If the imaginary part of the expressionz−1eθi+eθiz−1 be zero, then the locus of z can be

Letu=z−1eθi⇒eθiz−1=1u .Now (u+1u)−(u¯+1u)=0⇒(u−u¯)(1−1uu¯)=0If u is not purely real, then uu=1⇒|z−1eθi|=1⇒|z−1|=1

If the imaginary part of the expressionz−1eθi+eθiz−1 be zero, then the locus of z can be

Letu=z−1eθi⇒eθiz−1=1u .Now (u+1u)−(u¯+1u)=0⇒(u−u¯)(1−1uu¯)=0If u is not purely real, then uu=1⇒|z−1eθi|=1⇒|z−1|=1

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If the imaginary part of the expression $\frac{z - 1}{e^{θ i}} + \frac{e^{θ i}}{z - 1}$ be zero, then the locus of z can be

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A straight line parallel to $x -$ axis

None of these

A parabola

A circle of radius 1

answer is C.

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Let $u = \frac{z - 1}{e^{θ i}} \Rightarrow \frac{e^{θ i}}{z - 1} = \frac{1}{u}$ .

Now $(u + \frac{1}{u}) - (\bar{u} + \frac{1}{u}) = 0 \Rightarrow (u - \bar{u}) (1 - \frac{1}{u \bar{u}}) = 0$

If $u$ is not purely real, then $u u = 1 \Rightarrow | \frac{z - 1}{e^{θ i}} | = 1 \Rightarrow | z - 1 | = 1$

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If the imaginary part of the expressionz−1eθi+eθiz−1 be zero, then the locus of z can be

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If the imaginary part of the expression $\frac{z - 1}{e^{θ i}} + \frac{e^{θ i}}{z - 1}$ be zero, then the locus of z can be