How do I check that this function is analytic using the derivatives?

Question

I have the following problem.

We define $\Omega=\Bbb{C}\setminus \{x:x\in (-\infty 0]\}$. For $z=x+iy\in \Omega$ we define the curve $\gamma_z$ going first from $1$ to $1+iy$ and then from $1+iy$ to $z$. Then we define $$f(z)=\int_{\gamma_z} \frac{1}{w}dw$$ for $z\in \Omega$. I want to calculate the derivative of $f$ to show that $f$ is analytic.

Now I have decomposed the curve into two linear curves and then written the integral over $\gamma_z$ as the integral over these two curves. I obtained the following curves: $$\gamma_{z1}(t)=(1+iy)t+(1-t)=1+tiy$$ and $$\gamma_{z2}(t)=(x+iy)t+(1-t)(1+iy)=t(x-1)+1+iy$$ for $t\in [0,1]$. Then I calculated the following integrals $$\int_{\gamma_{z1}}\frac{1}{w}dw=\left(\frac{1}{iy}+1\right) \log(1+iy)$$ and $$\int_{\gamma_{z2}} \frac{1}{w}dw=\log(x+iy)-\log(1+iy)$$ It follows $$\int_{\gamma_z}\frac{1}{w}dw=\frac{1}{iy}\log(1+iy)+\log(x+iy)$$.

But is this even true up to this point? Now I thought I'd just derive these complex functions and I'd be done but I'm not quite sure if I'm allowed to do it this way.

Could someone help me here?

Thank you very much!

Answer 1

We are going to show that $f'(z_0)={1\over z_0}$ for $z_0\notin (-\infty,0].$ To this end we need to prove $$\lim_{z\to z_0}{f(z)-f(z_0)\over z-z_0}={1\over z_0}$$ Let $d={\rm dist}(z_0,(-\infty, 0])=\displaystyle \min_{t\le 0}|z_0-t|.$ Assume $|z-z_0|<{d\over 2}.$ Then $z\notin (-\infty, 0].$ Intuitively, it means that if $z$ is close to $z_0,$ then $z$ does not belong to $(-\infty, 0].$ Observe (by drawing a picture) that the line segment $[z_0,z]$ connecting $z_0$ and $z$ does not intersect the half-line $(-\infty, 0].$ More precisely, the distance of every point $w$ in this segment to $(-\infty, 0]$ is greater than ${d\over 2}.$

While calculating the derivative we can restrict to $z$ close to $z_0,$ for example to $|z-z_0|<{d\over 2}.$ Let $z=x+iy$ and $z_0=x_0+iy_0.$ By definition $$f(z)-f(z_0)=\int\limits_{\gamma_z}{1\over w}\,dw -\int\limits_{\gamma_{z_0}}{1\over w}\,dw =\int\limits_{\gamma_{z_0,z}}{1\over w}\,dw$$ where $\gamma_{z_0,z}$ consists of three line segments: the first one from $z_0$ to $1+iy_0,$ the second one from $1+iy_0$ to $1+iy$ and the third one from $1+iy$ to $z.$ As the function ${1\over w}$ is holomorphic in the domain, the integrals over $\gamma_{z_0,z}$ and the line segment $[z_0,z]$ are equal, as the region between them does not intersect $(-\infty,0]$ (again the picture could be helpful). Therefore $$f(z)-f(z_0)=\int\limits_{[z_0,z]}{1\over w}\,dw$$ As $z$ is an antiderivative of $1$ we get $$z-z_0=\int\limits_{[z_0,z]}\,dw$$ hence $${1\over z_0}={1\over z-z_0}\int\limits_{[z_0,z]}{1\over z_0}\,dw$$ Furthermore $${f(z)-f(z_0)\over z-z_0}-{1\over z_0}={1\over z-z_0}\int\limits_{[z_0,z]}\left [{1\over w}-{1\over z_0}\right ]\,dw$$ We obtain $$\left |{f(z)-f(z_0)\over z-z_0}-{1\over z_0}\right |\le {1\over |z-z_0|} l([z_0,z])\max_{w\in [z_0,z]}{|w-z_0| \over |w|\,|z_0|}$$ Observe that the length $l([z_0,z])$ is equal $|z-z_0|$ and that $|w-z_0|\le |z-z_0|$ for $w\in [z_0,z].$ Moreover as the distance of $w$ to $(-\infty, 0]$ is greater than ${d\over 2},$ we get $|w|\ge {d\over 2}.$ Summarizing we obtain $$ \left |{f(z)-f(z_0)\over z-z_0}-{1\over z_0}\right |\le {2|z-z_0|\over d|z_0|}$$ Therefore $f'(z)={1\over z}$ for any $z\notin (-\infty, 0].$ Hence the function $f(z)$ is holomorphic.

Remark

1. A similar reasoning can be performed replacing the function $z^{-1}$ by any holomorphic function in order to show that it admits antiderivative in an appriopriate region.
2. I intentionally haven't used the fact that $f(z)=\log z, $ (the main branch of the complex logarithm) as afterwards one has to show (also a known fact) that $\log z$ is holomorphic in the described region. Moreover the complex logarithm can be actually defined in terms of the integral by choosing $\gamma_z$ consisting of the line from $1$ to $|z|$ and part of the circle with radius $|z|,$ from $|z|$ to $z.$ In this way we get $\log z=\log|z|+i{\rm Arg }\,z.$

Answer 2

I believe your first integral is in error. In general, given any branch of a logarithm and any contour, $\gamma$, within a simply connected region of it's domain, that starts at $z_1$ and ends at $z_2$, you have

$$ \int_\gamma \frac{1}{w} dw = \log(z_2) - \log(z_1) $$

Likewise, as expected your mutual point should cancel out. This is because the path is irrelevant when you are integrating a differentiable function. This should look like the fundamental theorem of calculus.

$$ \int_{\gamma_1} \frac{1}{w} dw = \log(1+yi) - \log(1)$$

$$ \int_{\gamma_2} \frac{1}{w} dw = \log(x+yi) - \log(1+yi)$$

therefore,

$$ \int_{\gamma} \frac{1}{w} dw = \log(x+yi) - \log(1)$$

Even if you did $u$-substitution and did the contour integrals directly, you would still get the same answers.

Answer 3

Your approach is circular. To compute the integral $\int_\gamma \frac 1 w dw$ you use the fact that the function $\frac 1 w$ has a (complex!) antiderivative which you correctly denote by $\log$. But if you use this fact, you essentially invoke what are supposed to prove.

In fact, it is well-known that on the sliced plane $\Omega$ there exist infinitely many branches of the complex logarithm, any two differing by an integral multiple of $2 \pi i$. Using any branch and any differentiable path $\gamma : [a, b] \to \Omega$ you get $$\int_\gamma \frac 1 w dw = \int_a^b \frac{1}{\gamma(t)}\gamma'(t)dt = \int_a^b \log'(\gamma(t))dt = \log(b) - \log(a) .$$ You may take the branch with $\log(1) = 0$ and get $$\int_{\gamma_{z1}} \frac 1 w dw = \log(1+iy) $$ which shows that you made a mistake in your computation. Similarly you get $$\int_{\gamma_{z2}} \frac 1 w dw = \lg(x+iy) - \log(1+iy) =\log(z) - \log(1+iy)$$ and therefore $$\int_{\gamma_{z}} \frac 1 w dw = \log(z) .$$

A non-circular proof has been given by Ryszard Szwarc.