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.

  • 2
    $\begingroup$ 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. $\endgroup$ Aug 10, 2020 at 18:00
  • $\begingroup$ @IzaakvanDongen But, is there any other way I can do? $\endgroup$
    – Ankita Pal
    Aug 10, 2020 at 18:03

1 Answer 1


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 $cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$.


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