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If I m((z-1)/(e^(thetai))+(e^(thetai))/(...

If `I m((z-1)/(e^(thetai))+(e^(thetai))/(z-1))=0` , then find the locus of `zdot`

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Verified by Experts

The correct Answer is:
Circle having centre at `1 + i0` and radius 1

Let `u = (z -1)/(e^(thetai)) rArr (e^(thetai))/(z-1) = (1)/(u)`
Now `(u +(1)/(u)) - (baru +(1)/(baru))= 0`
` rArr (u- baru)(1(1)/(ubaru))=0`
If u is not purely real, then `ubaru = 1`
`rArr |(z-1)/(e^(thetai))| = 1 rArr |z-1|=1`
Hence, z lies on circle haivng centre at 1+ i0 and radius 1.
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