I have to do this:
Find the value of $\sqrt[3]{\dfrac{1-i}{1+i}}$ in binomial and polar form.
I have arrived to this point:
$$z=\sqrt[3]{\dfrac{1-i}{1+i}}=\sqrt[3]{\dfrac{1-i}{1+i}\cdot\dfrac{1-i}{1-i}}=\sqrt[3]{\dfrac{1-i-i-1}{1-i+i+1}}=\sqrt[3]{\dfrac{-2i}{2}}=\sqrt[3]{-i}$$
Then my guess is that I have to calculate the cube root of $-i$ ? I'm completly lost, can someone help me in solving this?
Thank you.
The cube root is a multiple-valued function on $\mathbb{C}$, since for any $z\in\mathbb{C}\setminus\{0\}$, the three different numbers $z,z\cdot e^{2\pi i/3},z\cdot e^{4\pi i /3}$ have the same cube. Anyway, since you noticed that: $$ \frac{1-i}{1+i}= -i = e^{-\frac{\pi}{2}i}$$ a cube root of such number is given by: $$ e^{-\frac{\pi}{6}i} = \cos\left(\frac{\pi}{6}\right)-i\,\sin\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}-i}{2}.$$