# Solving complex expression

The value of $$(-8\sqrt{-1})^{\left(\frac 16\right)}+(8\sqrt{-1})^{\left(\frac 16\right)}$$ is a pure real number. Can you find the real number?

My try: $$(-8\sqrt{-1})^{\left(\frac 16\right)}+(8\sqrt{-1})^{\left(\frac 16\right)}\\ =i^{\frac 16}(\sqrt2 i+\sqrt 2)$$ But, how can I find that this is a real number. Please help.

• You're going too fast! With complex numbers it's not always true that $(ab)^c = a^c b^c$. Carefully check what definition of exponentiation you're using, and evaluate each of the two terms you're adding separately. Aug 10, 2020 at 18:00
• @IzaakvanDongen But, is there any other way I can do? Aug 10, 2020 at 18:03

We have $$-8i=8e^{\frac{-i\pi}{2}}$$ and $$8i=8e^{\frac{i\pi}{2}}$$ using Euler's formula, so
$$(-8i)^{\frac{1}{6}}+(8i)^{\frac{1}{6}}=8^{\frac{1}{6}}(e^{\frac{-i\pi}{12}}+e^{\frac{i\pi}{12}})=2 \times 8^{\frac{1}{6}}\times\frac{e^{\frac{-i\pi}{12}}+e^{\frac{i\pi}{12}}}{2}=2 \times 8^{\frac{1}{6}}cos(\frac{\pi}{12})$$ where we have used that $$cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$$.