Question

Find the multiplicative inverse of the following complex numbers:
(i) 1 − i
(ii) $(1+i\sqrt{3}{)}^2$
(iii) 4 − 3i
(iv) $\sqrt{5}+3i$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}z=1-i.\mathrm{Then}, \\ \frac{1}{z}=\frac{1}{1-i} \\ =\frac{1}{1-i}\times \frac{1+i}{1+i} \\ =\frac{1+i}{1-i^2} \\ =\frac{1}{2}\left(1+i\right) \\ =\frac{1}{2}+\frac{1}{2}i \\ \left(\mathrm{ii}\right)z={\left(1+\sqrt{3}i\right)}^2 \\ =1+3i^2+2\sqrt{3}i \\ =-2+2\sqrt{3}i \\ \mathrm{Then},\frac{1}{z}=\frac{1}{-2+2\sqrt{3}i}\times \frac{-2-2\sqrt{3}i}{-2-2\sqrt{3}i} \\ =\frac{-2-2\sqrt{3}i}{4-12i^2} \\ =\frac{-2-2\sqrt{3}i}{16} \\ =\frac{-1}{8}-\frac{\sqrt{3}}{8}i \\ \left(\mathrm{iii}\right)z=4-3i \\ \mathrm{Then},\frac{1}{z}=\frac{1}{4-3i}\times \frac{4+3i}{4+3i} \\ =\frac{4+3i}{16-9i^2} \\ =\frac{4+3i}{25} \\ =\frac{4}{25}+\frac{3}{25}i \\ \left(\mathrm{iv}\right)z=\sqrt{5}+3i \\ \mathrm{Then},\frac{1}{z}=\frac{1}{\sqrt{5}+3i}\times \frac{\sqrt{5}-3i}{\sqrt{5}-3i} \\ =\frac{\sqrt{5}-3i}{5-9i^2} \\ =\frac{\sqrt{5}-3i}{5+9} \\ =\frac{\sqrt{5}-3i}{14} \\ =\frac{\sqrt{5}}{14}-\frac{3}{14}i \end{gathered}$$