If z_1=2-i, z_2=1+i, Find | z_1+z_2+1 z_1-z_2+i |. - Vidyadip

Question

If $\text{z}_1=2-\text{i}, \ \text{z}_2=1+\text{i},$ Find $\Big|\frac{\text{z}_1+\text{z}_2+1}{\text{z}_1-\text{z}_2+\text{i}}\Big|.$

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Answer

If $\text{z}=\text{x}+\text{iy}$ then $|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$
We have,
$\text{z}_1=2-\text{i}, \ \text{z}_2=1+\text{i}$
$\text{z}_1+\text{z}_2=2-\text{i}+1+\text{i}$
$=3$
And $\text{z}_1-\text{z}_2=2-\text{i}-1-\text{i}$
$=1-2\text{i}$
$\Big|\frac{\text{z}_1+\text{z}_2+1}{\text{z}_1-\text{z}_2+\text{i}}\Big|=\frac{3+1}{1-2\text{i}+\text{i}}$
$=\frac{4}{1-\text{i}}$
$=\frac{4}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$
$=\frac{4(1+\text{i})}{1^2+1^2}$
$=\frac{4(1+\text{i})}{2}$
$=2{(1+\text{i})}$
$\therefore\Big|\frac{\text{z}_1+\text{z}_2+1}{\text{z}_1-\text{z}_2+\text{i}}\Big|=|2(1+\text{i})| $
$=|2||1+\text{i}| \ \big(\therefore|\text{z}_1\text{z}_2|=|\text{z}_1|\times|\text{z}_2|\big)$
$=2\times\sqrt{1^2+1^2}$
$=2\times\sqrt{2}$
$=2\sqrt{2}$

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