# Math Help - Help with Sketching Region in Complex Plane

1. ## Help with Sketching Region in Complex Plane

Hey guys,

I am unsure about the steps needed to sketch the following region in a complex plane. It's the only question regarding the sketching of complex planes which I am unsure of the methodology use to approach.

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

I have done questions like |z - 2i| = 2 etc, never done a question with 2 modulus equal to each other. Any help would be greatly appreciated.

2. ## Re: Help with Sketching Region in Complex Plane

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*}.

3. ## Re: Help with Sketching Region in Complex Plane

Originally Posted by 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*}.
Thanks mate, I see how much easier it now becomes converting it into Cartesian form. Once again thank you for the in depth explanation.