Q55 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
If $-1 + \sqrt{-3} = r e^{i\theta}$, then the value of $\theta$ is
- A$-\dfrac{2\pi}{3}$
- B$\dfrac{\pi}{3}$
- C$-\dfrac{\pi}{3}$
- Correct answerD$\dfrac{2\pi}{3}$
Solution
Given complex number is $z = -1 + \sqrt{-3} = -1 + i\sqrt{3}$.
Let $z = r e^{i\theta}$, where $r = |z|$ and $\theta = \arg(z)$.
The real part $x = -1$ and the imaginary part $y = \sqrt{3}$.
Since $x < 0$ and $y > 0$, the complex number lies in the second quadrant.
The principal argument $\theta$ is given by $\pi - \tan^{-1}\left(\left|\dfrac{y}{x}\right|\right)$.
$\theta = \pi - \tan^{-1}(\sqrt{3}) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$.
Answer: $\dfrac{2\pi}{3}$
Let $z = r e^{i\theta}$, where $r = |z|$ and $\theta = \arg(z)$.
The real part $x = -1$ and the imaginary part $y = \sqrt{3}$.
Since $x < 0$ and $y > 0$, the complex number lies in the second quadrant.
The principal argument $\theta$ is given by $\pi - \tan^{-1}\left(\left|\dfrac{y}{x}\right|\right)$.
$\theta = \pi - \tan^{-1}(\sqrt{3}) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$.
Answer: $\dfrac{2\pi}{3}$
Hence, the correct answer is option: 'D'.