Let \(\alpha\) be the complex number \(-2+2\sqrt{3}i\). Express \(\alpha^3\) and the two square roots of \(\alpha\) in the standard form \(a+bi\).
Solution
The number \[ \alpha=-2+2\sqrt3\,i \] has modulus \(4\) and argument \(2\pi/3\), so \[ \alpha=4e^{2\pi i/3}. \] Hence \[ \alpha^3=4^3e^{2\pi i}=64. \] Its square roots are \[ \pm 2e^{\pi i/3} =\pm(1+\sqrt3\,i). \] Thus the two square roots are \[ 1+\sqrt3\,i,\qquad -1-\sqrt3\,i. \]