Complex transformations and loci

By Hal 2001 (P3046) on Thursday, March 15, 2001 - 12:24 pm :

Hello NRich,

For the transformation w = z² ,
show that as z moves once round the circle centre O and radius 2, w moves twice round the circle centre O and radius 4.

I am not too sure of how to put this down in writing, but here goes...

|z| = 2 (gives us a circle with origin O, and r=2)
If z = x + iy,
We have: |(x + iy)| = 2 ===> x² + y² = 4

Is it done answering the question? Or do I have to do something more? Help appreciated.

Hal.

By Dan Goodman (Dfmg2) on Thursday, March 15, 2001 - 03:42 pm :

Not quite done, all you've shown is that the image under the transformation is contained in a circle of radius 4 centred at 0. What you should do is to parameterise the curve going round the circle centred at O with radius 2 as

z (t) = 2 e^{2 π i t}

where

0 \leq t \leq 1

. Then the transformation gives another curve

w (t) = z (t)^{2} = 4 e^{4 π i t}

where

0 \leq t \leq 1

. Clearly this curve goes around the circle of radius 4 twice, at

t = 0.5

, it has been round once, and so by

t = 1

it has been round twice. You're happy with what

e^{i t}

is aren't you?

By Hal 2001 (P3046) on Thursday, March 15, 2001 - 05:06 pm :

Thanks Dan for your help. Is e^it = cos(t) + isin(t)? When you say 'parameterise', I am not completly clear on what it means? But I understand what you are getting at. I've not come across that terminology before.

How did you come to the conclusion of restricting the parameter t value? Why did you choose 2pi for the index of the e?

Hal.

By Dan Goodman (Dfmg2) on Thursday, March 15, 2001 - 05:20 pm :

Hal, yes,

e^{i t} = \cos (t) + i \sin (t)

. If you haven't come across parameterising the circle as

e^{2 π i t}

it probably needs a bit more explaining. Basically, a curve in the complex plane is defined as a continuous function from the interval [0,1] to the complex plane. For a curve

f

(a function from [0,1] to

C

), a good way to understand the parameterisation is to imagine a particle which at time

t

is in position

f (t)

. So, if

f (t) = e^{2 π i t}

, at time 0 the particle is at 1, at time 1/4, the particle is at

i

, at time 1/2, particle is at -1, ... until at time 1 the particle is back at 1 again, it has moved around the circle once. The

2 π

is in there so that

0 \leq t \leq 1

, you could have used the function

f (t) = e^{i t}

and

0 \leq t \leq 2 π

, the argument would have worked just as well. I'm not sure what you mean by ''restricting the parameter

t

value''. Do you mean why did I define

0 \leq t \leq 1

? If so, that's because with that restriction, the particle moves around the circle exactly once after time 1. Hope that's a bit clearer now, my brain doesn't seem to be working very well at the moment and that doesn't look as clear as it could have been. Sorry about that.

By Hal 2001 (P3046) on Thursday, March 15, 2001 - 10:41 pm :

Thank you Dan for that explanation. I think I managed to understand bits of it. I'll try and find out more from the text books.

Hal.