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Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.1

Solution

For each part, use the same inverse formula.

If z=a+ib (z≠0), thenz−1=1a+ib=a−iba2+b2\begin{aligned} & \boxed{\text{If } z=a+ib \ (z\ne 0), \text{ then}} \\ \\ & \boxed{z^{-1}=\frac{1}{a+ib}=\frac{a-ib}{a^2+b^2}} \end{aligned}

Complete part-wise solutions:

  • Part (i): see p-i.mdx
  • Part (ii): see p-ii.mdx
  • Part (iii): see p-iii.mdx
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