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Consider the equation az + b bar(z) + c ...

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

A

an ellipse

B

a circle

C

a point

D

a straight line

Text Solution

Verified by Experts

The correct Answer is:
D
`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.
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