Evaluation |az+b/cz+d|

Question

I am looking at a proof form some complex analysis notes, and the following is claimed, without explanation: $$ \left|\frac{az+b}{cz+d}\right|=\frac{|a||z-(b/a)|}{|c||z-(d/c)|} $$ Here, all the letters are complex numbers. Can someone explain to me why this is true?

Answer 1

$$\left|\frac{az+b}{cz+d}\right|=\left|\frac{a(z+\tfrac{b}{a})}{c(z+\tfrac{d}{c})}\right|=\left|\frac{a}{c}\cdot\frac{z+\tfrac{b}{a}}{z+\tfrac{d}{c}}\right|=\left|\frac{a}{c}\right|\cdot\left|\frac{z+\tfrac{b}{a}}{z+\tfrac{d}{c}}\right|=\frac{|a|}{|c|}\cdot\frac{|z+\tfrac{b}{a}|}{|z+\tfrac{d}{c}|}.$$

Answer 2

The first part is the same as that from @Servaes:

$$\left|\frac{az+b}{cz+d}\right|=\left|\frac{a(z+\tfrac{b}{a})}{c(z+\tfrac{d}{c})}\right|=\left|\frac{a}{c}\cdot\frac{z+\tfrac{b}{a}}{z+\tfrac{d}{c}}\right|=\left|\frac{a}{c}\right|\cdot\left|\frac{z+\tfrac{b}{a}}{z+\tfrac{d}{c}}\right|=\frac{|a|}{|c|}\cdot\frac{|z+\tfrac{b}{a}|}{|z+\tfrac{d}{c}|}=$$ $$\frac{|a|}{|c|}\cdot\frac{\left|\frac{az}{a}+\frac{b}{a}\right|}{\left|\frac{zc}{c}+\frac{d}{c}\right|}=\frac{|a|}{|c|}\cdot\frac{\left|\frac{az+b}{a}\right|}{\left|\frac{zc+d}{c}\right|}=\frac{|a|}{|c|}\cdot\frac{\frac{\left|az+b\right|}{\left|a\right|}}{\frac{\left|zc+d\right|}{\left|c\right|}}=\frac{|a|}{|c|}\cdot\frac{|az+b|}{|a|}\cdot\frac{|c|}{|zc+d|}=$$ $$\frac{|a||c||az+b|}{|a||c||zc+d|}=\frac{|az+b|}{|zc+d|}$$