Q47 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Mathematics

If $Z=\alpha+i \beta$ satisfies the equation $|Z|-Z=1+2 i$ and $|Z|=\sqrt{\alpha^2+\beta^2}$ then $Z \bar{Z}=$

  • A
    $\frac{5}{2}$
  • Correct answer
    B
    $\frac{25}{4}$
  • C
    $\frac{16}{9}$
  • D
    $\frac{36}{25}$

Solution

Given $Z=\alpha+i \beta$ Satisfies, $|\mathrm{Z}|-|\mathrm{z}|=1+2 \mathrm{i}$ $$ \begin{aligned} & \Rightarrow \sqrt{\alpha^2+\beta^2}-\alpha-i \beta=1+2 \mathrm{i} \\ & \Rightarrow \sqrt{\alpha^2+\beta^2}-\alpha=1, \beta=-2 \\ & \Rightarrow \alpha=\frac{3}{2} \end{aligned} $$ Now $Z \cdot \bar{Z}=\alpha^2+\beta^2=\frac{9}{4}+4=\frac{25}{4}$

Hence, the correct answer is option: 'B'.

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