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

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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:

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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 = beiθ, for all values of θ
 
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.