Mapping and Rotation of Complex Curves

[elimenohpee]

Homework Statement

Find the angle through which a curve drawn from the point z0 is rotated under the mapping w=f(z), and find the corresponding scale factor of the transformation.

z0 = -1, w=z^2

and

z0 = -1 + i, w = 1/z

Homework Equations

I honestly don't know how to begin

The Attempt at a Solution

I don't know where to start, these are 2 questions at the end of my complex variables book, and there is no example in the text. the answers for z0 = -1 are: arg w'(z0) = pi, |w'(z0)| = 2

and z0 = -1 + i are: arg w'(z0) = -pi/2, |w'(z0)| = 1/2

 

[Dick]

You want to expand f(z) into a taylor series around z=z0. Everything you want to know is then in the linear term. If f(z)=a*(z-z0) how can you find the scaling and rotation from the factor a?

 

[elimenohpee]

'a' would be the scaling factor would it not? Can you use the equation f(z) = a*(z-z0) to solve for the scaling factor somehow?

How would you find the arguement?

 

[elimenohpee]

I see if I just evaluate |w'(z0)|, I see that for the first question I get |-2| or 2, and 1/2 for the second.

I guess my question is really understanding what the question is asking. Like you stated, it looks as if its a taylor series expansion about a point, but what is the significance?

 

[Dick]

'a' would be the scaling factor would it not? Can you use the equation f(z) = a*(z-z0) to solve for the scaling factor somehow?

How would you find the arguement?

No, |a| would be the scaling. arg(a) would be the rotation, yes? Simplify it another notch. Just take f(z)=a*z. What's the rotation and what's the scaling? f(z)=i*z has scaling 1, it rotates by pi/2, right?

 

[Dick]

I see if I just evaluate |w'(z0)|, I see that for the first question I get |-2| or 2, and 1/2 for the second.

I guess my question is really understanding what the question is asking. Like you stated, it looks as if its a taylor series expansion about a point, but what is the significance?

The first term in the taylor series is just a translation, it doesn't scale or rotate. The powers of (z-z0) higher than 1 are negligible close to z0 compared with the first power. The coefficient of (z-z0) tells you everything.

 

[elimenohpee]

Ok perfect, makes sense now. Thanks for the help!