Question

If ${\mathrm{z}}_1=2-\mathrm{i}$ and${\mathrm{z}}_2=1+\mathrm{i}$ then $\left|\frac{\left({\mathrm{z}}_1-{\mathrm{z}}_2+1\right)}{\left({\mathrm{z}}_1+{\mathrm{z}}_2-\mathrm{i}\right)}\right|$ equals

• A. $\sqrt{\frac{5}{3}}$
• B. $\sqrt{\frac{3}{5}}$
• C. $\sqrt{\frac{4}{5}}$
• D. $\sqrt{\frac{5}{4}}$

Answer 1

The correct option is C

$\sqrt{\frac{4}{5}}$

Explanation for the correct option:

Finding the value of $\left|\frac{\left({\mathrm{z}}_1-{\mathrm{z}}_2+1\right)}{\left({\mathrm{z}}_1+{\mathrm{z}}_2-\mathrm{i}\right)}\right|$:

The given points are ${\mathrm{z}}_1=2-\mathrm{i}$ and${\mathrm{z}}_2=1+\mathrm{i}$

As we are all aware, $\left|\frac{\mathrm{a}}{\mathrm{b}}\right|=\frac{\left|\mathrm{a}\right|}{\left|\mathrm{b}\right|}$

$\begin{array}{c}\left|\frac{\left({\text{z}}_1-{\text{z}}_2+1\right)}{\left({\text{z}}_1+{\text{z}}_2-\text{i}\right)}\right|=\left|\frac{2-\mathrm{i}+1+\mathrm{i}+1}{2-\mathrm{i}-1-\mathrm{i}+\mathrm{i}}\right|\\ =\left|\frac{4}{\left(1-\mathrm{i}\right)}\times \frac{\left(1+\mathrm{i}\right)}{\left(1+\mathrm{i}\right)}\right|\\ =\left|\frac{4\left(1+\mathrm{i}\right)}{2}\right|\\ =\left|2+2\mathrm{i}\right|...\left(\mathrm{i}\right)\end{array}$

As we all know, $\left|\mathrm{a}+\mathrm{ib}\right|$ is $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

From $\left(\mathrm{i}\right)$, we get

$\begin{array}{c}\left|2+2\mathrm{i}\right|=\sqrt{{\left(2\right)}^2+{\left(2\right)}^2}\\ =\sqrt{8}\\ =2\sqrt{2}\end{array}$

Hence, the correct option is C.