Complex Mapping

  • #1

Homework Statement


Find the image of the circle |z| = 3 in the complex plane under the mapping

a) w = [itex]\frac{6}{z}[/itex]

b) w = [itex]\frac{6}{z}[/itex] + 2i


The Attempt at a Solution



a) w = [itex]\frac{6}{3}[/itex] = 2

So this is a circle in the w-plane of radius 2, centered on the origin?

b) w = [itex]\frac{6}{3}[/itex] + 2i

So this is a line in the w-plane extending from the origin to the point 2 + 2i ?

I'm not entirely sure of how this works, is w an absolute? like |w| = 2?
 

Answers and Replies

  • #2
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Hmmmm.

For the first question:
Let w= u + iv
Let z = x + iy
Hence |z|=x[itex]^{2}[/itex] + y[itex]^{2}[/itex] = 3 ( I hope you understand how this works. It's basically finding the distance of the point (x,y) from origin on a cartesian plane which is called an Argand plane for complex numbers)

Now w=[itex]\frac{6}{z}[/itex]

=[itex]\frac{6}{x+iy}[/itex]

=[itex]\frac{6(x-iy)}{(x+iy)(x-iy)}[/itex]

=[itex]\frac{6(x-iy)}{(x^{2} + y^{2})}[/itex]

=[itex]\frac{6x}{(x^{2} + y^{2})}[/itex] - i[itex]\frac{6y}{(x^{2} + y^{2})}[/itex]

=2x-i2y (Because x[itex]^{2}[/itex] + y[itex]^{2}[/itex] is always 3.

Hence u+iv=2x-i2y
Hence u=2x , v=-2y
So for any (x,y) you might choose on the circle x[itex]^{2}[/itex] + y[itex]^{2}[/itex]=3, (u,v) = (2x,-2y).
Now I leave that up to you to try to imagine the figure that should form.

Also try to do the next sum in a similar fashion.
 
  • #3
Hmmmm.

For the first question:
Let w= u + iv
Let z = x + iy
Hence |z|=x[itex]^{2}[/itex] + y[itex]^{2}[/itex] = 3 ( I hope you understand how this works. It's basically finding the distance of the point (x,y) from origin on a cartesian plane which is called an Argand plane for complex numbers)

Now w=[itex]\frac{6}{z}[/itex]

=[itex]\frac{6}{x+iy}[/itex]

=[itex]\frac{6(x-iy)}{(x+iy)(x-iy)}[/itex]

=[itex]\frac{6(x-iy)}{(x^{2} + y^{2})}[/itex]

=[itex]\frac{6x}{(x^{2} + y^{2})}[/itex] - i[itex]\frac{6y}{(x^{2} + y^{2})}[/itex]

=2x-i2y (Because x[itex]^{2}[/itex] + y[itex]^{2}[/itex] is always 3.

Hence u+iv=2x-i2y
Hence u=2x , v=-2y
So for any (x,y) you might choose on the circle x[itex]^{2}[/itex] + y[itex]^{2}[/itex]=3, (u,v) = (2x,-2y).
Now I leave that up to you to try to imagine the figure that should form.

Also try to do the next sum in a similar fashion.
Do you mean x[itex]^{2}[/itex] + y[itex]^{2}[/itex]=3[itex]^{2}[/itex] ? Because if you choose the point (3,0) on the circle, that will be equal to 9, not 3. Also |z| = √x[itex]^{2}[/itex] + y[itex]^{2}[/itex]

I'm still a little confused. If I start taking values on the circle from the z-plane. Say (3,0);(0,3);(-3,0);(0,-3) and substitute that in:

u(x,y) = (2x, -2y)

...it means I will get a circle of radius 6 in the w-plane.

I plotted this in MAPLE and I get a circle of radius 2.
 
Last edited:
  • #4
NVM, I just realised the equation was meant to be:

[itex]{\frac {6x}{{x}^{2}+{y}^{2}}}-{\frac {6\,iy}{{x}^{2}+{y}^{2}}}[/itex]

where the denominator is 9, and not 3 because of that earlier error, so now it all works out and the circle of radius 3 does indeed transform to a circle of radius 2.

Thanks for that method, now to move on to the second part!
 

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