# Let . Find

Question:

Let $z_{1}=2-i, z_{2}=-2+i$. Find

(i) $\operatorname{Re}\left(\frac{\mathrm{z}_{1} \mathrm{z}_{2}}{\overline{\mathrm{z}}_{1}}\right)$,

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

Solution:

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

(i) $\mathrm{z}_{1} \mathrm{z}_{2}=(2-\mathrm{i})(-2+\mathrm{i})=-4+2 \mathrm{i}+2 \mathrm{i}-\mathrm{i}^{2}=-4+4 \mathrm{i}-(-1)=-3+4 \mathrm{i}$

$\bar{z}_{1}=2+i$

$\therefore \frac{\mathrm{z}_{1} \mathrm{z}_{2}}{\overline{\mathrm{z}}_{1}}=\frac{-3+4 \mathrm{i}}{2+\mathrm{i}}$

On multiplying numerator and denominator by $(2-i)$, we obtain

$\frac{z_{1} z_{2}}{\bar{z}_{1}}=\frac{(-3+4 i)(2-i)}{(2+i)(2-i)}=\frac{-6+3 i+8 i-4 i^{2}}{2^{2}+1^{2}}=\frac{-6+1 l i-4(-1)}{2^{2}+1^{2}}$

$\frac{\mathrm{z}_{1} \mathrm{z}_{2}}{\mathrm{z}_{1}}=\frac{(-3+4 \mathrm{i})(2-\mathrm{i})}{(2+\mathrm{i})(2-\mathrm{i})}=\frac{-6+3 \mathrm{i}+8 \mathrm{i}-4 \mathrm{i}^{2}}{2^{2}+1^{2}}=\frac{-6+1 \mathrm{li}-4(-1)}{2^{2}+1^{2}}$

$=\frac{-2+11 \mathrm{i}}{5}=\frac{-2}{5}+\frac{11}{5} \mathrm{i}$

On comparing real parts, we obtain

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

(ii) $\frac{1}{z_{1} \bar{z}_{1}}=\frac{1}{(2-i)(2+i)}=\frac{1}{(2)^{2}+(1)^{2}}=\frac{1}{5}$

On comparing imaginary parts, we obtain

$\operatorname{Im}\left(\frac{1}{z_{1} \bar{z}_{1}}\right)=0$