Question

The complex number with argument $\frac{5\pi}{6}$ at a distance of 2 units from the origin is

• A. $\sqrt{3}-i$
• B. $\sqrt{3}+i$
• C. $-\sqrt{3}-i$
• D. $-\sqrt{3}+i$

Hint:

You can often solve this type of question in 5 seconds just by checking signs! An argument of $\frac{5\pi}{6}$ places the point firmly in the 2nd quadrant. In the 2nd quadrant, the real part ($a$) must be negative and the imaginary part ($b$) must be positive. Only $-\sqrt{3}+i$ fits this $(-, +)$ coordinate pattern!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the polar coordinates of a complex number: its distance from the origin (modulus, $r$) and the angle it makes with the positive real axis (argument, $\theta$). We need to convert it into the standard algebraic form $a + bi$.

Step 2: Key Formula or Approach:
The conversion from polar to algebraic form is achieved using Euler's relation mapped to the complex plane:
$$z = r(\cos \theta + i \sin \theta)$$ Where $r$ is the modulus and $\theta$ is the principal argument.

Step 3: Detailed Explanation:
We are given:
Modulus $r = 2$
Argument $\theta = \frac{5\pi}{6}$
Substitute these values into the polar form equation:
$$z = 2 \left( \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) \right)$$ The angle $\frac{5\pi}{6}$ lies in the second quadrant, where cosine is negative and sine is positive.
Using reference angles ($\pi - \frac{\pi}{6}$):
$$\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$ $$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$ Now, substitute the exact trigonometric values back into the expression:
$$z = 2 \left( -\frac{\sqrt{3}}{2} + i \left(\frac{1}{2}\right) \right)$$ Distribute the $2$:
$$z = -\sqrt{3} + i$$

Step 4: Final Answer:
The complex number in standard form is $-\sqrt{3}+i$, corresponding to option (D).