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

[ericm1234]

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.

 

[tiny-tim]

hi ericm1234!

(try using the X² button just above the Reply box )

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^iθ, for all values of θ

 

[ericm1234]

Awesome.

 

[HallsofIvy]

Note that, taking z= x+ iy, $|z- 2i|= |x+ (y-2)i|= r$ is the same as $\sqrt{x^23+ (y- 2)^2}= r$ so that $x^2+ (y- 2)^2= r^2$, the circle of radius r with center at (0, 2) or 2i in the complex plane.

 

[blue_raver22]

Yes, it is valid to write z=r*e^(i*theta)+2i in order to shift the center of the circle to 2i. This is known as a translation in complex numbers. However, it is important to note that the equation z=r*e^(i*theta) represents a circle with radius r centered at the origin (0,0), so adding 2i to the equation will shift the center to 2i, but the radius will remain the same. This means that the equation abs(z-2i)=r will still hold true for all points on the circle. If you want to solve for z, you can simply subtract 2i from both sides of the equation to get z=r*e^(i*theta). This will give you the complex numbers that lie on the shifted circle centered at 2i.