Revision as of 18:26, 20 May 2015

The $\mathrm{arctan}$ function is the inverse function of the tangent function.

Arctan.png

Graph of $\mathrm{arctan}$ on $[-1,1]$.
Complex arctan.jpg

Domain coloring of the analytic continuation of $\mathrm{arctan}$.

Properties

Proposition: $\dfrac{d}{dz} \mathrm{arctan}(z) = \dfrac{1}{z^2+1}$

Proof: If $\theta=\mathrm{arctan}(z)$ then $\tan \theta = z$. Now use implicit differentiation with respect to $z$ yields $$\sec^2(\theta)\theta'=1.$$ The following triangle shows that $\sec^2(\mathrm{arctan}(z))=z^2+1$:

Substituting back in $\theta=\mathrm{arccos(z)}$ yields the formula $$\dfrac{d}{dz} \mathrm{arccos(z)} = \dfrac{1}{\sec^2(\mathrm{arctan(z)})} = \dfrac{1}{z^2+1}. █$$

Proposition: $\displaystyle\int \mathrm{arctan}(z) = z\mathrm{arctan}(z) - \dfrac{1}{2}\log(1+z^2)+C$

Proof: █

Proposition: $\mathrm{arctan}(z) = \mathrm{arccot}\left( \dfrac{1}{z} \right)$

Proof: █

Relationship between arctan and hypergeometric 2F1