Find all cube roots of \(2-2i\) and express them in the standard form \(a + bi\).
<1>2. Every cube root therefore has the form \[ z_k=\sqrt2\,e^{i(-\pi/12+2\pi k/3)},\qquad k=0,1,2. \]
<1>3. For \(k=0\), \[ z_0=\frac{1+\sqrt3}{2}-\frac{\sqrt3-1}{2}i. \]
<1>4. For \(k=1\), \[ z_1=\frac{1-\sqrt3}{2}+\frac{1+\sqrt3}{2}i. \]
<1>5. For \(k=2\), \[ z_2=-1-i. \]
<1>6. These are all the cube roots of \(2-2i\).
Solution
1. Write \(2-2i\) in polar form as\[ 2-2i=2\sqrt2\,e^{-i\pi/4}. \]
The modulus of a cube root is \((2\sqrt2)^{1/3}=\sqrt2\), and its argument is one third of \(-\pi/4+2\pi k\).
Using \(\cos 15^\circ=(\sqrt6+\sqrt2)/4\) and \(\sin 15^\circ=(\sqrt6-\sqrt2)/4\), \[ \sqrt2(\cos15^\circ-i\sin15^\circ) =\frac{1+\sqrt3}{2}-\frac{\sqrt3-1}{2}i. \]
Here the argument is \(105^\circ\), so \[ \sqrt2(\cos105^\circ+i\sin105^\circ) =\frac{1-\sqrt3}{2}+\frac{1+\sqrt3}{2}i. \]
Here the argument is \(225^\circ\), and \[ \sqrt2(\cos225^\circ+i\sin225^\circ)=-1-i. \]
The equation \(z^3=2-2i\) has degree \(3\), and 2 gives three distinct roots.