Find the multiplicative inverse of the following complex numbers:

Solution:

(i) Let $z=1-i$. Then,

$\frac{1}{z}=\frac{1}{1-i}$

$=\frac{1}{1-i} \times \frac{1+i}{1+i}$

$=\frac{1+i}{1-i^{2}}$

$=\frac{1}{2}(1+i)$

$=\frac{1}{2}+\frac{1}{2} i$

(ii) $z=(1+\sqrt{3} i)^{2}$

$=1+3 i^{2}+2 \sqrt{3} i$

$=-2+2 \sqrt{3} i$

Then, $\frac{1}{z}=\frac{1}{-2+2 \sqrt{3} i} \times \frac{-2-2 \sqrt{3} i}{-2-2 \sqrt{3} i}$

$=\frac{-2-2 \sqrt{3} i}{4-12 i^{2}}$

$=\frac{-2-2 \sqrt{3} i}{16}$

$=\frac{-1}{8}-\frac{\sqrt{3}}{8} i$

(iii) $z=4-3 i$

Then, $\frac{1}{z}=\frac{1}{4-3 i} \times \frac{4+3 i}{4+3 i}$

$=\frac{4+3 i}{16-9 i^{2}}$

$=\frac{4+3 i}{25}$

$=\frac{4}{25}+\frac{3}{25} i$

(iv) $z=\sqrt{5}+3 i$

Then, $\frac{1}{z}=\frac{1}{\sqrt{5}+3 i} \times \frac{\sqrt{5}-3 i}{\sqrt{5}-3 i}$

$=\frac{\sqrt{5}-3 i}{5-9 i^{2}}$

$=\frac{\sqrt{5}-3 i}{5+9}$

$=\frac{\sqrt{5}-3 i}{14}$

$=\frac{\sqrt{5}}{14}-\frac{3}{14} i$