renegade05
Full Member
- Joined
- Sep 10, 2010
- Messages
- 260
Problem:
Show that \(\displaystyle f(z) = ln(|z|) + iArg(z)\) is analytic on \(\displaystyle \mathbb{C} \backslash \mathbb{R}\_ = \mathbb{C} \backslash \{x \epsilon \mathbb{R} : x \le0\} \)
Agh, this one is giving me a headache.
I'm not sure how to prove this is analytic. I know I must show the Cauchy-Riemann equations hold true.
I think I need to have special cases for this one:
Let
\(\displaystyle u(x,y) = ln(|z|) = ln(|x+iy|)\)
\(\displaystyle v(x,y) = Arg(z) = Arg(x+iy) \)
for:
\(\displaystyle x>0; Arg(z)=arctan\frac{y}{x}\)
\(\displaystyle y>0; Arg(z)=arccot\frac{x}{y}\)
\(\displaystyle y<0; Arg(z)=arccot\frac{x}{y} - \pi\)
Should I take:
\(\displaystyle \frac{\partial v}{\partial y}\) and \(\displaystyle \frac{\partial v}{\partial x}\) in each case?
How can I take \(\displaystyle \frac{\partial u}{\partial y}\) and \(\displaystyle \frac{\partial u}{\partial x}\) ?
To show:
\(\displaystyle \frac{\partial u}{\partial x}\) = \(\displaystyle \frac{\partial v}{\partial y}\)
and
\(\displaystyle \frac{\partial v}{\partial x}\) = \(\displaystyle -\frac{\partial u}{\partial y}\)
Show that \(\displaystyle f(z) = ln(|z|) + iArg(z)\) is analytic on \(\displaystyle \mathbb{C} \backslash \mathbb{R}\_ = \mathbb{C} \backslash \{x \epsilon \mathbb{R} : x \le0\} \)
Agh, this one is giving me a headache.
I'm not sure how to prove this is analytic. I know I must show the Cauchy-Riemann equations hold true.
I think I need to have special cases for this one:
Let
\(\displaystyle u(x,y) = ln(|z|) = ln(|x+iy|)\)
\(\displaystyle v(x,y) = Arg(z) = Arg(x+iy) \)
for:
\(\displaystyle x>0; Arg(z)=arctan\frac{y}{x}\)
\(\displaystyle y>0; Arg(z)=arccot\frac{x}{y}\)
\(\displaystyle y<0; Arg(z)=arccot\frac{x}{y} - \pi\)
Should I take:
\(\displaystyle \frac{\partial v}{\partial y}\) and \(\displaystyle \frac{\partial v}{\partial x}\) in each case?
How can I take \(\displaystyle \frac{\partial u}{\partial y}\) and \(\displaystyle \frac{\partial u}{\partial x}\) ?
To show:
\(\displaystyle \frac{\partial u}{\partial x}\) = \(\displaystyle \frac{\partial v}{\partial y}\)
and
\(\displaystyle \frac{\partial v}{\partial x}\) = \(\displaystyle -\frac{\partial u}{\partial y}\)