Solving and graphing all values of $z$. 
Question: Find and graph all values of $z$ such that $$|z-3|=|z+2i|\tag1$$


I'm not sure how to find $z$. Here's my attempt:

Let $z=a+bi$. Plugging that into $(1)$ gives us $$|a+bi-3|=|a+bi+2i|\tag2$$
And using the magnitude definition, we have $$\sqrt{(a-3)^2+b^2}=\sqrt{a^2+(2+b)^2}\tag3$$
Squaring both sides and moving all the constants to one side, we get the linear equation $$6a+4b=5$$
Which takes care of the graphing part. But I'm not sure how to find the value of $z$ at this point.

EDIT: The book states the answer as $5$. How did they get $5$??
 A: Say it in words and perhaps some hidden knowledge of geometry will kick in: the set of all complex numbers (=points in the plane) that fulfill $\;|z-3|=|z+2i|\;$ is the set of all points in the plane whose distance from $\;3\;$ equals their distance from $\;-2i\;$ ...some bell ringing?

This is just the perpendicualr bisector of the segment of line joining the points $\;3\,,\,\,-2i\;$ in the complex plane.

Added on request: Putting $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ , we get:
$$|z-3|^2=|z+2i|^2\implies(x-3)^2+y^2=x^2+(y+2)^2\implies$$
$$-6x+9=4y+4\implies\;\;\ell:\; 6x+4y-5=0$$
The above is a straight line with slope equal to $\;m=-\frac32\;$ . Observe that the slope between $\;3\sim(3,0)\;$ and $\;2i\sim (0,-2)\;$ (here we identify complex numbers with points in the plane, as usual) is $\;\cfrac{0-(-2)}{3-0}=\cfrac23\;$  , so $\;\cfrac23\cdot m=-1\;$ and $\;\ell\;$ indeed is perpendicular to the line segment whose extremes are $\;(3,0),\,(0,-2)\;$, as expected.
Finally, to see $\;\ell\;$ is a bisector of the line segment, we show the distance of iether extreme point to the line is the same:
$$\begin{align*}&(3,0)\;\text{from}\;\ell:\;\;\frac{|6\cdot3+4\cdot0-5|}{\sqrt{6^2+4^2}}=\frac{13}{\sqrt{52}}=\frac{\sqrt{13}}2\\{}\\
&(0,-2)\;\text{from}\;\ell:\;\;\frac{|6\cdot0+4\cdot(-2)-5|}{\sqrt{6^2+4^2}}=\frac{\sqrt{13}}2\end{align*}$$
A: Geometrically, it is the set of points equidistant from the images $A$ and $B$ of  $-2i$  and $3$, i.e. the perpendicular bisector of the segment $[AB]$.
So we have to find:
1) the affix of the middle point $I$ of the segment: $\;z_I=\dfrac32+2i$.
2) a complex number $z_0$ with image orthogonal to $[AB]$. As the affix of $\overrightarrow{AB}$ is $3+2i$, $z_0=2-3i$ is a solution.
Hence the values of $z$ we seek is
$$z=\dfrac32+2i+t(2-3i)=\dfrac32+2t+(2-3t)i\qquad(t\in\mathbf R).$$
