Question
Find the multiplicative inverse of the following complex numbers:
$1-\text{i}$

Answer

If $\text{z}=\text{x}+\text{iy}$ is a complex number, then the multiplicative inverse of z, denoted by $\text{z}^{-1}$ or $\frac{1}{\text{z}}$ is defined as $\text{z}^{-1}=\frac{1}{\text{z}}$
$=\frac{1}{\text{x}+\text{iy}}$
$=\frac{1}{\text{x}+\text{iy}}\times\frac{\text{x}-\text{iy}}{\text{x}-\text{iy}}$
$=\frac{\text{x}-\text{iy}}{\text{x}^2+\text{y}^2}$
$=\frac{\text{x}}{\text{x}^2+\text{y}^2}-\frac{\text{y}}{\text{x}^2+\text{y}^2}\text{i}$
Given
$\text{z}=1-\text{i}$
$\therefore \ \text{z}^{-1}=\frac{1}{1^2+1^2}-\frac{(-1)}{1^2+1^2}\times\text{i}$

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