If $z_{1}=2-i, z_{2}=1+i,$ find $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|$
If $z_{1}=2-i, z_{2}=1+i,$ find $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|$
$z_{1}=2-i, z_{2}=1+i$
$\therefore\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|=\left|\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}\right|$
$=\left|\frac{4}{2-2 i}\right|=\left|\frac{4}{2(1-i)}\right|$
$=\left|\frac{2}{1-i} \times \frac{1+i}{1+i}\right|=\left|\frac{2(1+i)}{\left(1^{2}-i^{2}\right)}\right|$
$=\left|\frac{2(1+i)}{1+1}\right| \quad\left[i^{2}=-1\right]$
$=\left|\frac{2(1+i)}{2}\right|$
$=|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2}$
Thus, the value of $\left|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|$ is $\sqrt{2}$
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