Question
Express the following complex numbers in the standard form a + ib: $\frac{3-4\text{i}}{(4-2\text{i})(1+\text{i})}$

Answer

$\frac{3-4\text{i}}{(4-2\text{i})(1+\text{i})}=\frac{3-4\text{i}}{4(1+\text{i})-2\text{i}(1+\text{i})}$ $=\frac{3-4\text{i}}{4+4\text{i}-2\text{i}+2}$ $=\frac{3-4\text{i}}{6+2\text{i}}$ $=\frac{3-4\text{i}}{6+2\text{i}}\times\frac{6-2\text{i}}{6-2\text{i}}$ $=\frac{3(6-2\text{i})-4\text{i}(6-2\text{i})}{6^2+2^2}$ $=\frac{18-6\text{i}-24\text{i}-8}{36+4}$ $=\frac{10(1-3\text{i})}{40}$ $=\frac{1-3\text{i}}{4}$ $=\frac{1}{4}-\frac{3}{4}\text{i}$ $\therefore \ \frac{3-4\text{i}}{(4-2\text{i})(1+\text{i})}=\frac{1}{4}-\frac{3}{4}\text{i}$

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