$\operatorname{Arg}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)$ lies in the interval

$\operatorname{Arg}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)$ lies in the interval
  1. $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
  2. $\left(-\pi, \frac{-\pi}{2}\right)$
  3. $\left(\frac{-\pi}{2}, 0\right)$
  4. $\left(0, \frac{\pi}{4}\right)$

Solution

$ \begin{aligned} & \text {}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right) \\ & =\frac{(4+2 i)(2+3 i)+(3+4 i)(1-2 i)}{(1-2 i)(2+3 i)} \\ & =\frac{(2+16 i)+(11-2 i)}{(8-i)}=\frac{(13+14 i)}{(8-i)} \\ & =\frac{90+109 i}{64+1}=\frac{90}{65}+\frac{109}{65} i \\ & \arg \left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)=\arg \left(\frac{90}{65}+\frac{109}{65} i\right) \\ & =\tan ^{-1}\left(\frac{109}{90}\right)=\tan ^{-1}(1.21) \end{aligned} $ Since $\tan \frac{\pi}{4}=1$ and $\tan \frac{\pi}{2}=\infty$ Therefore, $ \frac{\pi}{4} < \arg \left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right) < \frac{\pi}{2} $

Asked in: AP EAMCET 2023 (19 May Shift 1)

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