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 $\sqrt{2}$
Similar Questions
If the equation, $x^{2}+b x+45=0(b \in R)$ has conjugate complex roots and they satisfy $|z+1|=2 \sqrt{10},$ then
If the equation, $x^{2}+b x+45=0(b \in R)$ has conjugate complex roots and they satisfy $|z+1|=2 \sqrt{10},$ then
- [JEE MAIN 2020]
If ${z_1},{z_2} \in C$, then $amp\,\left( {\frac{{{{\rm{z}}_{\rm{1}}}}}{{{{{\rm{\bar z}}}_{\rm{2}}}}}} \right) = $
If ${z_1},{z_2} \in C$, then $amp\,\left( {\frac{{{{\rm{z}}_{\rm{1}}}}}{{{{{\rm{\bar z}}}_{\rm{2}}}}}} \right) = $
Let $z$be a purely imaginary number such that ${\mathop{\rm Im}\nolimits} \,(z) > 0$. Then $arg(z)$ is equal to
Let $z$be a purely imaginary number such that ${\mathop{\rm Im}\nolimits} \,(z) > 0$. Then $arg(z)$ is equal to
If $z_1, z_2 $ are any two complex numbers, then $|{z_1} + \sqrt {z_1^2 - z_2^2} |$ $ + |{z_1} - \sqrt {z_1^2 - z_2^2} |$ is equal to
If $z_1, z_2 $ are any two complex numbers, then $|{z_1} + \sqrt {z_1^2 - z_2^2} |$ $ + |{z_1} - \sqrt {z_1^2 - z_2^2} |$ is equal to
Amplitude of $\left( {\frac{{1 - i}}{{1 + i}}} \right)$ is
Amplitude of $\left( {\frac{{1 - i}}{{1 + i}}} \right)$ is