# Thread: Principal branch

1. ## Principal branch

We define the principal branch of the inverse tangent function by,

$\displaystyle Arctan(z) = \frac{i}{2}\log(\frac{i+z}{i-z}$)

For which values of z is Arctan(z) defined?

2. The function $\displaystyle \tan^{-1} z = \frac{i}{2} \ln \frac{i+z}{i-z}$ is defined for all $\displaystyle z \in \mathbb{C}$ with the only exceptions of the points $\displaystyle z=i$ and $\displaystyle z=-i$, that represent the 'singularities' of the function ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by Richmond
We define the principal branch of the inverse tangent function by,

$\displaystyle Arctan(z) = \frac{i}{2}\log(\frac{i+z}{i-z}$)

For which values of z is Arctan(z) defined?
Its defined for any value that doesn't cause $\displaystyle \log\left(\frac{i+z}{i-z}\right)$ to be undefined.

Can you try to take it from here?

EDIT: chisigma beat me...

4. Thanks

So as long as $\displaystyle \frac{i}{2} \log \frac{i+z}{i-z}$ is defined,
we can say that $\displaystyle Arctan(z)$ is defined too which are all values except $\displaystyle z=i$ and $\displaystyle z=-i$

5. Originally Posted by Richmond
Thanks

So as long as $\displaystyle \frac{i}{2} \log \frac{i+z}{i-z}$ is defined,
we can say that $\displaystyle Arctan(z)$ is defined too which are all values except $\displaystyle z=i$ and $\displaystyle z=-i$
No, for beeing log z well defined, and analytic, we need eliminate a branch
(line beginning from 0). So, if nothing else is especified, logz is defined in
C\[0,\infty). You need to solve the equatiom

$\displaystyle \arg\frac{i+z}{i-z}=\pi$
to obtain the correct domain of definition.