Question
Express the following complex numbers in the standard form a + ib: $(1-2\text{i})^{-3}$

Answer

$(1-2\text{i})^{-3}=\frac{1}{(1+2\text{i}^3)} \ \Big(\therefore \ \text{z}^{-3}=\frac{1}{\text{z}^3}\Big)$ $=\frac{1}{1^3+(2\text{i})^3+3\times1\times2\text{i}(1+2\text{i})}$ $=\frac{1}{1^3+2^3\times\text{i}^3+6\text{i}(1+2\text{i})}$ $=\frac{1}{1-8\text{i}+6\text{i}-12} \ \big(\therefore \ \text{i}^3=-\text{i} \ \text{and} \ \text{i}^2=-1\big)$ $=\frac{1}{-11-2\text{i}}$ $=\frac{1}{-11-2\text{i}}\times\frac{(-11+2\text{i})}{(-11+2\text{i})}$ $=\frac{-11+2\text{i}}{121+4}$ $=\frac{-11}{125}+\frac{2}{125}\text{i}$ $\therefore \ (1-2\text{i})^{-3}=\frac{-11}{125}+\frac{2}{125}\text{i}$

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