Question

If one of the values of \[ \sqrt{-1-\sqrt{3}i} \] is \(\alpha+i\beta\), where \(\alpha<0\) and \(\beta>0\), then \(\alpha=\)

• A. \(-\frac{1}{\sqrt{2}}\)
• B. \(-\frac{\sqrt{3}}{\sqrt{2}}\)
• C. \(x-\sqrt{3}\)
• D. \(x-\sqrt{2}\) \bigskip

Hint:

For square roots of complex numbers, first convert the number into polar form and then divide the argument by 2.

Answer 1

The Correct Option is A

Concept: To find the square root of a complex number, it is convenient to convert the number into polar form and then apply De Moivre's theorem. If \[ z=r(\cos\theta+i\sin\theta), \] then \[ \sqrt{z} = \sqrt{r} \left( \cos\frac{\theta}{2} + i\sin\frac{\theta}{2} \right). \] The sign is chosen according to the conditions given in the problem.

Step 1: Express the complex number in polar form.
Given \[ z=-1-\sqrt3\,i. \] Its modulus is \[ |z| = \sqrt{(-1)^2+(-\sqrt3)^2} = \sqrt{1+3} = 2. \] Hence, \[ r=2. \]

Step 2: Determine the argument.
Since both real and imaginary parts are negative, the point lies in the third quadrant. Also, \[ \tan\theta = \frac{-\sqrt3}{-1} = \sqrt3. \] Therefore the reference angle is \[ \frac{\pi}{3}. \] Hence \[ \theta = \pi+\frac{\pi}{3} = \frac{4\pi}{3}. \] Thus \[ -1-\sqrt3 i = 2\left( \cos\frac{4\pi}{3} +i\sin\frac{4\pi}{3} \right). \]

Step 3: Find the square roots.
Applying De Moivre's theorem, \[ \sqrt{-1-\sqrt3 i} = \sqrt2 \left( \cos\frac{2\pi}{3} +i\sin\frac{2\pi}{3} \right). \] Substituting the values, \[ = \sqrt2 \left( -\frac12 +i\frac{\sqrt3}{2} \right). \] Therefore \[ = -\frac{\sqrt2}{2} + i\frac{\sqrt6}{2}. \]

Step 4: Identify \(\alpha\).
Comparing with \[ \alpha+i\beta, \] we get \[ \alpha = -\frac{\sqrt2}{2} = -\frac{1}{\sqrt2}. \] Since \(\alpha<0\) and \(\beta>0\), this root satisfies the given condition. Hence, \[ \boxed{\alpha=-\frac{1}{\sqrt2}}. \] Therefore the correct option is \[ \boxed{(A)}. \]