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

Answer

$\frac{(1+\text{i})(1+\sqrt{3}\text{i})}{1-\text{i}}=\frac{1(1+\sqrt{3}\text{i})+\text{i}(1+\sqrt{3}\text{i})}{1-\text{i}}$ $=\frac{(1+\sqrt{3}\text{i}+\text{i}-\sqrt{3})}{1-\text{i}} \ \big(\therefore \ \text{i}^2=-1\big)$ $=\frac{(1-\sqrt{3})+\text{i}(1+\sqrt{3})}{1-\text{i}}\times\frac{(1+\text{i})}{1+\text{i}}$ (Rationalising the denominator) $=\frac{(1-\sqrt{3})(1+\text{i})+\text{i}(1+\sqrt{3})(1 +\text{i})}{1^2+1^2}$ $=\frac{1+\text{i}-\sqrt{3}(1+\text{i})+\text{i}(1+\text{i}+\sqrt{3}(1+\text{i}))}{2}$ $=\frac{1+\text{i}-\sqrt{3}-\sqrt{3}\text{i}+\text{i}(1+\text{i}+\sqrt{3}+\sqrt{3}\text{i})}{2}$ $=\frac{1-\sqrt{3}+\text{i}-\sqrt{3}\text{i}+\text{i}-1+\sqrt{3}\text{i}+\sqrt{3}}{2}$ $=\frac{-2\sqrt{3}+2\text{i}}{2}$ $=\sqrt{3}+\text{i}$ $\therefore \ \frac{(1+\text{i})(1+\sqrt{3}\text{i})}{1-\text{i}}=-\sqrt{3}+\text{i}$

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