Question
If $\text{z}_1=2-\text{i}, \ \text{z}_2=-2+\text{i},$ Find
$\text{Re}\Big(\frac{\text{z}_1\text{z}_2}{\text{z}_1}\Big)$
$\text{Re}\Big(\frac{\text{z}_1\text{z}_2}{\text{z}_1}\Big)$
✓
Answer
$\frac{\text{z}_1\text{z}_2}{\text{z}_1}=\frac{\text{z}_1\text{z}_2}{\text{z}_1}\times\frac{\text{z}_1}{\text{z}_1}$ (rationalising the denominator)
$=\frac{(\text{z}_1)^2\text{z}_2}{\text{z}_1\text{z}_1}$
$=\frac{(2-\text{i})^2(-2-\text{i})}{|\text{z}_1|^2} \ \big(\therefore \ \text{z}\bar{\text{z}=|\text{z}|^2}\big)$
$=\frac{(2^2+\text{i}^2-2\times2\times\text{i})(-2+\text{i})}{|2-\text{i}|^2}$
$=\frac{(4-1-4\text{i}^2)(-2+\text{i})}{2^2+(-1)^2}$
$=\frac{(3-4\text{i})(-2+\text{i})}{4+\text{i}}$
$=3(-2+\text{i})-4\text{i}(-2+\text{i})$
$=\frac{-6+3\text{i}+8\text{i}+4}{5}$
$=\frac{-2+11\text{i}}{5}$
$\therefore \ \text{Re}\Big(\frac{\text{z}_1\text{z}_2}{\text{z}_1}\Big)=\text{Re}\Big(\frac{-2}{5}+\frac{11}{5}\text{i}\Big)$
$=\frac{-2}{5}$
$=\frac{(\text{z}_1)^2\text{z}_2}{\text{z}_1\text{z}_1}$
$=\frac{(2-\text{i})^2(-2-\text{i})}{|\text{z}_1|^2} \ \big(\therefore \ \text{z}\bar{\text{z}=|\text{z}|^2}\big)$
$=\frac{(2^2+\text{i}^2-2\times2\times\text{i})(-2+\text{i})}{|2-\text{i}|^2}$
$=\frac{(4-1-4\text{i}^2)(-2+\text{i})}{2^2+(-1)^2}$
$=\frac{(3-4\text{i})(-2+\text{i})}{4+\text{i}}$
$=3(-2+\text{i})-4\text{i}(-2+\text{i})$
$=\frac{-6+3\text{i}+8\text{i}+4}{5}$
$=\frac{-2+11\text{i}}{5}$
$\therefore \ \text{Re}\Big(\frac{\text{z}_1\text{z}_2}{\text{z}_1}\Big)=\text{Re}\Big(\frac{-2}{5}+\frac{11}{5}\text{i}\Big)$
$=\frac{-2}{5}$
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