First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightComplex Numberschevron_rightExercise 1.5

Solution

Let $z=x+iy$ .

Given:

|3z-2+i|=|3z+i|

Compute each term:

3z-2+i=3x+3iy-2+i=(3x-2)+i(3y+1)

3z+i=3x+3iy+i=3x+i(3y+1)

Square both sides:

(3x-2)^2+(3y+1)^2=(3x)^2+(3y+1)^2

Cancel $(3y+1)^2$ :

(3x-2)^2=(3x)^2

Expand:

9x^2-12x+4=9x^2\Rightarrow -12x+4=0\Rightarrow x=\frac{1}{3}

So the locus is the vertical line $x=\frac{1}{3}$ :

\boxed{x=\frac{1}{3}}