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**Prove It** I would convert everything to Cartesians:

$\displaystyle \begin{align*} \left| z - 3 \right| &= \left| z + 2i \right| \\ \left| \left( x - 3 \right) + y\,i \right| &= \left| x + \left( y + 2 \right)\, i \right| \\ \sqrt{ \left( x - 3 \right) ^2 + y^2 } &= \sqrt{ x^2 + \left( y + 2 \right) ^2 } \\ \left( x - 3 \right) ^2 + y^2 &= x^2 + \left( y + 2 \right) ^2 \\ x^2 - 6x + 9 + y^2 &= x^2 + y^2 + 4y + 4 \\ -6x + 9 &= 4y + 4 \\ -6x + 5 &= 4y \\ y &= -\frac{3}{2}x + \frac{5}{4} \end{align*}$

So the graph will be the line with gradient $\displaystyle \begin{align*} -\frac{3}{2} \end{align*}$ and y intercept $\displaystyle \begin{align*} \left( 0, \frac{5}{4} \right) \end{align*}$.