1.

If the imaginery part of `(z-3)/(e^(itheta))+(e^(itheta))/(z-3)` is zero, then `z` can lie onA. a circle with unit radiusB. a circle with radius `3` unitsC. a straight line through the point `(3,0)`D. a parabola with the vertex `(3,0)`

Answer» Correct Answer - A::C
`(a,c)` Let `z-3=re^(iphi)`
`:. (z-3)/(e^(itheta))+(e^(itheta))/(z-3)=re^(i(phi-theta))+(1)/(r )e^(i(theta-phi))`
Imaginargy part of the above `=r sin (phi-theta)-(1)/(r )sin(phi-theta)`
Given that `r sin(phi-theta)-(1)/(r )sin(phi-theta)=0`
`impliesr-(1)/(r )=0` or `sin(phi-theta)=0`
`impliesr-(1)/(r )=0impliesr=1`
`implies|z-3|=1`
`impliesz` lies on a circle with unit radius.
`sin(phi-theta)=0impliesphi=theta`
`:.z-3=re^(itheta)`
`z-3=rcostheta`, `y=r sin theta`
`impliesx-3=rcostheta`, `y=rsin theta`
`(x-3)/(costheta)=(y)/(sintheta)=r`
This represents a straight line through `(3,0)`.


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