If the imaginery part of `(z-3)/(e^(itheta))+(e^(itheta))/(z-3)` is zero, then `z` can lie on

Options

A. a circle with unit radius
B. a circle with radius `3` units
C. a straight line through the point `(3,0)`
D. a parabola with the vertex `(3,0)`

Answer

A, C

Explanation

`(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)`.