Why is the principal value of $\tan^{−1}θ$ always in $(\frac{\pi}{2}, \frac{\pi}{2})$? Also, why is the principal value of $\arg z$ always in the interval $(−π,π]$?

Is this a convention? Why these intervals only?

Thank you for answering.

  • $\begingroup$ Convention. But is a good convention, having the inverse tan of 0 being 0 is somewhat more intuitive than making it 2*pi. $\endgroup$ – Peter Webb Apr 2 '15 at 15:26
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    $\begingroup$ Yes, they are conventions. For the arctangent, it uses the branch passing through $(0,0)$. For the argument it uses the a detemination of the complex logarithm that includes the positive reals. $\endgroup$ – egreg Apr 2 '15 at 15:34

They are both conventions in a way. One can make a good argument that there is a good canonical choice for the function $\arctan$, namely the usual one which takes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, but there is no such choice for the argument function $\arg$.

The function $x \mapsto \tan x$ is periodic, with asymptotes at $\left(k + \frac{1}{2}\right) \pi$, $k \in \mathbb{Z}$. Now, the restriction of $x \mapsto \tan x$ to each interval $$I_k := \left(\left(k - \frac{1}{2}\right) \pi, \left(k + \frac{1}{2}\right) \pi)\right)$$ is bijective, so we can define a right inverse to $\tan$ for any of these intervals. But one of these is special, namely the interval $I_0$, which is symmetric about $0$; we define $\arctan$ to be the inverse of the restriction to this interval: $$\arctan := (\tan\!\vert_{I_0})^{-1}.$$ By construction, the function satisfies $\arctan 0 = 0$, and this is the only such inverse for which this holds.

On the other hand, there is no analogous line of reasoning for the argument function: A choice of branch of the complex logarithm is by definition a right inverse of the complex exponential function $\exp$, and such a choice determines the an argument function (namely, the imaginary part of that logarithm), but unlike the real tangent function, the domain $\mathbb{C}$ of the complex exponential is connected, and so there is no component of the domain on which the $\exp$ is bijective.

So, one can pick out particularly classes of branch cuts for $\arg$, e.g., those that take values in intervals $[\theta_0, \theta_0 + 2 \pi)$ or $(\theta_0, \theta_0 + 2 \pi]$, and as egreg points out, another natural restriction is choosing a branch cut so that $\arg$ is $0$ on the positive real half-axis. The particular choice that takes values in $(-\pi, \pi]$, however, is not preferred in the way that the usual $\arctan$ function is, and it's certainly always not the appropriate one.

For example, often an integral of a real function $f$ over $[0, \infty)$ can be handled by integrating a suitable complex extension of $f$ over a suitable closed contour and applying the Cauchy Integral Formula. If $f$ involves a logarithm, often the best contour is a keyhole contour, with key-slot along the positive real axis. In order to apply C.I.F., one should choose a logarithm function with branch cut outside the contour, namely, along the positive real axis, e.g., the one whose argument function takes values in $[0, 2 \pi)$.

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