Find the modulus and argument of the complex number: $\frac{1+i}{1-i}$

(N/A) Given the complex number $z = \frac{1+i}{1-i}$.
Multiply the numerator and denominator by the conjugate of the denominator $(1+i)$:
$z = \frac{1+i}{1-i} \times \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i$.
We can write $z = 0 + i$.
Let $z = r(\cos \theta + i \sin \theta)$,where $r$ is the modulus and $\theta$ is the argument.
Comparing $0 + i$ with $r \cos \theta + i r \sin \theta$:
$r \cos \theta = 0$ and $r \sin \theta = 1$.
Squaring and adding: $r^2(\cos^2 \theta + \sin^2 \theta) = 0^2 + 1^2 \implies r^2 = 1 \implies r = 1$ (since $r > 0$).
Now,$\cos \theta = 0$ and $\sin \theta = 1$.
This implies $\theta = \frac{\pi}{2}$.
Thus,the modulus is $1$ and the argument is $\frac{\pi}{2}$.