If $|z|=1$ ,$z \ne 1$,then maximum value of $\arg\left(\dfrac{1}{1-z}\right)$
$$z=\cos \theta + i\sin \theta ,\qquad -\pi \lt \theta \lt \pi $$ $$\omega=\dfrac{1}{1-z}=\dfrac{1}{1-\cos \theta -i\sin \theta }$$ $$=\dfrac{1-\cos \theta +i\sin \theta}{2(1-\cos \theta)}$$ $$=\frac12 +\frac{i}{2}\tan\left(\frac{\pi}{2} -\frac{\theta}{2}\right)$$
I want to know how I will coclude maximum value of argument from this.
Your equation already shows that the $\omega$ is a straight line running downwards at $x = \frac{1}2$ with $\theta$ running from 0 to $\frac{\pi}2$ (there's the minus sign in the tangent).
Any complex number $Z$ can actually be written as $Z = \text{Re}[Z] + i \text{Re}[Z] \tan\phi$, where $\phi = \text{Arg}[Z]$.
The argument in your tangent function in the imaginary part IS the desired argument.
As $z = e^{i\theta}$ traces out the unit circle, $z_s =1-z$ traces out a circle centered at $x = 1$, one positive unit on the real line.
For any complex number $Z$, the algebraic inverse $\frac{1}Z$ is also the geometric inverse with respect to the unit circle.
The inverse of a circle with respect to another center is a straight line (that necessarily passes through the intersections of the two circles).
The blue curve is $z$ the unit circle, the yellow curve is $z_s = 1-z$, and the green curve is the inverse $\omega = \frac{1}{z_s}$.
Thus as $z_s = 1-z$ traces its circle, the inverse $\omega$ traces out the straightline at $x = \frac{1}2$, with the argument runs from $\frac{\pi}2$ to $-\frac{\pi}2$
