If $$\dfrac{\left(w-\overline wz\right)}{1-z}\;\mathrm{is}\;\mathrm{purely}\;\mathrm{real}\;\mathrm{where}\;w=\alpha+i\beta,\;\beta\neq0\;and\;z\neq1$$, then set of values of z is-
$$A.\;z:\left|z\right|=1\\B.\;z:z=\overline z\\C.\;z:z\neq1\\D.\;z:\left|z\right|=1,\;z\neq1$$

Hint: One should have knowledge of complex numbers to solve this problem. The conjugate of a purely real number is the same as the number. It can be converted by changing the sign of the iota term.

Complete step by step answer:
Now, the given expression is purely real hence-
$$\dfrac{w-\overline wz}{1-z}=\dfrac{\overline{w-\overline wz}}{\overline{1-z}}\\\dfrac{w-\overline wz}{1-z}=\dfrac{\overline w-w\overline z}{1-\overline z}$$
Cross multiplying both sides we get-
$$\left(w-\overline wz\right)\left(1-\overline z\right)=\left(\overline w-w\overline z\right)\left(1-z\right)\\w-w\overline z-\overline wz+wz\overline z=\overline w-\overline wz-w\overline z+\overline wz\overline z\\w-\overline w=\left|z\right|^2\left(w-\overline w\right)\\\left|z\right|^2=1\\\left|z\right|=1,\;z\neq1\;$$

So, the correct answer is “Option D”.

Note: The value of w is given in the question, so students may substitute that value and try to solve the question. But it is a very lengthy method to do. Instead we should use the properties that have been given to us to solve the problem. One should also know that the product of a complex number and its conjugate is equal to the square of their modulus.