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 $\frac{\pi}{2}$.
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- [JEE MAIN 2013]