Complex Numbers: How do I evaluate $(−1/2 + j1/2) ^{−2/3}$ in the form of $z = re^{jθ}$?

Question

How do I evaluate $(−\frac 1 2 + j \frac 1 2) ^{ -\frac 2 3}$ in the form of $z = re^{jθ}$? I am struggling to evaluate this equation in the form of $z = re^{jθ}$. So far, I found that the modulus for $z = \sqrt{\frac 1 2}$ and that $θ = \frac{3π} 4$. How do I continue from here?

Answer 1

This comes down to how you choose to define $z^{-2/3}$. The "obvious" answer that $(re^{i\theta})^{-2/3}=r^{-2/3}e^{-2i\theta/3}$ overlooks the invariance $\theta\mapsto\theta+2\pi$, which can multiply the result by $e^{-4ik\pi/3}$ for an integer $k$, without loss of generality $k\in\{0,\,1,\,2\}$. In your example, the result would be $(2^{-1/2}e^{3i\pi/4})^{-2/3}=\sqrt[3]{2}e^{-i\pi/2-4ik\pi/3}$. This gives us $-\sqrt[3]{2}i$ if $k=0$, $\frac{-\sqrt{3}+i}{\sqrt[3]{4}}$ if $k=1$, or $\frac{-\sqrt{3}-i}{\sqrt[3]{4}}$ if $k=2$.