Question

$z$ is a complex number such that $|z|=3$ then $1-z$ lies on

• A. a circle centre $(0,0)$ radius $\frac{1}{3}$
• B. straight line passing through origin
• C. circle with centre $(-1,0)$ radius $2$
• D. circle with centre $(1,0)$ radius $3$

Answer 1

The correct option is D circle with centre $(1,0)$ radius $3$

Let $z=x+iy$

$|z|=3$

$\Rightarrow x^2+y^2=9$

$1-z=1-x-iy$

Let $X=1-x$ , $Y=-y$

Hence,

$(1-X{)}^2+Y^2=9$

$\Rightarrow (X-1{)}^2+(Y-0{)}^2=9$

This is the equation of a circle with centre $(1,0)$ and radius $3$.