Question:
Express $(1-2 i)^{-3}$ in the form $(a+i b)$
Solution:
We have, $(1-2 i)^{-3}$
$\Rightarrow \frac{1}{(1-2 i)^{3}}=\frac{1}{1-8 i^{3}-6 i+12 i^{2}}=\frac{1}{1+8 i-6 i-12}=\frac{1}{2 i-11}$
$\Rightarrow \frac{1}{-11+2 i}$
$=\frac{1}{-11+2 i} \times \frac{-11-2 i}{-11-2 i}$
$=\frac{-11-2 i}{(-11)^{2}-(2 i)^{2}}=\frac{-11-2 i}{121+4}$
$=\frac{-11-2 i}{125}$
$=\frac{-11}{125}-\frac{2 i}{125}$
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