MHT CET · Maths · Complex Number
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
- 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 \(\tan ^{-1}\left(\frac{19}{22}\right)\)
Answer & Solution
Correct Answer
(A) \(\tan ^{-1}\left(\frac{22}{19}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{z_1+z_2}{z_1-z_2}=\frac{8-i}{2-3 i} \)
\( =\frac{8-i}{2-3 i} \times \frac{2+3 i}{2+3 i} \)
\( =\frac{19+22 i}{13}=\frac{19}{13}+\frac{22 i}{13} \)
\( \therefore \arg \left(\frac{z_1+z_2}{z_1-z_2}\right)=\tan ^{-1}\left(\frac{22}{19}\right)\)
\( =\frac{8-i}{2-3 i} \times \frac{2+3 i}{2+3 i} \)
\( =\frac{19+22 i}{13}=\frac{19}{13}+\frac{22 i}{13} \)
\( \therefore \arg \left(\frac{z_1+z_2}{z_1-z_2}\right)=\tan ^{-1}\left(\frac{22}{19}\right)\)
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