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If z = (3)/( 2 + cos theta + I sin thet...

If `z = (3)/( 2 + cos theta + I sin theta)`, then prove that z lies on the circle.

Text Solution

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Given `z = (3)/(2 + cos theta + i sintheta)`
` rArr cos theta + i sin theta = (3)/(z) - 2 = (3-2z)/(z)`
`rArr 1= (|3-2z|)/(|z|)" "["taking modulus"]`
`rArr (|z-(3)/(2)|)/(|z|) = (1)/(2)`
Therefore, locus of z is circle.
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