Q54 / 150 · MHT CET 2024 (16 May Shift 2) · Mathematics

If $z_1=5-2 i$ and $z_2=3+i$, where $i=\sqrt{-1}$, then $\arg \left(\frac{z_1+z_2}{z_1-z_2}\right)$ is

  • Correct answer
    A
    $\tan ^{-1}\left(\frac{22}{19}\right)$
  • B
    $\tan ^{-1}\left(\frac{22}{13}\right)$
  • C
    $\tan ^{-1}\left(\frac{21}{19}\right)$
  • D
    $\tan ^{-1}\left(\frac{19}{22}\right)$

Solution

$\begin{aligned} & \frac{z_1+z_2}{z_1-z_2}=\frac{8-i}{2-3 i} \\ &=\frac{8-i}{2-3 i} \times \frac{2+3 i}{2+3 i} \\ &=\frac{19+22 i}{13}=\frac{19}{13}+\frac{22 i}{13} \\ & \therefore \quad \arg \left(\frac{z_1+z_2}{z_1-z_2}\right)=\tan ^{-1}\left(\frac{22}{19}\right)\end{aligned}$

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

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