Question

One of the values of the square root of \[ \left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right) \] is

• A. \(\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}\)
• B. \(\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\)
• C. \(-\dfrac{\sqrt{3}}{2}+\dfrac{i}{2}\)
• D. \(-\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}\)

Hint:

For square roots of complex numbers, first convert the number into polar form. Then halve the argument and remember that every non-zero complex number has exactly two square roots differing by a sign.

Answer 1

The Correct Option is D

Concept: If a complex number is written in polar form as \[ z=r(\cos\theta+i\sin\theta), \] then its square roots are \[ \sqrt{r}\left(\cos\frac{\theta+2k\pi}{2} +i\sin\frac{\theta+2k\pi}{2}\right), \qquad k=0,1. \]

Step 1: Express the given complex number in polar form. Given \[ z=\frac{1}{2}+\frac{\sqrt{3}}{2}i. \] Its modulus is \[ |z| = \sqrt{\left(\frac12\right)^2+ \left(\frac{\sqrt3}{2}\right)^2} =1. \] Also, \[ \cos\theta=\frac12, \qquad \sin\theta=\frac{\sqrt3}{2}. \] Therefore, \[ \theta=\frac{\pi}{3}. \] Hence, \[ z=\cos\frac{\pi}{3} +i\sin\frac{\pi}{3}. \]

Step 2: Find the square roots. The square roots are \[ \cos\frac{\pi}{6} +i\sin\frac{\pi}{6} = \frac{\sqrt3}{2} +\frac{i}{2}, \] and \[ \cos\left(\frac{\pi}{6}+\pi\right) +i\sin\left(\frac{\pi}{6}+\pi\right). \] Thus, \[ = -\frac{\sqrt3}{2} -\frac{i}{2}. \]

Step 3: Match with the given options. One of the square roots is \[ -\frac{\sqrt3}{2} -\frac{i}{2}. \] Therefore, \[ \boxed{-\frac{\sqrt3}{2}-\frac{i}{2}} \]