0
$\begingroup$

Let $I \subseteq \mathbb{R}$ be an interval and $g: I \to \mathbb{C}$ continuous. Define $f: \mathbb{C} \backslash \overline{Im(f)} \to \mathbb{C}$ by

$f(z) := \int_I \frac{1}{g(x) - z} dx$

(with $\overline{Im(f)}$ being the closure of $Im(f)$.)

I now want to show that $f$ is analytic, and rewrite it into a power series that's (locally) defined for each $z_0 \in \mathbb{C} \backslash \overline{Im(f)}$.

Now I must admit that I don't really know how to get started. I haven't dealt much with analytic functions before. I know that a complex function is per definition analytic iff it can be written as a power series (therefore, by completing the second part of the task, the first one would follow, although I don't really know how I could write the function a), and iff it is differentiable once (hence differentiable infinitely often).

Therefore, it would also be sufficient for the first part to show that $f$ is differentiable, I think? How do I show that though? I would need to differentiate by $z$, whereas $f$ is defined as an integral with respect to $x$. I'm rather confused by this function.

$\endgroup$
  • $\begingroup$ Just a soft comment: $Im( f)$ is not great notation because in complex analysis, the first thing that comes to mind is the imaginary part of $f.$ $\endgroup$ – zhw. Oct 21 '15 at 23:35
  • $\begingroup$ OP: Did you actually check the approach in the answer you accepted? If you did so seriously, you probably met some problems... $\endgroup$ – Did Oct 28 '15 at 23:28
  • $\begingroup$ @Did Well I must admit that I just read the first part briefly, as the Cauchy-Riemann equations weren't introduced to me at that point, and seeing as the second part of the post answered my question in a shorter way (it showed that $f$ is analytic, since it has a power series representation). The Cauchy-Riemann equations were introduced to me this week and I planned to also work through the first part of the answer, but seeing as the second part was sufficient to answer the question, I didn't want to keep the poster waiting with accepting the answer. Thanks for pointing that out, though. $\endgroup$ – moran Oct 29 '15 at 18:35
  • $\begingroup$ @moran, I fix the problem and now it is perfectly right. $\endgroup$ – Math Wizard Nov 5 '15 at 1:33
1
$\begingroup$

Let $g(t)=g_1(t)+ig_2(t)$ $$ f(z) = \int_I \frac{1}{g(t) - z} dt=\int_I \frac{1}{g(t) - (x+iy)} dt=u+vi $$ where $$ u=\int_I \frac{g_1(t) - x}{(g_1(t) - x)^2+(g_2(t) - y)^2} dt,\quad v=-\int_I \frac{g_2(t) - y}{(g_1(t) - x)^2+(g_2(t) - y)^2} dt $$ By Cauchy-Riemann equation $$ u_x=v_y=\int_I\frac{(g_1(t) - x)^2-(g_2(t) - y)^2}{((g_1(t) - x)^2+(g_2(t) - y)^2)^2}dt $$ $$ u_y=-v_x=\int_I\frac{2(g_1(t) - x)(g_2(t) - y)}{((g_1(t) - x)^2+(g_2(t) - y)^2)^2}dt $$ So $f$ is analytic. And $$ f(z) = \int_I \frac{1}{g(t)(1 - z/g(t))} dt=\int_I\sum_{n=0}^{\infty} \frac{z^n}{g(t)^{n+1}} dt=\sum_{n=0}^{\infty}\int_I \frac{dt}{g(t)^{n+1}} z^n $$

$\endgroup$
  • $\begingroup$ First, this assumes that $g$ is real-valued to identify the real part and the imaginary part of $f$, although the question states that $g$ is complex-valued. Second, none of the four derivatives $u_x$, $v_y$, $v_x$, $u_y$ is what is written (and what would be $t$ in these formulas anyway?) $\endgroup$ – Did Oct 28 '15 at 23:27
  • $\begingroup$ ((Gotta love the sound of silent edits in the evening.)) $\endgroup$ – Did Oct 29 '15 at 10:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.