Post by Admin on Dec 22, 2015 14:32:29 GMT
Gary
I just looked at your solution to this exercise and was a bit confused by your answer. $|f(z)|$ is in general not a constant. Take $f(z)=z$ for example. $|f(z)|=|z|=\sqrt{x^2+y^2}$ which is not a constant.
Vasco
I just looked at your solution to this exercise and was a bit confused by your answer. $|f(z)|$ is in general not a constant. Take $f(z)=z$ for example. $|f(z)|=|z|=\sqrt{x^2+y^2}$ which is not a constant.
Vasco
