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Question: If \[\dfrac{\left(w-\overline wz\right)}{1-z}\;\mathrm{is}\;\mathrm{purely}\;\mathrm{real}\;\mathrm{...

If (w−w‾z)1−z  is  purely  real  where  w=α+iβ,  β≠0  and  z≠1\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:∣z∣=1\B.  z:z=z‾C.  z:z≠1\D.  z:∣z∣=1,  z≠1A.\;z:\left|z\right|=1\\\B.\;z:z=\overline z\\\C.\;z:z\neq1\\\D.\;z:\left|z\right|=1,\;z\neq1

Explanation

Solution

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-
w−w‾z1−z=w−w‾z‾1−z‾w−w‾z1−z=w‾−wz‾1−z‾\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-
(w−w‾z)(1−z‾)=(w‾−wz‾)(1−z)\w−wz‾−w‾z+wzz‾=w‾−w‾z−wz‾+w‾zz‾\w−w‾=∣z∣2(w−w‾)∣z∣2=1∣z∣=1,  z≠1  \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.