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FE Sample Question 4 (a)

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hanyu Qi:
Hello, I am wondering whether we can have different function f(z) for this question.

In the posted solution, Yilin set $ f(z) = \lambda \frac{z-a}{1-\bar{a}z}$ $| \lambda | = 1 $ and get final result as $ f(z) = \frac{5+z}{1+5z}$.

Based on a hint in Textbook 3.3 Example 1, I try to set $ f(z) = \lambda \frac{a-z}{1-\bar{a}z}$ and $| \lambda |=1 $then I did the following computation.

let $\lambda = e^{it}$ , $ a = re^{i \theta}$

$f(0) = 5$ -> $\lambda a = 5$ and $ |\lambda a| = |a| = 5 $ so $ a = 5e^{i\theta}$

$\lambda a = 5e^{it} e^{i\theta} =5$ so $e^{-it} = e^{i\theta}$

$ f(-1) = \lambda \frac{a+1}{1+\bar{a}} = -1$ so $e^{-i\theta} = -1$ ---> $e^{it} =1$ and $ \theta = \pi $

so $\lambda = 1$ and $a = -5$

$f(z) = \frac{-5-z}{1-5z}$

I am not sure if there is a computation mistake for my solution or we could set f(z) in many forms.

Thank you.

Ye Jin:
Hanyu，I think you made a mistake when you calculated $e^{it}$,

since $e^{it}=e^{-i\theta}=-1$

then $\lambda=-1$

And you can also check the value of $\lambda$ and a since $\lambda$a=5 in your third line of calcualtion

Min Gyu Woo:
Can you approach this method another way without the hint?

I.e) using the three fixed points (choose the last one to be any point that matches the conditions)

Xingyu Wang:

--- Quote from: Min Gyu Woo on December 08, 2018, 02:14:56 PM ---Can you approach this method another way without the hint?

I.e) using the three fixed points (choose the last one to be any point that matches the conditions)

--- End quote ---

Can we choose the third point on the boundary of the domain and image (in this case, both are the unit circle itself), like f(1) = 1?

Ende Jin:

--- Quote from: Xingyu Wang on December 08, 2018, 03:37:19 PM ---
--- Quote from: Min Gyu Woo on December 08, 2018, 02:14:56 PM ---Can you approach this method another way without the hint?

I.e) using the three fixed points (choose the last one to be any point that matches the conditions)

--- End quote ---

Can we choose the third point on the boundary of the domain and image (in this case, both are the unit circle itself), like f(1) = 1?

--- End quote ---

I don't think you can because in that way you will get a mobius transformation that
$1 \mapsto 1, 0 \mapsto 5, -1 \mapsto -1$, this set of constraints can also be interpreted as the circle passing $1,0,-1$ is mapped to a circle passing $1, 5, -1$, which doesn't satisfy the question?

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