Question

If $z_1=2-i,z_2=-2+i,$ find
(i) Re $\left(\frac{z_1{z}_2}{z_1}\right)$
(ii) Im $\left(\frac{1}{z_1{\overline{z}}_1}\right)$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)z_1=2-i,z_2=-2+i,\overline{z_1}=2+i \\ \therefore \left(\frac{z_1{z}_2}{\overline{z_1}}\right)=\left(\frac{\left[2-i\right]\left[-2+i\right]}{2+i}\right) \\ =\left(\frac{-4+2i+2i-i^2}{2+i}\right) \\ =\left(\frac{-3+4i}{2+i}\right) \\ =\left[\frac{-3+4i}{2+i}\times \left(\frac{2-i}{2-i}\right)\right] \\ =\left(\frac{-6+3i+8i-4i^2}{2^2-i^2}\right) \\ =\left(\frac{-2+11i}{4-\left(-1\right)}\right) \\ =\left(\frac{-2+11i}{5}\right) \\ \mathrm{Re}\left(\frac{z_1{z}_2}{\overline{z_1}}\right)=\frac{-2}{5} \\ \left(\mathrm{ii}\right)\left(\frac{1}{z_1\overline{z_1}}\right)=\frac{1}{\left(2-i\right)\left(2+i\right)} \\ =\frac{1}{2^2-i^2} \\ =\frac{1}{5} \\ \mathrm{Im}\left(\frac{1}{z_1\overline{z_1}}\right)=0\left(\mathrm{Since}\mathrm{no}\mathrm{term}\mathrm{containing}\mathit{}i\mathrm{is}\mathrm{present}\right) \end{gathered}$$