Question

Find the Cartesian equation of the locus of the point P representing the complex
number z. Sketch the locus of P in each case.

(a) 2|z + 1| = |z − 2|

(b) |(z + i )/ z - 5 - 2i | = 1

(c) Im (z + (9/z ) ) = 0

Answer 1

(a) $2|z + 1| = |z − 2|$

Solution:

Let $z=x+iy$

Substitute z in (a)

$2|x+iy+1|=|x+iy-2|$

$2\sqrt{\smash[b]{x+iy+1}}=\sqrt{\smash[b]{x+iy-2}}$

$2\sqrt{\smash[b]{x+1+iy}}=\sqrt{\smash[b]{x-2+iy}}$

$2\sqrt{\smash[b]{(x+1)^2+y^2}}=\sqrt{\smash[b]{(x-2)^2+y^2}}$

$2((x+1)^2+y^2)=(x-2)^2+y^2$

$2x^2+4x+2+2y^2=x^2-4x+4+y^2$

$2x^2-x^2+2y^2-y^2+4x+4x=4-2$

Answer: $x^2+y^2+8x=2$

(b) $|(z + i )/ z - 5 - 2i | = 1$

Solution:

Cross multiply equation to get;

$|z+i|=|z-5-2i|$

Let $z=x+iy$

Substitute for z;

$x+iy+i=x+iy-5-2i$

$x+i(y+1)=x-5+i(y-2)$

$\sqrt{\smash[b]{x^2+(y+1)^2}}=\sqrt{\smash[b]{(x-5)^2+(y-2)^2}}$

$x^2+y^2+2y+1=x^2-10x+25+y^2-4y+4$

Simplify;

$x^2-x^2+y^2-y^2+10x+2y+4y=25+4-1$

Answer : $10x+6y=28$

(c) $Im (z + \frac{9}{z}) = 0$

Solution:

Let $z =x+iy$

Substitute for z in the expression;

$x+iy+\frac{9}{x+iy}$

Simplify; $\frac{x+iy(x+iy)}{x+iy}+\frac{9}{x+iy}$

$\frac{(x+iy)(x+iy)+9}{x+iy}$

$\frac{x^2+2ixy-y^2+9}{x+iy}$

Rationalize the denominator;

$\frac{x^2+2ixy-y^2+9}{x+iy}*\frac{x-iy}{x-iy}$

$\frac{x^2(x-iy)+2ixy(x-iy)-y^2(x-iy)+9(x-iy)}{x^2+y^2}$

$\frac{x^3-ix^2y+2ix^2y+2xy^2-y^2x+iy^3+9x-9iy}{x^2+y^2}$

Put the like terms together;

$\frac{x^3+2xy^2-y^2x+9x-ix^2y+2ix^2y+iy^3-9iy}{x^2+y^2}$

$\frac{x^3+2xy^2-y^2x+9x}{x^2+y^2} +\frac{i(x^2y+y^3-9y)}{x^2+y^2}$

$Im$ part states;

$\frac{(x^2y+y^3-9y)}{x^2+y^2}=0$

$x^2y+y^3-9y=0$

$y(x^2+y^2-9)=0$

$x^2+y^2-9=0$

Answer: $x^2+y^2=9$