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Question

A complex number $$z$$ is said to be unimodular if $$|z|=1$$. Suppose $$z_1$$ and $$z_2$$ are complex numbers such that $$\dfrac {z_1-2z_2}{2-z_1\overline z_2}$$ is unimodular and $$z_2$$ is not unimodular. Then the point $$z_1$$ lies on a


A
straight line parallel to x-axis
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B
straight line parallel to y-axis
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C
circle of radius 2
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D
circle of radius 2
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Solution

The correct option is D circle of radius $$2$$
$$\displaystyle \left |\frac {z_1-2z_2}{2-z_1\overline z_2}\right |=1$$

$$\Rightarrow |z_1-2z_2|^2=|2-z_1\overline{z_2}|^2$$

Using the property, $$|a|^2 = a \times \overline{a} $$

$$\Rightarrow (z_1-2z_2)(\overline{z}_1-2\overline{z_2})=(2-z_1\overline{z}_2)(2-\overline z_1z_2)$$

$$\Rightarrow |z_1|^2+4|z_2|^2 - 2z_1\overline{z_2} -2\overline{z_1}z_2  = 4 -2z_1\overline{z_2} -2\overline{z_1}z_2 + |z_1|^2|z_2|^2 $$

$$\Rightarrow |z_1|^2+4|z_2|^2-|z_1|^2|z_2|^2-4=0$$

$$\Rightarrow |z_1|^2(1-|z_2|^2)-4(1-|z_2|^2)=0$$

$$\Rightarrow (|z_1|^2-4)(1-|z_2|^2)=0$$

$$\Rightarrow |z_1|^2=4 \Longrightarrow |z_1|=2$$ Clearly this is locus of circle with radius 2

Mathematics

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