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
Let $S$ be the set of all complex numbers $z$ satisfying $\left|z^2+z+1\right|=1$. Then which of the following statements is/are $TRUE$?
$(A)$ $\left|z+\frac{1}{2}\right| \leq \frac{1}{2}$ for all $z \in S$ $(B)$ $|z| \leq 2$ for all $z \in S$
$(C)$ $\left|z+\frac{1}{2}\right| \geq \frac{1}{2}$ for all $z \in S$ $(D)$ The set $S$ has exactly four elements
Let $S$ be the set of all complex numbers $z$ satisfying $\left|z^2+z+1\right|=1$. Then which of the following statements is/are $TRUE$?
$(A)$ $\left|z+\frac{1}{2}\right| \leq \frac{1}{2}$ for all $z \in S$ $(B)$ $|z| \leq 2$ for all $z \in S$
$(C)$ $\left|z+\frac{1}{2}\right| \geq \frac{1}{2}$ for all $z \in S$ $(D)$ The set $S$ has exactly four elements
- [IIT 2020]
Let $a = lm\left( {\frac{{1 + {z^2}}}{{2iz}}} \right)$, where $z$ is any non-zero complex number. The set $A = \{ a:\left| z \right| = 1\,and\,z \ne \pm 1\} $ is equal to
Let $a = lm\left( {\frac{{1 + {z^2}}}{{2iz}}} \right)$, where $z$ is any non-zero complex number. The set $A = \{ a:\left| z \right| = 1\,and\,z \ne \pm 1\} $ is equal to
- [JEE MAIN 2013]
The argument of the complex number $\frac{{13 - 5i}}{{4 - 9i}}$is
The argument of the complex number $\frac{{13 - 5i}}{{4 - 9i}}$is
Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+ i \sin \left(\frac{2 k \pi}{10}\right) ; k =1,2, \ldots 9$.
List $I$
List $II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$
$1.$ True
$Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers.
$2.$ False
$R.$ $\frac{\left|1-z_1\right|\left|1-z_2\right| \ldots . .\left|1-z_9\right|}{10}$ equals
$3.$ $1$
$S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals
$4.$ $2$
Codes: $ \quad P \quad Q \quad R \quad S$
Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+ i \sin \left(\frac{2 k \pi}{10}\right) ; k =1,2, \ldots 9$.
| List $I$ | List $II$ |
| $P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$ | $1.$ True |
| $Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers. | $2.$ False |
| $R.$ $\frac{\left|1-z_1\right|\left|1-z_2\right| \ldots . .\left|1-z_9\right|}{10}$ equals | $3.$ $1$ |
| $S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals | $4.$ $2$ |
Codes: $ \quad P \quad Q \quad R \quad S$
- [IIT 2014]
Let $z_1, z_2 \in C$ such that $| z_1 + z_2 |= \sqrt 3$ and $|z_1| = |z_2| = 1,$ then the value of $|z_1 - z_2|$ is
Let $z_1, z_2 \in C$ such that $| z_1 + z_2 |= \sqrt 3$ and $|z_1| = |z_2| = 1,$ then the value of $|z_1 - z_2|$ is