Question

Find the conjugates of the following complex numbers:
(i) 4 − 5 i
(ii) $\frac{1}{3+5i}$
(iii) $\frac{1}{1+i}$
(iv) $\frac{(3-i{)}^2}{2+i}$
(v) $\frac{(1+i)(2+i)}{3+i}$
(vi) $\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}z=4-5i \\ \therefore \overline{z}=4+5i\left(z=a+ib,\mathrm{so}\overline{z}=a-ib\right) \\ \left(\mathrm{ii}\right)\mathrm{Let}z=\frac{1}{3+5i} \\ =\frac{1}{3+5i}\times \frac{3-5i}{3-5i} \\ =\frac{3-5i}{9-25i^2} \\ =\frac{3-5i}{9+25} \\ =\frac{3-5i}{34} \\ \therefore \overline{z}=\frac{3+5i}{34} \\ \left(\mathrm{iii}\right)\mathrm{Let}z=\frac{1}{1+i} \\ =\frac{1}{1+i}\times \frac{1-i}{1-i} \\ =\frac{1-i}{1-i^2} \\ =\frac{1-i}{2} \\ \Rightarrow \overline{z}=\frac{1+i}{2} \\ \left(\mathrm{iv}\right)\mathrm{Let}z=\frac{{\left(3-i\right)}^2}{2+i} \\ =\frac{\left(9-6i-1\right)}{2+i} \\ =\frac{8-6i}{2+i}\times \frac{2-i}{2-i} \\ =\frac{16-8i-12i+6i^2}{4-i^2} \\ =\frac{10-20i}{5} \\ =2-4i \\ \therefore z=2+4i \\ \left(\mathrm{v}\right)\mathrm{Let}z=\frac{\left(1+i\right)\left(2+i\right)}{3+i} \\ =\frac{2+i+2i+i^2}{3+i} \\ =\frac{1+3i}{3+i} \\ =\frac{1+3i}{3+i}\times \frac{3-i}{3-i} \\ =\frac{3-i+9i-3i^2}{9-i^2} \\ =\frac{6+8i}{10} \\ =\frac{3+4i}{5} \\ \Rightarrow \overline{z}=\frac{3-4i}{5} \\ \left(\mathrm{vi}\right)\mathrm{Let}z=\frac{\left(3-2i\right)\left(2+3i\right)}{\left(1+2i\right)\left(2-i\right)} \\ =\frac{6+9i-4i-6i^2}{2-i+4i-2i^2} \\ =\frac{6+6+5i}{2+2+3i} \\ =\frac{12+5i}{4+3i}\times \frac{4-3i}{4-3i} \\ =\frac{48-36i+20i-15i^2}{16-9i^2} \\ =\frac{63-16i}{25} \\ \therefore \overline{z}=\frac{63+16i}{25} \end{gathered}$$