MHT CET · Maths · Complex Number
If \(Z_1=2+i\) and \(Z_2=3-4 i\) and \(\frac{\overline{Z_1}}{Z_1}=a+b i\), then the value of \(-7 a+b\) is (where \(i=\sqrt{-1}\) and \(\mathrm{a}, \mathrm{b} \in \mathbb{R})\)
- A 1
- B -1
- C \(\frac{-3}{25}\)
- D \(\frac{-9}{25}\)
Answer & Solution
Correct Answer
(B) -1
Step-by-step Solution
Detailed explanation
\(\mathrm{Z}_1=2+\mathrm{i} \quad \mathrm{Z}_2=3-4 \mathrm{i} \)
\( \overline{Z_1}=2-\mathrm{i} \quad \overline{Z_2}=3+4 \mathrm{i} \)
\( \frac{\overline{Z_1}}{\overline{Z_2}}=\frac{2-\mathrm{i}}{3+4 \mathrm{i}} \)
\( =\frac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)} \)
\( =\frac{6-8 i-3 i+4 i^2}{(3)^2-(4 i)^2} \)
\( =\frac{6-1 \mathrm{li}-4}{9+16} \quad \ldots\left[\mathrm{i}^2=-1\right] \)
\( \mathrm{a}+\mathrm{bi}=\frac{2-1 \mathrm{li}}{25} \)
\( \therefore a=\frac{2}{25}, b=\frac{-11}{25}\)
Now \(-7 \mathrm{a}+\mathrm{b} \)
\( =-7\left(\frac{2}{25}\right)-\frac{11}{25} \)
\( =\frac{-14-11}{25} \)
\( =\frac{-25}{25}=-1\)
\( \overline{Z_1}=2-\mathrm{i} \quad \overline{Z_2}=3+4 \mathrm{i} \)
\( \frac{\overline{Z_1}}{\overline{Z_2}}=\frac{2-\mathrm{i}}{3+4 \mathrm{i}} \)
\( =\frac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)} \)
\( =\frac{6-8 i-3 i+4 i^2}{(3)^2-(4 i)^2} \)
\( =\frac{6-1 \mathrm{li}-4}{9+16} \quad \ldots\left[\mathrm{i}^2=-1\right] \)
\( \mathrm{a}+\mathrm{bi}=\frac{2-1 \mathrm{li}}{25} \)
\( \therefore a=\frac{2}{25}, b=\frac{-11}{25}\)
Now \(-7 \mathrm{a}+\mathrm{b} \)
\( =-7\left(\frac{2}{25}\right)-\frac{11}{25} \)
\( =\frac{-14-11}{25} \)
\( =\frac{-25}{25}=-1\)
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