The product of two complex numbers each of unit modulus is also a complex number, of
The product of two complex numbers each of unit modulus is also a complex number, of
- A
Unit modulus
- B
Less than unit modulus
- C
Greater than unit modulus
- D
None of these
Similar Questions
If the conjugate of $(x + iy)(1 - 2i)$ be $1 + i$, then
If the conjugate of $(x + iy)(1 - 2i)$ be $1 + i$, then
If $|z|\, = 1$ and $\omega = \frac{{z - 1}}{{z + 1}}$ (where $z \ne - 1)$, then ${\mathop{\rm Re}\nolimits} (\omega )$ is
If $|z|\, = 1$ and $\omega = \frac{{z - 1}}{{z + 1}}$ (where $z \ne - 1)$, then ${\mathop{\rm Re}\nolimits} (\omega )$ is
- [IIT 2003]
The moduli of two complex numbers are less than unity, then the modulus of the sum of these complex numbers
The moduli of two complex numbers are less than unity, then the modulus of the sum of these complex numbers
The set of all $\alpha \in R$, for which $w = \frac{{1 + \left( {1 - 8\alpha } \right)z}}{{1 - z}}$ is a purely imaginary number, for all $z \in C$ satisfying $\left| z \right| = 1$ and ${\mathop{\rm Re}\nolimits} \,z \ne 1$, is
The set of all $\alpha \in R$, for which $w = \frac{{1 + \left( {1 - 8\alpha } \right)z}}{{1 - z}}$ is a purely imaginary number, for all $z \in C$ satisfying $\left| z \right| = 1$ and ${\mathop{\rm Re}\nolimits} \,z \ne 1$, is
- [JEE MAIN 2018]
Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0, z \in C$. Then $4\left(\alpha^2+\beta^2\right)$ is equal to :
Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0, z \in C$. Then $4\left(\alpha^2+\beta^2\right)$ is equal to :
- [JEE MAIN 2024]