Q154 / 270 · NDA 2018 (Phase 1) · Mathematics

What is the principal argument of $(-1-i)$, where $i=\sqrt{-1}$ ?

  • A
    $\frac{\pi}{4}$
  • B
    $-\frac{\pi}{4}$
  • Correct answer
    C
    $-\frac{3 \pi}{4}$
  • D
    $\frac{3 \pi}{4}$

Solution

$\begin{aligned} & \mathrm{z}=\mathrm{x}+\mathrm{iy}=(-1-\mathrm{i}) \\ & \therefore \mathrm{x}=-1, \mathrm{y}=-1 \end{aligned}$ This lies in $3^{\text {rd }}$ Quadrant. $\therefore \arg (\mathrm{z})=\theta-\pi$ $\begin{aligned} & =\tan ^{-1}\left(\frac{y}{x}\right)-\pi \\ & =\tan ^{-1}\left(\frac{-1}{-1}\right)-\pi=\tan ^{-1}(1)-\pi \\ & =\frac{\pi}{4}-\pi=\frac{-3 \pi}{4}\end{aligned}$

Hence, the correct answer is option: 'C'.

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