Find a complex number that satisfies both $|z-3+i|=3$ and $|z+2-i|=|z-3+2i|$.

Question

What I did:

Let $z=x+yi$, Then:

$$|(x-3)+(y+1)i|=3$$ $$|(x+2)+(y-1)i|=|(x-3)+(y+2)i|$$

Since $|x+iy|=\sqrt{x^2+y^2}$:

$$(x-3)^2+(y+1)^2=9$$ $$y=\frac{5x-4}{3}$$ Hence:

$$(x-3)^2+\bigg(\frac{5x-1}{3}\bigg)^2=9$$ Or $$34x^2-64x+1=0$$ So $$x_{1,2}=\frac{32\pm3\sqrt{110}}{34}$$ And

$$y_{1,2}=\frac{8\pm5\sqrt{110}}{34}$$ Hence

$$z_1=\frac{32+3\sqrt{110}}{34}+\frac{8+5\sqrt{110}}{34}i$$ $$z_2=\frac{32-3\sqrt{110}}{34}+\frac{8-5\sqrt{110}}{34}i$$ But, when I looked up the answer in the book:

I think there is a misprint in the book as this is the question:

I think that in part c they meant $|z+2-i|=|z-1+2i|$ and not $|z+2-i|=|z-3+2i|$, becuase then the answer would make sense, but I thought I would double check. Thanks for any help

Answer 1

Geometrically u are looking for the intersection of the circle centered at $(3,-1)$ with radius 3 and the perpendicular bisector of the line segment ralating the points $(-2,1)$ and $(3,-2)$