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$\operatorname{Arg}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)$ lies in the interval
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Solution:
1051 Upvotes
Verified Answer
The correct answer is:
$\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
$$
\begin{aligned}
& \text {}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right) \\
& =\frac{(4+2 i)(2+3 i)+(3+4 i)(1-2 i)}{(1-2 i)(2+3 i)} \\
& =\frac{(2+16 i)+(11-2 i)}{(8-i)}=\frac{(13+14 i)}{(8-i)} \\
& =\frac{90+109 i}{64+1}=\frac{90}{65}+\frac{109}{65} i \\
& \arg \left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)=\arg \left(\frac{90}{65}+\frac{109}{65} i\right) \\
& =\tan ^{-1}\left(\frac{109}{90}\right)=\tan ^{-1}(1.21)
\end{aligned}
$$
Since $\tan \frac{\pi}{4}=1$ and $\tan \frac{\pi}{2}=\infty$
Therefore,
$$
\frac{\pi}{4} < \arg \left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right) < \frac{\pi}{2}
$$
\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}
$$
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