Question

If \(z=\dfrac{1-i\sqrt{3}}{1+i\sqrt{3}}\), then \(\arg(z)\) is

• A. \(60^\circ\)
• B. \(120^\circ\)
• C. \(240^\circ\)
• D. \(300^\circ\)

Hint:

For complex division: \(\arg\left(\frac{z_1}{z_2}\right)=\arg(z_1)-\arg(z_2)\). Then convert negative angle to \(0^\circ\) to \(360^\circ\).

Answer 1

The Correct Option is C

Step 1: Express numerator and denominator in polar form.
Numerator:
\[ 1-i\sqrt{3} \] Its modulus:
\[ \sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2 \] Argument:
\[ \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right)=-60^\circ \] So:
\[ 1-i\sqrt{3} = 2(\cos(-60^\circ)+i\sin(-60^\circ)) \] Denominator:
\[ 1+i\sqrt{3} \] Modulus = 2, argument = \(+60^\circ\).
\[ 1+i\sqrt{3} = 2(\cos 60^\circ+i\sin 60^\circ) \] Step 2: Divide in polar form.
\[ z=\frac{2(\cos(-60^\circ)+i\sin(-60^\circ))}{2(\cos 60^\circ+i\sin 60^\circ)} \] \[ z = \cos(-120^\circ)+i\sin(-120^\circ) \] Step 3: Convert \(-120^\circ\) to positive coterminal angle.
\[ -120^\circ = 240^\circ \] Final Answer: \[ \boxed{240^\circ} \]