I am interested in the value of $\arctan (\cosh u)$ as $u\to -\infty $ $$\arctan (\cosh u)= \dfrac i 2 \log \left| \dfrac {1-i\cosh u}{1+i\cosh u} \right|$$
and since $$\cosh u= \dfrac {e^u+e^{-u}}{2}$$ we know it tends to positive infinity. So $$\dfrac i 2 \log \left| \dfrac {\dfrac {1}{\cosh u} - i} {\dfrac {1}{\cosh u} + i}\right| = \dfrac i 2 \log \left| \dfrac {-i}{i} \right| = - \dfrac i 2 \log \left| -1 \right|$$
Ok I am always confused with the modulus in the logs, whether it introduces modulus when we take $\log$ both sides of equation or not. If modulus does exist it becomes $0$, if not $\log(-1) = 1.36437635 i$ so $- \dfrac i 2 \log \left| -1 \right| = 0.68218817692$
But looking at the graph we can see $\arctan$ function tends to $\dfrac \pi 2$. So whats the reason for the difference when we use the log form?
Many thanks in advance
Fundamentally, your question seems to be about calculating the limit of the arctangent in complex form (and so the $\cosh$ is not actually important). First, let's work out what the form should be, and then decide how to interpret it.
In complex form, $y=\tan{x}$ is $$ y = \frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}, $$ using the definitions. We can rearrange this into a quadratic equation: $$ iy(e^{2ix}+1) = e^{2ix}-1 \\ (1-iy) e^{2ix} = (1+iy) \\ e^{2ix} = \frac{1+iy}{1-iy}. $$ Now, the problem comes when we take the logarithm: as a first guess, we'll get $$ x = \frac{1}{2i}\log{\left( \frac{1+iy}{1-iy} \right)}, $$ but since $e^{2\pi i}=1$, this is only one possibility: we could also add any multiple of $\pi$ to the right-hand side. So which one is the correct branch to choose? We have to take a continuous path from $y=0$, where we know $x=0$, so in fact the form above is right to start with. Now, we should look at the cases of $y$ large and positive real and $y$ large and negative real. For this, the form $$ \frac{1+iy}{1-iy} = \frac{1-y^2}{1+y^2} + i\frac{2y}{1+y^2} $$ is useful (and yes, does look rather like the $t$-formulae. Not a coincidence.). In particular, as $y \to +\infty$, the imaginary part is positive, so the argument of $e^{2ix}$ increases continuously, and in the complex plane, $-1$ is approached by passing above zero. This means that we must take the first value of $\log{(-1)} = i\arg{(-1)}$ with imaginary part/argument greater than zero: $i\pi$. Then the answer is that $\arctan{y} \to \pi/2$. Higher values ($3\pi/2, 5\pi/2, \dotsc$) would be reached by encircling the origin several times, but this cannot happen since the imaginary part of $\frac{1+iy}{1-iy}$ is always larger than zero.
On the other hand, if $y \to -\infty$, the imaginary part is always negative, so $-1$ is approached through decreasing values of the argument, and we end up at $-i\pi$, and $\arctan{y} \to -\infty$.