Toronto Math Forum
MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: hanyu Qi on December 07, 2018, 05:28:43 PM
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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.
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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
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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)
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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)
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?
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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)
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?
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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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)
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?
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?
Using $w_{3} = z_{3} = 1$ as the third point in the triple gives the same answer... maybe it's just luck?