Question

If $t$ and $c$ are two complex numbers such that $\left|t\right|\ne \left|c\right|,\left|t\right|=1$ and $z=\frac{(at+b)}{(t-c)}$, $z=x+iy$. Locus of $z$ is (where $a$, $b$ are complex number )

• A. line segment
• B. straight line
• C. circle
• D. none oh these

Answer 1

Answer: (C)

circle

$z=\frac{at+b}{t-c}$
$z(t-c)=(at+b)$
$t(z-a)=b+zc$
$t=\frac{zc+b}{z-a}$
$t=c\left(\frac{z+\frac{b}{c}}{z-a}\right)$
$|t|=|c||\frac{z+\frac{b}{c}}{z-a}|$
Now $|t|=1$
Therefore,
$|\frac{z+\frac{b}{c}}{z-a}|=\frac{1}{|c|}$
This represents a circle.
Thus,
$z$ lies on a circle given be
$|\frac{z+\frac{b}{c}}{z-a}|=\frac{1}{|c|}$
Hence, option 'C' is correct.