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

Solution:

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

$=\frac{4-3 i}{16+9}=\frac{4}{25}-\frac{3}{25} i$

$\Rightarrow \quad \bar{z}=\frac{4}{25}+\frac{3}{25} i$

(ii) Let $z=(2+3 i)^{2}=(2+3 i)(2+3 i)$

$=4+6 i+6 i+9 i^{2}$

$=4+12 i+9 i^{2}$

$=4+12 i-9$

$=-5+12 i$

$\bar{z}=-5-12 i$

(iii) Let $Z=\frac{(2-i)}{(1-2 i)^{2}}=\frac{(2-i)}{1+4 i^{2}-4 i}$

$=\frac{(2-i)}{1-4 i-4}=\frac{2-i}{-3-4 i}$

$=\frac{2-i}{-3-4 i} \times \frac{-3+4 i}{-3+4 i}=\frac{(2-i)(-3+4 i)}{9+16}$

$=\frac{-6+11 i-4 i^{2}}{25}=\frac{-2+11 i}{25}$

$=\frac{-2}{25}+\frac{11}{25} i$

$\bar{z}=\frac{-2}{25}-\frac{11}{25} i$

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

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

$=\frac{-1+3 i}{1+3 i} \times \frac{1-3 i}{1-3 i}=\frac{-1+3 i+3 i-9 i^{2}}{1-9 i^{2}}=\frac{-1+6 i+9}{1+9}=\frac{8+6 i}{10}$

$=\frac{8}{10}+\frac{6}{10} i$

$\bar{z}=\frac{8}{10}-\frac{6}{10} i$

(v) Let $Z=\left(\frac{1+2 i}{2+i}\right)^{2}=\frac{1+4 i^{2}+2 i}{4+i^{2}+4 i}=\frac{1-4+2 i}{4-1+4 i}=\frac{-3+2 i}{3+4 i}$

$=\frac{-3+2 i}{3+4 i} \times \frac{3-4 i}{3-4 i}$

$=\frac{-9+12 i+6 i-8 i^{2}}{9+16}=\frac{-9+18 i+8}{25}=\frac{-1+18 i}{25}$

$=\frac{-1}{25}+\frac{18}{25} i$

$\bar{z}=\frac{-1}{25}-\frac{18}{25} i$

(vi) Let $Z=\frac{(2+i)}{(3-i)(1+2 i)}=\frac{2+i}{3+6 i-1-2 i^{2}}$

$=\frac{2+i}{3+6 i-1+2}=\frac{2+i}{4+6 i}$

$=\frac{2+i}{4+6 i} \times \frac{4-6 i}{4-6 i}$

$=\frac{8-12 i+4 i-6 i^{2}}{16+36}$

$=\frac{8-8 i+6}{52}$

$=\frac{14-8 i}{52}$

$=\frac{14}{52}-\frac{8}{52} i$

$\bar{z}=\frac{14}{52}+\frac{8}{52} i$