I'm doing an exercise from the book "Algorithms for Computer Algebra" by Keith O. Geddes. I'm asked to show that if $u$ and $v$ are two relatively prime polynomials in $\mathbb{Q}[x]$ and $s$ and $t$ are polynomials such that $deg(s) < deg(v)$ and $deg(t) <deg(u)$ satisfying $us-vt = 1$, then
$\frac{d}{dx}(\log{\frac{u+iv}{u-iv}}) = -2i\frac{d}{dx}(\arctan(ut+vs)) + \frac{d}{dx}(\log{\frac{t+is}{t-is}})$
But if:
$u = x$, $v = 1$ and $s = 0$, $t = -1$
then:
$\frac{d}{dx}(\log{\frac{x+i}{x-i}}) = \frac{-2i}{x^2+1} $
and
$-2i\frac{d}{dx}(\arctan(-x)) + \frac{d}{dx}(\log{\frac{-1}{-1}}) = \frac{2i}{x^2+1}$
These two differ by a sign and are not equal. Is it posible that there is a mistake in the book (and if there is one, then what is the right formula for converting logarithms to inverse tangents?) or have I done something wrong?
One has $$ \left|\frac{z}{\bar z}\right|=1\text{ and } \arg\left(\frac{z}{\bar z}\right)=\arg\left(\frac{z^2}{|z|^2}\right)=2\arg(z) \pmod{2\pi} $$ so that $$ Ln\left(\frac{z}{\bar z}\right)=0+2i\arg(z)\pmod{2i\pi} $$ In your case, $z=(u+iv)(t-is)=ut+vs+i(vt-us)=ut+vs-i$ and $$\arg(z)=-\tfrac12\pi+\arg(iz)=-\tfrac12\pi+\arctan(ut+vs)$$ The only role of the differentiation is to allow to neglect the constants in these formulas.
So apparently, your observation that there should be no minus sign is right.