Single Choice

Let $$w(Im w\neq 0)$$ be a complex number. Then the set of all complex number z satisfying the equation $$w-\overline wz=k(1-z)$$, for some real number k, is :

A$$\left \{z:|z|=1 \right \}$$
B$$\left \{z:z=\overline z \right \}$$
C$$\left \{z:z\neq 1 \right \}$$
D$$\left \{z:|z|=1, z\neq 1 \right \}$$
Correct Answer

Solution

let 'w' be $$a+ib$$ and z be $$'x+iy'$$

$$a+ib-(a-ib)(x+iy)=k(1-x-iy)$$

$$a+ib-[ax+ayi-bxi+by]=k(1-x)-iky$$

$$a+ib-ax-ayi+bxi-by=k(1-x)-kyi$$

$$(a-ax-by)+i(b-ay+bx)=k(1-x)-kyi$$
.
comparing real and imaginary coefficients from both sides.

$$k=\dfrac {(a-ax-by)}{(1-x)}, k=\left (\dfrac {ay-bx-b}{y}\right )$$

$$\dfrac {a-ax-by}{1-x}=\dfrac {ay-bx-b}{y}$$

$$ay-axy-by^2=ay-bx-b-ayx+bx^2+bx$$

$$-by^2=bx^2-b$$

$$b=by^2+bx^2$$

$$x^2+y^2=1$$

$$|z|=1$$

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