Solve the following
Question:

If $z_{1}=2-i, z_{2}=-2+i$, find

(i) $\operatorname{Re}\left(\frac{z_{1} z_{2}}{z_{1}}\right)$

(ii) $\operatorname{lm}\left(\frac{1}{z_{1} \bar{z}_{1}}\right)$

Solution:

(i) $z_{1}=2-i, z_{2}=-2+i, \overline{z_{1}}=2+i$

$\therefore\left(\frac{z_{1} z_{2}}{z_{1}}\right)=\left(\frac{[2-i][-2+i]}{2+i}\right)$

$=\left(\frac{-4+2 i+2 i-i^{2}}{2+i}\right)$

$=\left(\frac{-3+4 i}{2+i}\right)$

$=\left[\frac{-3+4 i}{2+i} \times\left(\frac{2-i}{2-i}\right)\right]$

$=\left(\frac{-6+3 i+8 i-4 i^{2}}{2^{2}-i^{2}}\right)$

$=\left(\frac{-2+11 i}{4-(-1)}\right)$

$=\left(\frac{-2+11 i}{5}\right)$

$\operatorname{Re}\left(\frac{z_{1} z_{2}}{z_{1}}\right)=\frac{-2}{5}$

(ii) $\left(\frac{1}{z_{1} \overline{z_{1}}}\right)=\frac{1}{(2-i)(2+i)}$

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

$=\frac{1}{5}$

$\operatorname{Im}\left(\frac{1}{z_{1} \overline{z_{1}}}\right)=0$

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