Z-plane to w-plane mapping

  • Thread starter thomas49th
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Homework Statement


Consider the mapping w = 1/(z-1) from the z-plane to the w plane. Show that in the z plane the circle
(x-2)² + y² = 4
maps to a circle in the w-plane. What is the radius of this circle and where is it's centre.

So in the z-plane this is a circle with radius 2 at the point (1,0) in the z plane.

Homework Equations





The Attempt at a Solution


Hmmm. Well I know that w = 1/(z-1) => u² + v² = 1 / ((x-1)² +y²)

I presume that will help at some point

In the z-plane (x-1)² + y² = 4 what part is the imaginary part? The z plane has 2 axes:
x and y... am I right in thinking x = Re and y = Im? I recall f(z) = u(x,y) + iv(x,y) but does that mean u = (x-1)² + y² ??? How does the 4 come into it. What about the imaginary part?

I think I may of bodged it by getting

(x-1)² + y² = (1/2)^2

by sticking 1 / ((x-1)² +y²) = 4, but I don't think that is the correct method.

Thanks
Thomas
 

Answers and Replies

  • #2
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Homework Statement


Consider the mapping w = 1/(z-1) from the z-plane to the w plane. Show that in the z plane the circle
(x-2)² + y² = 4
maps to a circle in the w-plane. What is the radius of this circle and where is it's centre.

So in the z-plane this is a circle with radius 2 at the point (1,0) in the z plane.
The center of the circle would be at (2, 0), not (1, 0).

Homework Equations





The Attempt at a Solution


Hmmm. Well I know that w = 1/(z-1) => u² + v² = 1 / ((x-1)² +y²)

I presume that will help at some point

In the z-plane (x-1)² + y² = 4 what part is the imaginary part? The z plane has 2 axes:
x and y... am I right in thinking x = Re and y = Im? I recall f(z) = u(x,y) + iv(x,y) but does that mean u = (x-1)² + y² ??? How does the 4 come into it. What about the imaginary part?

I think I may of bodged it by getting

(x-1)² + y² = (1/2)^2

by sticking 1 / ((x-1)² +y²) = 4, but I don't think that is the correct method.

Thanks
Thomas
 

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