- Thread starter
- #1

$$

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.

- Thread starter dwsmith
- Start date

- Thread starter
- #1

$$

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.

- Moderator
- #2

- Feb 7, 2012

- 2,738

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

$$

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.

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.

- Thread starter
- #3

So $ad > bc$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.

- Moderator
- #4

- Feb 7, 2012

- 2,738

So $ad > bc$