Evaluate the Mobius transformation

[docnet]

Homework Statement

evaluate the quotient of two complex numbers

Relevant Equations

evaluate the quotient of two complex numbers

Let $|z|=1$ and $1-\bar{a}z\neq 0$.

Evaluate $\frac{|z-a|}{|1-\bar{a}z|}$. It should be a real number.

I read that $f=\frac{|z-a|}{|1-\bar{a}z|}$ is a mobious transformation, but I do not know what it means. @fresh_42$z=e^{i\theta_1}, a=r_2e^{i\theta_2}$

$\frac{|z-a|}{|1-\bar{a}z|}=\frac{|e^{i\theta_1}-r_2e^{i\theta_2}|}{|1-r_2e^{i(\theta_1-\theta_2)}|}$

 

[fresh_42]

Here is a short explanation with an interesting property (problem #8):
https://www.physicsforums.com/threads/math-challenge-august-2021.1005685/
(behind the 'solved by' note is a link to the solution @julian gave)
and here is a longer one:
https://en.wikipedia.org/wiki/Möbius_transformation

I would square the function:
$$
f^2=\dfrac{(z-a)(\bar{z}-\bar{a})}{(1-\bar{a}z)(1-a\bar{z})}
$$
and work with Cartesian coordinates: $z=x+i y\, , \,\bar{z}=x- i y\, , \,a=u+ i v\, , \,\bar{a}=u-iv$ and see if the expression simplifies. The polar coordinates might work as well, but I think it remains a mess. Of course, you could also use the sine and cosine functions, but I would definitely start with the Cartesian coordinates.

 

[docnet]

Here is a short explanation with an interesting property (problem #8):
https://www.physicsforums.com/threads/math-challenge-august-2021.1005685/
(behind the 'solved by' note is a link to the solution @julian gave)
and here is a longer one:
https://en.wikipedia.org/wiki/Möbius_transformation

I would square the function:
$$
f^2=\dfrac{(z-a)(\bar{z}-\bar{a})}{(1-\bar{a}z)(1-a\bar{z})}
$$
and work with Cartesian coordinates: $z=x+i y\, , \,\bar{z}=x- i y\, , \,a=u+ i v\, , \,\bar{a}=u-iv$ and see if the expression simplifies. The polar coordinates might work as well, but I think it remains a mess. Of course, you could also use the sine and cosine functions, but I would definitely start with the Cartesian coordinates.

I would have never guessed to try complex squaring the function... I tried multiplying the denominator and the numerator by $1-a\bar{z}$ and was unable to simplify.

By doing the squaring, and plugging in $z\bar{z}=1$, the function simplifies so that the denominator equals the numerator and it's just 1. (I didn't use cartesian coordinates and still got this result, and I really hope i didn't mess up the calculation because it really would be embarrassing)

 

[docnet]

$f^2=(\frac{|z-a|}{|1-\bar{a}z|} )^2=\frac{z\bar{z}-a\bar{z}-\bar{a}z+a\bar{a}}{1-\bar{a}z-a\bar{z}+a\bar{a}z\bar{z}}$

$\frac{1-a\bar{z}-\bar{a}z+a\bar{a}}{1-\bar{a}z-a\bar{z}+a\bar{a}}=1$

$\therefore \frac{|z-a|}{|1-\bar{a}z|} =1$

This question was on a midterm I took yesterday (so I'm not cheating!)

 

[Office_Shredder]

You can also multiply the numerator by $|\bar{z}|$ (which is 1) and then the numerator and denominator are complex conjugates of each other.

 

[docnet]

You can also multiply the numerator by $|\bar{z}|$ (which is 1) and then the numerator and denominator are complex conjugates of each other.

and that is a clue that the quotient should be complex squared, right?

 

[fresh_42]

and that is a clue that the quotient should be complex squared, right?

Squaring was only meant to get rid of the root in the definition of the norm. After you saw the simplifications you made, you could write a root above every term of your calculation, so squaring wasn't actually necessary. However, it removed the "distraction" by the root behind $|\, \text{.} \,|$ and allowed us to concentrate on the essential parts.

I have two major principles approaching an algebraic expression: Get rid of what disturbs, and use the definitions.

 

[Office_Shredder]

and that is a clue that the quotient should be complex squared, right?

Once the numerator and denominator are complex conjugates, then you know they have the same magnitude, so you're done.