value of $\arctan (\cosh u)$ as $u \to -\infty $ 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 
 A: 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 $\arctan{y} \to -\infty$.
