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Mathematics Question on argand plane

If z=1−i31+i3z=\frac{1-i \sqrt{3}}{1+i \sqrt{3}}, then arg⁡(z)\arg (z) is

A

60∘60^{\circ}

B

120∘120^{\circ}

C

240∘240^{\circ}

D

300∘300^{\circ}

Answer

240∘240^{\circ}

Explanation

Solution

z=(1−i31+i3)z = \left(\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\right)
⇒z=1−i31+i3×1−i31−i3\Rightarrow z = \frac{1-i\sqrt{3}}{1+i\sqrt{3}} \times\frac{1- i \sqrt{3}}{1-i\sqrt{3}}
⇒z=1−3−23i1+3=−2−23i4\Rightarrow z = \frac{1-3-2\sqrt{3}i}{1+3} = \frac{-2 -2\sqrt{3}i}{4}
⇒z=−12−32i\Rightarrow z = - \frac{1}{2} - \frac{\sqrt{3}}{2} i
⇒z=cos⁡240∘−isin⁡240∘\Rightarrow \sqrt{ z = \cos240^{\circ} } -i\sin240^{\circ}
Thus, arg (z)=240∘\left(z\right) =240^{\circ}