Can You Shift a Circle in the Complex Plane to Center at 2i?

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Discussion Overview

The discussion revolves around the representation of a circle in the complex plane, specifically focusing on how to express a circle centered at the point 2i. Participants explore different formulations and implications of this representation.

Discussion Character

  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions the validity of writing z = r*e^(i*theta) + 2i to represent a circle centered at 2i, expressing a desire to solve for z without using the absolute value formulation.
  • Another participant confirms that the expression z - a = b*e^(i*theta) is valid for a circle centered at a with radius b, suggesting that the proposed formulation is acceptable.
  • A later reply elaborates on the relationship between the Cartesian and polar forms, indicating that the equation |z - 2i| = r can be expressed in Cartesian coordinates as x^2 + (y - 2)^2 = r^2, which describes a circle centered at (0, 2) or 2i.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the proposed formulation for representing a circle centered at 2i, but there are different approaches to expressing the relationship between the polar and Cartesian forms.

Contextual Notes

The discussion does not resolve the implications of using different forms or the potential limitations of each representation in specific contexts.

ericm1234
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We typically have z=r*e^(i*theta). But let's say I want a circle centered at 2i.
Is it valid to write z=r*e^(i*theta)+2i ?

I ask this because I don't want to have abs(z-2i)=r; I want to solve for z.
 
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hi ericm1234! :smile:

(try using the X2 button just above the Reply box :wink:)
ericm1234 said:
We typically have z=r*e^(i*theta). But let's say I want a circle centered at 2i.
Is it valid to write z=r*e^(i*theta)+2i ?

yes, a circle centre a and radius b is z - a = be, for all values of θ
 
Awesome.
 
Note that, taking z= x+ iy, [itex]|z- 2i|= |x+ (y-2)i|= r[/itex] is the same as [itex]\sqrt{x^23+ (y- 2)^2}= r[/itex] so that [itex]x^2+ (y- 2)^2= r^2[/itex], the circle of radius r with center at (0, 2) or 2i in the complex plane.
 

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