Question

Find the locus of the complex number $z=x+iy$, satisfying relations $\arg (z-1)=\frac{\pi }{4}$ and $|z-2-3i|=2$. Illustrate the locus on the Argand plane.

Answer 1

We have, $z-1=x+iy-1=(x-1)+iy$
$arg(z-1)=\frac{\pi }{4}$
or, $arg\left[\right(x-1)+iy]=\frac{\pi }{4}$
or, ${\tan }^{-1}\left(\frac{y}{x-1}\right)=\frac{\pi }{4}$
or, $\frac{y}{x-1}=\tan \frac{\pi }{4}$
or, $\frac{y}{x-1}=1$
or, $y=x-1$

Equation $\left(i\right)$ represents a straight line intersecting $X$-axis at $A(1,0)$ and which is inclined at an angle of $\frac{\pi }{4}$ to $OX$.
$|z-2-3i|=2$
$|x+iy-2-3i|=2$
$|(x-2)+i(y-3)|=2$
$(x-2{)}^2+(y-3{)}^2=4$

This equation represents a circle of radius 2 and centre (2, 3).
$(x-2{)}^2+(x-1-3{)}^2=4$ [From (1)]
$(x-2{)}^2+(x-4{)}^2=4$
$x^2-4x+4+x^2-8x+16=4$
$2x^2-12x+16=0$
$x^2-6x+8=0$
$(x-4)(x-2)=0$
$x=4$ or $2$
$y=3$ or $1$
$\therefore$ The locus will be points $(2,1)$ and $(4,3)$