operatorname{Arg}left[frac{(1+i sqrt{3})(-sqr | vedclass

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Question

(B) Let $z = \frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}$Numerator: $(1+i \sqrt{3})(-\sqrt{3}-i) = -\sqrt{3} - i - 3i - i^2 \sqrt{3} = -\sqrt{3} - 4

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $z = \frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}$Numerator: $(1+i \sqrt{3})(-\sqrt{3}-i) = -\sqrt{3} - i - 3i - i^2 \sqrt{3} = -\sqrt{3} - 4i + \sqrt{3} = -4i$Denominator: $(1-i)(-i) = -i + i^2 = -i - 1 = -(1+i)$So,$z = \frac{-4i}{-(1+i)} = \frac{4i}{1+i}$Multiply numerator and denominator by the conjugate $(1-i)$:$z = \frac{4i(1-i)}{(1+i)(1-i)} = \frac{4i - 4i^2}{1 - i^2} = \frac{4i + 4}{1 + 1} = \frac{4+4i}{2} = 2+2i$Since $z = 2+2i$ lies in the $I^{st}$ quadrant,the argument is given by $	heta = 	an^{-1}\left(\frac{2}{2}ight) = 	an^{-1}(1) = \frac{\pi}{4}$.

$\operatorname{Arg}\left[\frac{(1+i \sqrt{3})(-\sqrt{3}-i)}{(1-i)(-i)}\right]=$

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