# If $\;z_{1}=2-i\;,z_{2}=1+i\;$ find $\;|\large\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}|$

$(a)\;0\qquad(b)\;1\qquad(c)\;\sqrt{3}\qquad(d)\;\sqrt{2}$

Answer : $\sqrt{2}$
Explanation :
$\;z_{1}=2-i\;,z_{2}=1+i\;$
Therefore ,
$\;|\large\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}|= | \large\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+i}|$
$= |\large\frac{4}{2-2i}|$
$= |\large\frac{4}{2(1-i)}|$
$= |\large\frac{2}{1-i} \times \large\frac{1+i}{1+i}|$
$= |\large\frac{2(1+i)}{1^{2}-i^{2}}|$
$= |\large\frac{2(1+i)}{1+1}| \quad [i^2=-1]$
$= |\large\frac{2(1+i)}{2}|$
$=|1+i| = \sqrt{1^{2}+1^{2}}$
$= \sqrt{2}$
Thus ,the value of $\;|\large\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}|\;$ is $\;\sqrt{2}$