Express each of the following in the form (a + ib) and find its multiplicative

Solution:

(i) Let $Z=\frac{1+2 i}{1-3 i}$

$=\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}=\frac{1+3 i+2 i+6 i^{2}}{1-9 i^{2}}$

$=\frac{1+5 i+6 i^{2}}{1+9}=\frac{-5+5 i}{10}$

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

$\Rightarrow \bar{z}=\frac{-1}{2}-\frac{1}{2} i$

$\Rightarrow|z|^{2}=\left(\frac{-1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

$\therefore$ The multiplicative inverse of $\frac{1+2 i}{1-3 i}$

$z^{-1}=\frac{\bar{z}}{|z|^{2}}=\frac{\frac{-1}{2}-\frac{1}{2} i}{\frac{1}{2}}=-1-i$

(ii) Let $Z=\frac{1+7 i}{(2-i)^{2}}$

$=\frac{1+7 i}{4+i^{2}-4 i}=\frac{1+7 i}{4-1-4 i}=\frac{1+7 i}{3-4 i}$

$=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}$

$=\frac{3+4 i+21 i+28 i^{2}}{9+16}=\frac{3+25 i-28}{25}=\frac{-25+25 i}{25}$

$z=-1+i$

$\Rightarrow \bar{Z}=-1-i$

$\Rightarrow|z|^{2}=(-1)^{2}+(1)^{2}=1+1=2$

$\therefore$ The multiplicative inverse of $\frac{1+7 i}{(2-i)^{2}}$

$z^{-1}=\frac{\bar{z}}{|z|^{2}}=\frac{-1-i}{2}=\frac{-1}{2}-\frac{1}{2} i$

(iii) Let $Z=\frac{-4}{(1+i \sqrt{3})}$

$=\frac{-4}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}$

$=\frac{-4+i 4 \sqrt{3}}{1+3}=\frac{-4+i 4 \sqrt{3}}{4}$

$=-1+i \sqrt{3}$

$Z=-1+i \sqrt{3}$

$\Rightarrow \bar{z}=-1-i \sqrt{3}$

$\Rightarrow|z|^{2}=(-1)^{2}+(\sqrt{3})^{2}=1+3=4$

$\therefore$ The multiplicative inverse of $\frac{-4}{(1+i \sqrt{3})}$

$z^{-1}=\frac{\bar{z}}{|z|^{2}}=\frac{-1+i \sqrt{3}}{4}=\frac{-1}{4}+\frac{i \sqrt{3}}{4}$