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Model Paper 4 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 41 Mark
MCQ
The multiplicative inverse of $(3+2 i)^2$ is
  • A
    $\left(\frac{-4}{169}+\frac{12}{169} i\right)$
  • B
    $\left(\frac{5}{169}-\frac{12}{169} i\right)$
  • C
    $\left(\frac{-5}{169}+\frac{12}{169} i\right)$
  • D
    $\left(\frac{5}{169}+\frac{12}{169} i\right)$
✓

Answer

(b) $\left(\frac{5}{165}-\frac{12}{100} i\right)$
Explanatien: $z=(3+2 i)^2=\left(9+4 i^2+12 i\right)=(9-4+12 i)=(5+120)$
$\begin{array}{l}\Rightarrow z^{-1}=\frac{1}{(5+12 i)} \times \frac{(5-12 i)}{(5-12 i)}=\frac{(5-12 i)}{\left(25-144 i^2\right)}=\frac{(5-12 i)}{(25+144)}=\frac{(5-12 i)}{(16 i)} \\ \Rightarrow z^{-1}=\left(\frac{5}{169}-\frac{12}{169} i\right)\end{array}$

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