If |z- bar z| + |z+ bar z| = 2, then z lies on (1) a circle (2) a square (3) an ellipse (4) a line

Solution:

Given \(\left | z-\bar{z} \right |+\left | z+\bar{z} \right |=2\)

Consider z = x+iy

\(\bar{z}=x-iy\) \(z+\bar{z}= 2x\) \(z-\bar{z}= 2iy\)

So |2iy|+|2x| = 2

2y+2x = 2

=> x+y = 1

z lies on a straight line.

Hence option (4) is the answer.

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