Consider the equation `az + bar(bz) + c =0`, where a,b,c `in`Z
If `|a| ne |b|`, then z represents

Options

A. circle
B. straight line
C. one point
D. ellispe

Answer

C

Explanation

`az + b barz + c=0" "(1)`
or `barabarz+ baraz+barc = 0" "(2)`
Eliminating `barz` form (1) and (2) we, get
`z = (cbara - b barc)/(|b|^(2)-|a|^(2))`
If `|a|ne|b|`, then z represents one point on the Argand Plane. If
`|a| = |b| and barac ne b barc`, then no such z exists. Adding (1) and (2),
`(bara + b)barz +(a+ barb) z + (c + barc) = 0`
This is of the form `Abarz + Abarz + B = 0` Where `B = c + barc ` is real.
Hence, locus of is a straight line.