Find the modulus and argument of the complex numbers:
$\frac{1+i}{1-i}$
Find the modulus and argument of the complex numbers:
$\frac{1+i}{1-i}$
We have, $\frac{1+i}{1-i}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}=\frac{1-1+2 i}{1+1}=i=0+i$
Now, let us put $0=r \cos \theta, \quad 1=r \sin \theta$
Squaring and adding, $r^{2}=1$ i.e., $r=1$ so that
$\cos \theta=0, \sin \theta=1$
Therefore, $\theta=\frac{\pi}{2}$
Hence, the modulus of $\frac{1+i}{1-i}$ is $1$ and the argument is $\frac{\pi}{2}$.
Similar Questions
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]
If $z = \cos \frac{\pi }{6} + i\sin \frac{\pi }{6}$ then
If $z = \cos \frac{\pi }{6} + i\sin \frac{\pi }{6}$ then
The amplitude of the complex number $z = \sin \alpha + i(1 - \cos \alpha )$ is
The amplitude of the complex number $z = \sin \alpha + i(1 - \cos \alpha )$ is
Find the modulus and the argument of the complex number $z=-\sqrt{3}+i$
Find the modulus and the argument of the complex number $z=-\sqrt{3}+i$
If $arg\, z < 0$ then $arg\, (-z)\, -arg(z)$ is equal to
If $arg\, z < 0$ then $arg\, (-z)\, -arg(z)$ is equal to