If $z=\frac{1}{2}-2 i$, is such that $|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$ and $\alpha, \beta \in R \quad$, then $\alpha+\beta$ is equal to
If $z=\frac{1}{2}-2 i$, is such that $|z+1|=\alpha z+\beta(1+i), i=\sqrt{-1}$ and $\alpha, \beta \in R \quad$, then $\alpha+\beta$ is equal to
- [JEE MAIN 2024]
- A
$-4$
- B
$3$
- C
$2$
- D
$-1$
Similar Questions
If $\frac{{2{z_1}}}{{3{z_2}}}$ is a purely imaginary number, then $\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right|$ =
If $\frac{{2{z_1}}}{{3{z_2}}}$ is a purely imaginary number, then $\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right|$ =
If ${z_1} = 10 + 6i,{z_2} = 4 + 6i$ and $z$ is a complex number such that $amp\left( {\frac{{z - {z_1}}}{{z - {z_2}}}} \right) = \frac{\pi }{4},$ then the value of $|z - 7 - 9i|$ is equal to
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- [IIT 1990]
Let $A =\left\{\theta \in(0,2 \pi): \frac{1+2 i \sin \theta}{1- i \sin \theta}\right.$ is purely imaginary $\}$. Then the sum of the elements in $A$ is
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- [JEE MAIN 2023]
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 ${z_1}$ be a complex number with $|{z_1}| = 1$ and ${z_2}$be any complex number, then $\left| {\frac{{{z_1} - {z_2}}}{{1 - {z_1}{{\bar z}_2}}}} \right| = $
Let ${z_1}$ be a complex number with $|{z_1}| = 1$ and ${z_2}$be any complex number, then $\left| {\frac{{{z_1} - {z_2}}}{{1 - {z_1}{{\bar z}_2}}}} \right| = $