Conic Sections on the Complex Plane (circle)

[miniradman]

Homework Statement

Describe the locus and determine the Cartesian Equation of:
$\left|z-3-5i\right|= 2$

Homework Equations

$\left|z-C\right|= r$ -----> formula for a circle on complex plane
Where
C = the centre
z = the moving point (locus)

$(x-h)^{2}+(y-k)^{2}=r^{2}$ -----> Formula for a circle on the cartesian plane

The Attempt at a Solution

Ok I think I've got the first section, describe the locus

Well if -C = -3-5i
that means C = 3+5i

So the centre of the circle will be at 3+5i on the complex plane.

But I get stuck when converting it into the cartesian form.

$z = x + yi$

$\left|(x + yi)-3-5i\right|= 2$
$\sqrt{(x-3)^{2}-i(y + 5)^{2}}$
$\uparrow$
But I don't know how to proceede from there because I can't figure out how to get rid of the $i$

Anyone know how to?

 

[Bread18]

Homework Statement

Describe the locus and determine the Cartesian Equation of:
$\left|z-3-5i\right|= 2$

Homework Equations

$\left|z-C\right|= r$ -----> formula for a circle on complex plane
Where
C = the centre
z = the moving point (locus)

$(x-h)^{2}+(y-k)^{2}=r^{2}$ -----> Formula for a circle on the cartesian plane

The Attempt at a Solution

Ok I think I've got the first section, describe the locus

Well if -C = -3-5i
that means C = 3+5i

So the centre of the circle will be at 3+5i on the complex plane.

But I get stuck when converting it into the cartesian form.

$z = x + yi$

$\left|(x + yi)-3-5i\right|= 2$
$\sqrt{(x-3)^{2}-i(y + 5)^{2}}$
$\uparrow$
But I don't know how to proceede from there because I can't figure out how to get rid of the $i$

Anyone know how to?

I'm not sure exactly what you have done, I don't do these questions this way.

The easy way to do it is to let $z=x + iy$ and then sub into $\left|z-3-5i\right|= 2$. Now you just find the modulus as if it were an complex number and you end up with an eqn for a circle.

 

[QED Andrew]

The modulus of a complex number $z = x + iy$ is defined by

$|x+iy| \equiv \sqrt{x^2 + y^2}$.​

Note the absence of $i$ on the right-hand side of the above equation.

 

[HallsofIvy]

3+5i, on the complex plane, corresponds to the point (3, 5) in the Cartesian plane. There is no "i" when writing points in the Cartesian plane. Yes, the equation |z- (a+ bi|= |(x- iy)- (a+ bi)|= r is a circle in the Complex plane with center at a+ bi which corresponds to (a, b) and radius r. It has equation $(x- a)^2+ (y- b)^2= r^2$. You do NOT include the "i" in the equation when converting to the Cartesian form.