The maximum value of $|z|$ where z satisfies the condition $\left| {z + \frac{2}{z}} \right| = 2$ is
The maximum value of $|z|$ where z satisfies the condition $\left| {z + \frac{2}{z}} \right| = 2$ is
- A
$\sqrt 3 - 1$
- B
$\sqrt 3 + 1$
- C
$\sqrt 3 $
- D
$\sqrt 2 + \sqrt 3 $
Similar Questions
If $|{z_1}|\, = \,|{z_2}|$ and $arg\,\,\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \pi $, then ${z_1} + {z_2}$ is equal to
If $|{z_1}|\, = \,|{z_2}|$ and $arg\,\,\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \pi $, then ${z_1} + {z_2}$ is equal to
The amplitude of $\frac{{1 + \sqrt 3 i}}{{\sqrt 3 + 1}}$ is
The amplitude of $\frac{{1 + \sqrt 3 i}}{{\sqrt 3 + 1}}$ 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]
If $\bar z$ be the conjugate of the complex number $z$, then which of the following relations is false
If $\bar z$ be the conjugate of the complex number $z$, then which of the following relations is false
If $|z|\, = 1,(z \ne - 1)$and $z = x + iy,$then $\left( {\frac{{z - 1}}{{z + 1}}} \right)$ is
If $|z|\, = 1,(z \ne - 1)$and $z = x + iy,$then $\left( {\frac{{z - 1}}{{z + 1}}} \right)$ is