Post by Admin on Dec 23, 2015 7:36:31 GMT
Gary
Comments on your published solution
1. $e^x\cos y$ could be the real or imaginary part of a function $f=u+iv$, since $u$ and $v$ are both real.
2. The result of Exercise 2 as I understand it is:
Given an analytic function $f=u+iv$ of $z$ then it follows that the real and imaginary parts of $f$ are harmonic.
So one way to show that a given real function of $x$ and $y$ is harmonic is to show that it is the real or imaginary part of an analytic function of $z$.
3. It does not follow that because $e^x\cos y$ is real that it is harmonic.
4. It does not follow that because a function of $x$ and $y$ is real that $\partial^2_x$ and $\partial^2_y$ are zero.
$\partial^2_x[e^x\cos y]=e^x\cos y\neq 0$ and $\partial^2_y[e^x\cos y]=-e^x\cos y\neq 0$
Vasco
Comments on your published solution
1. $e^x\cos y$ could be the real or imaginary part of a function $f=u+iv$, since $u$ and $v$ are both real.
2. The result of Exercise 2 as I understand it is:
Given an analytic function $f=u+iv$ of $z$ then it follows that the real and imaginary parts of $f$ are harmonic.
So one way to show that a given real function of $x$ and $y$ is harmonic is to show that it is the real or imaginary part of an analytic function of $z$.
3. It does not follow that because $e^x\cos y$ is real that it is harmonic.
4. It does not follow that because a function of $x$ and $y$ is real that $\partial^2_x$ and $\partial^2_y$ are zero.
$\partial^2_x[e^x\cos y]=e^x\cos y\neq 0$ and $\partial^2_y[e^x\cos y]=-e^x\cos y\neq 0$
Vasco