Answer
$\sqrt[6]{2}cis105^\circ$, $\sqrt[6]{2}cis225^\circ$ and $\sqrt[6]{2}cis345^\circ$ are the three cube roots of $1 - i$.
Work Step by Step
$1 - i$ is at $315^\circ$ with absolute value $\sqrt{1^2 + (-1)^2} = \sqrt{2}$. The equivalent trigonometric form of $1 - i$ is $\sqrt{2}cis315^\circ$.
Suppose the cube root of $1 - i$ is represented by $rcis\alpha$,
$(rcis\alpha)^3$
= $1 - i$
= $\sqrt{2}cis315^\circ$
By De Moivre’s theorem, this equation becomes
$r^3cis3\alpha = \sqrt{2}cis315^\circ$
By equating, $r^3 = \sqrt{2}$, $r = \sqrt[6]{2}$, and $cos3\alpha = cos315^\circ, sin3\alpha = sin315^\circ$
For these equations to be satisfied, $3\alpha$ must represent an angle that is coterminal with $315^\circ$.
Therefore, we must have
$\alpha = \frac{315^\circ + 360^\circ\cdot k}{3}$ and k take on the integer values 0, 1, and 2.
When $k = 0, \alpha = 105^\circ$,
$k = 1, \alpha = 225^\circ$, and
$k = 2, \alpha = 345^\circ$
Hence, when $k = 0,$ the root is $\sqrt[6]{2}cis105^\circ$,
$k = 1,$ the root is $\sqrt[6]{2}cis225^\circ$,
$k = 2,$ the root is $\sqrt[6]{2}cis345^\circ$,
$\sqrt[6]{2}cis105^\circ$, $\sqrt[6]{2}cis225^\circ$ and $\sqrt[6]{2}cis345^\circ$ are the three cube roots of $1 - i$.
