Question
Express $(1-2 i)^{-3}$ in the form of $(a+i b)$.

Answer

Let $z=(1-2 i)^{-3}$
$=\frac{1}{(1-2 i)^3}=\frac{1}{1-8 i^3-6 i+12 i^2}\left[\because(a-b)^3=a^3-b^3-3 a^2 b+3 a b^2\right]$
$=\frac{1}{1-8 i^2 \cdot i-6 i+12(-1)}$
$=\frac{1}{1+8 i-6 i-12}\left[\because i^2=-1\right]$
$=\frac{1}{-11+2 i}=\frac{1}{-11+2 i} \times \frac{-11-2 i}{-11-2 i}[\text { multiplying numerator and denominator by }-11-2 i]$
$=\frac{-11-2 i}{(-11)^2-(2 i)^2}=\frac{-11-2 i}{121+4}\left[\because(a-b)(a+b)=a^2-b^2\right]$
$=\frac{-11-2 i}{125}=\frac{-11}{125}-\frac{2 i}{125}=a+i b[\text { say }]$
where, $a=\frac{-11}{125}$ and $b =\frac{-2}{125}$

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