If `ta n dc` are two complex numbers such that `|t|!=|c|,|t|=1a n dz=(a t+b)//(t-c), z=x+i ydot` Locus of `z` is (where a, b are complex numbers) a. line segment b. straight line c. circle d. none of these

Options

A. line segment
B. straight line
C. circle
D. none of these

Answer

C

Explanation

`z=(at+b)/(t-c)rArrt=(b+cz)/(z-a)`
Now, `|t|=1`
`rArr |(b+cz)/(z-a)|=1`
`rArr |(z+(b)/(c))/(z-a)|=(1)/(|c|)" " (ne 1" as "|c|ne|t|)`
`rArr` locus of z is a circle