Conic Sections on the Complex Plane (circle)

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Homework Statement


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


Homework Equations


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

[itex](x-h)^{2}+(y-k)^{2}=r^{2}[/itex] -----> 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.

[itex]z = x + yi[/itex]

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

Anyone know how to?
 

Answers and Replies

  • #2
100
0

Homework Statement


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


Homework Equations


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

[itex](x-h)^{2}+(y-k)^{2}=r^{2}[/itex] -----> 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.

[itex]z = x + yi[/itex]

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

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 [itex]z=x + iy[/itex] and then sub into [itex]\left|z-3-5i\right|= 2[/itex]. Now you just find the modulus as if it were an complex number and you end up with an eqn for a circle.
 
  • #3
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The modulus of a complex number [itex]z = x + iy[/itex] is defined by

[itex]|x+iy| \equiv \sqrt{x^2 + y^2}[/itex].​

Note the absence of [itex]i[/itex] on the right-hand side of the above equation.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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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 [itex](x- a)^2+ (y- b)^2= r^2[/itex]. You do NOT include the "i" in the equation when converting to the Cartesian form.
 

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