Question

Using the fact that if z = r(cosθ +isinθ) then z² = r²(cosθ +isinθ), find the square root of

2√3 -2i

Answer 1

$\displaystyle \textsf{Let}\,\,\, z = 2\sqrt{3} - 2i\\ \arg{z} = \arctan\left(\frac{-2}{2\sqrt{3}}\right) = -\frac{\pi}{6}\\ |z| = \sqrt{12 + 4} = 4\\ \implies z = 4e^{-\frac{i\pi}{6}}\\ \begin{aligned} z^{\frac{1}{2}} &= \pm 2e^{-\frac{i\pi}{12}} \\&= \pm 2\cos\left(\frac{\pi}{12}\right) - \left(\pm i2\sin\left(\frac{\pi}{12}\right)\right) \\&= \pm 2\cos\left(\frac{\pi}{12}\right) \mp i2\sin\left(\frac{\pi}{12}\right) \end{aligned}\\ \textsf{It implies that the square root is}\\2\cos\left(\frac{\pi}{12}\right) - i2\sin\left(\frac{\pi}{12}\right) \,\,\textup{or}\\ -2\cos\left(\frac{\pi}{12}\right) + i2\sin\left(\frac{\pi}{12}\right)$