Fractional linear transformation--conformal mapping

[Dustinsfl]

Find necessary and sufficient conditions on the real numbers $a$, $b$, $c$, and $d$ such that the fractional linear transformation
$$
f(z) = \frac{az + b}{cz + d}
$$
maps the upper half plane to itself.

I just need some guidance on starting this one since I am not sure on how to begin.

 

[Opalg]

Find necessary and sufficient conditions on the real numbers $a$, $b$, $c$, and $d$ such that the fractional linear transformation
$$
f(z) = \frac{az + b}{cz + d}
$$
maps the upper half plane to itself.

I just need some guidance on starting this one since I am not sure on how to begin.

Begin by multiplying top and bottom by the complex conjugate of the denominator: $$f(z) = \dfrac{az + b}{cz + d} = \dfrac{(az + b)(c\overline{z} + d)}{(cz + d)(c\overline{z} + d)}.$$

The denominator in that last fraction is real, so you need to find the imaginary part of the numerator and investigate what makes it positive whenever $z$ has positive imaginary part.

 

[Dustinsfl]

Begin by multiplying top and bottom by the complex conjugate of the denominator: $$f(z) = \dfrac{az + b}{cz + d} = \dfrac{(az + b)(c\overline{z} + d)}{(cz + d)(c\overline{z} + d)}.$$

The denominator in that last fraction is real, so you need to find the imaginary part of the numerator and investigate what makes it positive whenever $z$ has positive imaginary part.

So $ad > bc$

 

[Opalg]

So $ad > bc$

(Yes)

 

[blue_raver22]

Sure, I can help you get started on this problem. First, let's define the upper half plane, denoted by $\mathbb{H}$, as the set of all complex numbers $z$ such that $z = x + iy$ with $y > 0$. This means that the upper half plane is the region above the real axis on the complex plane.

Now, in order for the fractional linear transformation $f(z)$ to map $\mathbb{H}$ to itself, we need to find conditions on the real numbers $a$, $b$, $c$, and $d$ such that $f(z)$ maps any point $z \in \mathbb{H}$ to another point in $\mathbb{H}$. In other words, we want to ensure that $f(z)$ keeps all points in the upper half plane.

To do this, we can use the fact that a conformal mapping preserves angles. This means that if we take any two intersecting curves in the upper half plane, their images under $f(z)$ should also intersect at the same angle in the upper half plane. Let's consider two curves, $y = mx$ and $y = nx$, with $m,n \in \mathbb{R}$, that intersect at a point $z_0 \in \mathbb{H}$. We can then find the angle between these two curves at $z_0$ by taking the derivative of $y$ with respect to $x$ and evaluating it at $z_0$. This gives us the slope of the curve at $z_0$, which we can use to calculate the angle.

Now, let's take the images of these two curves under $f(z)$, denoted by $y' = m'x$ and $y' = n'x$, and find the angle between them at $f(z_0)$. Using the chain rule, we can show that the derivative of $y'$ with respect to $x$ is equal to the derivative of $y$ with respect to $x$, multiplied by a factor of $f'(z_0)$, i.e. the derivative of $f(z)$ evaluated at $z_0$. Therefore, the angle between the two images at $f(z_0)$ is the same as the angle between the original curves at $z_0$ if and only if $f'(z_0)$ is