If $x=\frac{5}{1-2 \mathrm{i}}, \mathrm{i}=\sqrt{-1}$, then the value of $x^3+x^2-x+22$ is

If $x=\frac{5}{1-2 \mathrm{i}}, \mathrm{i}=\sqrt{-1}$, then the value of $x^3+x^2-x+22$ is
  1. $7$
  2. $9$
  3. $17$
  4. $39$

Solution

$\begin{aligned} & x=\frac{5}{1-2 \mathrm{i}}=\frac{5(1+2 \mathrm{i})}{1+4}=1+2 \mathrm{i} \\ & \therefore \quad x^2=(1+2 \mathrm{i})^2=1-4+4 \mathrm{i}=-3+4 \mathrm{i} \\ & \therefore \quad x^3=(-3+4 \mathrm{i})(1+2 \mathrm{i}) \\ & =-3-6 i+4 i-8 \\ & =-11-2 \mathrm{i} \\ & \therefore \quad x^3+x^2-x+22 \\ & =(-11-2 \mathrm{i})+(-3+4 \mathrm{i})-(1+2 \mathrm{i})+22 \\ & =-11-2 \mathrm{i}-3+4 \mathrm{i}-1-2 \mathrm{i}+22 \\ & =7 \\ & \end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 2)

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