Let $\left|\mathrm{z}\right|=2\text{ and }\mathrm{w}=\frac{\mathrm{z}+1}{\mathrm{z}-1}$where z, $\mathrm{w}\in \mathrm{C}$ (where C is the set of complex numbers). Then product of least and greatest value of modulus of $\mathrm{\omega }$ is ______

Let$\mathrm{z}=\mathrm{a}+\mathrm{ib}$
Given $\left|\mathrm{z}\right|=2\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=4\Rightarrow \mathrm{a},\mathrm{b}\in [-2,2]$
Now, $\mathrm{w}=\frac{(\mathrm{a}+1)+\mathrm{ib}}{(\mathrm{a}-1)+\mathrm{ib}}$
$\begin{array}{l}\Rightarrow \left|\mathrm{w}\right|=\sqrt{\frac{(\mathrm{a}+1{)}^2+{\mathrm{b}}^2}{(\mathrm{a}-1{)}^2+{\mathrm{b}}^2}}\\ =\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{a}+1}{{\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{a}+1}}=\sqrt{\frac{5+2\mathrm{a}}{5-2\mathrm{a}}}\\ |\mathrm{w}{|}_{max}=\sqrt{\frac{5+4}{1}}=3(\text{ when }\mathrm{a}=2)\\ |\mathrm{w}{|}_{min}=\sqrt{\frac{5-4}{9}}=\frac{1}{3}(\text{ when }\mathrm{a}=-2)\end{array}$
Hence, required product is 1.