Complex Numbers Locus Question

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The discussion focuses on finding the Cartesian equation of the locus defined by the equation |z + 3 + 2i| = Re(z). Participants clarify that z can be expressed as x + iy, leading to the realization that Re(z) equals x. The equation is transformed into x = √{(x + 3)² + (y + 2)²}, which is then squared to eliminate the square root. After simplification, the resulting equation is 6x + 4y + y² = -13, which is noted as being in canonical form. The conversation emphasizes the importance of expressing the final result in a standard Cartesian format.
Amaru58
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Homework Statement


Find the cartesian equation of the locus |z+3+2i|=Re(z)


Homework Equations





The Attempt at a Solution


You let z= x +iy
therefore Re(z)=x?
 
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Amaru58 said:

Homework Statement


Find the cartesian equation of the locus |z+3+2i|=Re(z)


Homework Equations





The Attempt at a Solution


You let z= x +iy
therefore Re(z)=x?
Indeed. So you now have

x = \left| x + 3 +iy + 2i\right|

Can you write that in canonical form?
 
OK. I'm not too sure what you mean by canonical form?
Would x = {(x+3)^2 + (y+2)^2}^1/2
Therefore: x^2 = x^2 +6x + 9 +y^2 +4y+4
6x+4y+y^2=-13
I'm not quite sure what to do now
Thanks for the help
 
Amaru58 said:
OK. I'm not too sure what you mean by canonical form?
Would x = {(x+3)^2 + (y+2)^2}^1/2
Therefore: x^2 = x^2 +6x + 9 +y^2 +4y+4
6x+4y+y^2=-13
I'm not quite sure what to do now
Thanks for the help
You're there. The canonical form in Cartesian coordinates is usually f(x,y) = 0 or f(x,y) = const., which is what you have there.
 

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