Let's $z_0=-e^{i\theta_0}$ and $z$ not in the line $(Oz_0)$

How can I compute the following integral

$$\int_0^1 \frac{dt}{z_0 + t(z-z_0)}$$

without talking about complex logarithm and holomorphic functions?

The value is obviously $\frac{1}{z-z_0}(\log(z) - \log(z_0))$, where $\log$ is a determination of logarithm everywhere except on $(Oz_0)$ but I'd like to compute this integral without all the theory of complex logarithms.


1 Answer 1


If $z$ and $z_0$ are any distinct, nonzero complex numbers which are not on the same line through the origin, then the number $z_0/(z - z_0)$ has a nonzero imaginary part. Let $a$ and $b$ denote the real and imaginary parts of $z_0/(z - z_0),$ respectively.

The integral is

$$\begin{align} \frac{1}{z - z_0}\int_0^1 \frac{dt}{(a + bi) + t} &= \frac{1}{z - z_0}\int_0^1 \frac{(a - bi + t)\,dt}{\lvert a + bi + t\rvert^2}\\ &= \frac{1}{z - z_0}\biggl[\int_0^1 \frac{(t + a)\,dt}{(t + a)^2 + b^2} + i\int_0^1 \frac{(-b)\,dt}{(t + a)^2 + b^2}\biggr]. \end{align}$$

The first integrand has $\log\sqrt{(t + a)^2 + b^2}$ as an antiderivative. This is the usual logarithm defined on the positive reals, so there is no complex logarithm being used here.

The second integrand has $\cot^{-1}[(t + a)/b]$, with range in $(0,\pi),$ as an antiderivative. So, the result is

$$ \frac{1}{z - z_0}\biggl[\log \sqrt{\frac{(1 + a)^2 + b^2}{a^2 + b^2}} + i\Bigl(\cot^{-1}\frac{1 + a}{b} - \cot^{-1}\frac{a}{b}\Bigr)\biggr]. $$

This simplifies to

$$ \frac{1}{z - z_0}\Bigl[\log\frac{\lvert z\rvert}{\lvert z_0\rvert}{} + i \Bigl(\cot^{-1}\frac{1 + a}{b} - \cot^{-1}\frac{a}{b}\Bigr)\Bigr], $$

using the definition of $a$ and $b$.

Edit: I would like to point out that the imaginary part of the expression in brackets is one way to calculate the difference of the arguments of $z$ and $z_0$, so this (expectedly) agrees with a complex logarithm.


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