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

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 $(3,-2)$
