z_2)^2 = 64$$$$\implies z_1 \cdot z_2 = \boxed{4},$$which is the product of the two complex solutions which have $a>0$.