I have been fascinated (perhaps I am in love too with it) by Mathematics since I was a child. As a teenager in High School, I was a very curious student ( I am curious indeed yet) and I tried to understand some weird results I got in the classroom. This entry is devoted to some of those problems that made me wonder and think a lot out of class, at home. It took me some years and to learn complex variable function theory to understand one of the issues I could not understand before I learned complex variable. 3 years to understand a simple derivate! Yes, it is too much time. However, understanding stuff deeply carries time. Sometimes more, sometimes less.
This is the problem. A very simple problem indeed! The following “weird” (real) trigonometric-like function has a null derivative:
![\[ g(x)= \tan^{-1}((1+x)/(1-x))-\tan ^{-1} (x)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-719e6955c65398cf2aa69fce60b55fd6_l3.png "Rendered by QuickLaTeX.com")
where the 
is the arctangent function 
( the inverse of the tangent function 
).
Proof:
![\[ g'(x) = \dfrac{dg}{dx}=\dfrac{\dfrac{(1-x)+(1+x)}{(1-x)^2}}{1+\left(\dfrac{(1+x)}{(1-x)}\right)^2}-\dfrac{1}{1+x^2}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-8ee7ed283d83f22984482ca8c8b9c5a1_l3.png "Rendered by QuickLaTeX.com")
![\[ g'(x) = \dfrac{\dfrac{2}{(1-x)^2}}{1+\left(\dfrac{1+x}{1-x}\right)^2}-1/(1+x^2)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-920983684f411d1789755684edc435f9_l3.png "Rendered by QuickLaTeX.com")
![\[ g'(x) = \dfrac{2}{\left(1-x\right)^2+\left(1+x\right)^2}-1/(1+x^2)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-18e6ad0e91647a3076984db49d167b2d_l3.png "Rendered by QuickLaTeX.com")
![\[ g'(x) = \dfrac{2}{2+2x^2}-1/(1+x^2)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-19dcef8fb8fc60a39c5e70603c731f1b_l3.png "Rendered by QuickLaTeX.com")
![\[ g'(x) = \dfrac{1}{1+x^2}-\dfrac{1}{1+x^2}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-a007a6610139922a2ba76063c80f9c86_l3.png "Rendered by QuickLaTeX.com")
![\[ g'(x)=0 \]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-75464c12581634aa8d22492958b8439d_l3.png "Rendered by QuickLaTeX.com")
q.e.d.
Since the derivative 
is zero, the two functions must differ by a constant. We can guess that constant with usual real variable calculus. It’s quite simple:
![\[ g(x)=constant.\rightarrow g(x=0)= constant = \tan^{-1}((1+0)/(1-0))-\tan ^{-1} (0)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-c7ee460c60da1e57cca42b1e910078a4_l3.png "Rendered by QuickLaTeX.com")
![\[ g(0)= \tan^{-1}(1)-\tan ^{-1} (0)=\pi /4\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-2e914e83de8a301be3b2853cb4416652_l3.png "Rendered by QuickLaTeX.com")
So, the difference is 
. However, this calculation does not explain why the two functions differ by a constant. The secret lies in the complex function origin of the arctangent. In complex variable function theory, it can be proved that
![\[ \tan^{-1}(x)=\dfrac{1}{2i} \ln ((1+ix)/(1-ix))\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-1a53e223cc1718d08de77b929c0dbc6c_l3.png "Rendered by QuickLaTeX.com")
since
![\[ \tan(x)=y=\dfrac{\left[\dfrac{1}{2i} (\exp (ix)-\exp (-ix))\right]}{\left[\dfrac{1}{2} \left(\exp (ix)+\exp (-ix)\right)\right]}=\dfrac{1}{i}\dfrac{\exp (ix)-\exp (-ix)}{\exp (ix)+\exp (-ix)}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-0cd53b16e62aea5675f408a8489bf096_l3.png "Rendered by QuickLaTeX.com")
then 
and thus 
and therefore 
Q.E.D.
Now, we can calculate the functions in terms of complex variables:
![\[ \tan^{-1}\left((1+x)/(1-x)\right)=\arctan \left((1+x)/(1-x)\right)=\dfrac{1}{2i} \ln \left(\dfrac{(1+i\left(\frac{1+x}{1-x}\right)}{(1-i\left(\frac{1+x}{1-x}\right)}\right)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-69a4b7c63c6ff4af0bd5b7df9b5fb58c_l3.png "Rendered by QuickLaTeX.com")
We can make some algebra inside the logarithm function to get:
![\[ \arctan \left((1+x)/(1-x)\right)=\dfrac{1}{2i} \ln \left( \dfrac{(1-x)+i(1+x)}{(1-x)-i(1+x)} \right)= \dfrac{1}{2i} \ln \left( \dfrac{(1+ix)+i(1+ix)}{(1-ix)-i(1-ix)} \right)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-f55d71182f1088b8a5cbfa9902541cc2_l3.png "Rendered by QuickLaTeX.com")
By the other hand, we also have
![\[ \tan^{-1}(x)=\arctan (x)=\dfrac{1}{2i} \ln \left( \dfrac{1+ix}{1-ix} \right)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-bbb3a60f315893af95f8e108ed1990de_l3.png "Rendered by QuickLaTeX.com")
Thus,
![\[ g(x) = \dfrac{1}{2i} \left( \arctan ((1+x)/(1-x)) - \arctan (x) \right) = \dfrac{1}{2i} \left( \ln \left [ \dfrac{\dfrac{(1+ix)+i(1+ix)}{(1-ix)-i(1-ix)}}{ \dfrac{1+ix}{1-ix}} \right]\right)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-c64f319cd612e0a301d99ae38fd260b5_l3.png "Rendered by QuickLaTeX.com")
i.e.,the terms depending on x cancel to get a pure complex number! The number is
![\[ g(x)=number=\dfrac{1}{2i} \ln \left( \dfrac{1+i}{1-i}\right)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-64b908419fed0901aafd391103ff18e2_l3.png "Rendered by QuickLaTeX.com")
We have to calculate the logarith of the complex number
![\[ z=\dfrac{1+i}{1-i}=\dfrac{(1+i)(1+i)}{(1-i)(1+i)}=\dfrac{2i}{2}=i=exp(i\pi/2)\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-f7c101376ad675de0d3d91e212dc047e_l3.png "Rendered by QuickLaTeX.com")
Then, 
. Of course, that is we were expecting to get since in this case
![\[ g(x)=\dfrac{1}{2i} \dfrac{i\pi}{2}=\dfrac{\pi}{4}\]](http://www.thespectrumofriemannium.com/wp-content/ql-cache/quicklatex.com-87191e6c1edbfee94a04e60fc910c95c_l3.png "Rendered by QuickLaTeX.com")
as before! That is, we recover the phase difference we also got with real calculus. The origin of the cancellation was in the complex origin of the arctangent function! Beautiful mathematics! The complex world is fascinating but very enlightening even for real functions!
