complex tangent and cotangent

The tangent and the cotangent function for complex values of the $z$ are defined with the equations

$$\tan z:=\frac{\sin z}{\cos z},\cot z:=\frac{\cos z}{\sin z}.$$

Using the Euler’s formulae (http://planetmath.org/ComplexSineAndCosine), one also can define

$\tan z:=-i{\displaystyle \frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}},\cot z:=i{\displaystyle \frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}}.$ (1)

The subtraction formulae of cosine and sine (http://planetmath.org/ComplexSineAndCosine) yield an additional between the cotangent and tangent:

$$\cot (\frac{\pi }{2}-z)=\frac{\cos (\frac{\pi }{2}-z)}{\sin (\frac{\pi }{2}-z)}=\frac{\cos \frac{\pi }{2}\cos z+\sin \frac{\pi }{2}\sin z}{\sin \frac{\pi }{2}\cos z-\cos \frac{\pi }{2}\sin z}=\frac{\sin z}{\cos z}=\tan z.$$

Thus the properties of the tangent are easily derived from the corresponding properties of the cotangent.

Because of the identic equation  ${\cos }^{2}z+{\sin }^{2}z=1$  the cosine and sine do not vanish simultaneously, and so their quotient $\cot z$ is finite in all finite points $z$ of the complex plane except in the zeros  $z=n\pi$  ($n=0,\pm 1,\pm 2,\mathrm{\dots }$) of $\sin z$, where $\cot z$ becomes infinite.  We shall see that these multiples of $\pi$ are simple poles of $\cot z$.

If one moves from $z$ to $z+\pi$, then both $\cos z$ and $\sin z$ change their signs (cf. antiperiodic function), and therefore their quotient remains unchanged.  Accordingly, $\pi$ is a period of $\cot z$.  But if $\omega$ is an arbitrary period of $\cot z$, we have  $\cot (z+\omega )=\cot z$,  and especially  $z=0$ gives  $\cot \omega =\mathrm{\infty }$;  then (1) says that  $e^{i\omega }=e^{-i\omega }$,  i.e.  $e^{2i\omega }=1$.  Since the prime period of the complex exponential function is $2i\pi$, the last equation is valid only for the values  $\omega =n\pi$  ($n=0,\pm 1,\pm 2,\mathrm{\dots }$).  Thus we have shown that the prime period of $\cot z$ is $\pi$.

We know that

$$\frac{\sin z}{z}=\frac{\sin z-\sin 0}{z}\to \cos 0=1\mathit{\hspace{1em}}\mathrm{as}\mathit{\hspace{1em}}z\to 0;$$

therefore

$$z\cot z=\frac{z}{\sin z}\cdot \cos z\to 1\cdot \cos 0=1\mathit{\hspace{1em}}\mathrm{as}\mathit{\hspace{1em}}z\to 0.$$

This result, together with

$$\cot z\to \mathrm{\infty }\mathit{\hspace{1em}}\mathrm{as}\mathit{\hspace{1em}}z\to 0,$$

means that  $z=0$  is a simple pole of $\cot z$.

Because of the periodicity, $\cot z$ has the simple poles in the points $z=0,\pm \pi ,\pm 2\pi ,\mathrm{\dots }$.  Since one has the derivative

$$\frac{d\cot z}{dz}=-\frac{1}{{\sin }^{2}z},$$

$\cot z$ is holomorphic in all finite points except those poles, which accumulate only to the point  $z=\mathrm{\infty }$.  Thus the cotangent is a meromorphic function.  The same concerns naturally the tangent function.

As all meromorphic functions, the cotangent may be expressed as a series with the partial fraction (http://planetmath.org/PartialFractionsOfExpressions) terms of the form $\frac{a_{jk}}{{(z-p_j)}^k}$, where $p_j$’s are the poles — see this entry (http://planetmath.org/ExamplesOfInfiniteProducts).

The real (http://planetmath.org/CmplexFunction) and imaginary parts of tangent and cotangent are seen from the formulae

$$\tan (x+iy)=\frac{\sin x\cos x+i\sinh y\cosh y}{{\cos }^{2}x+{\sinh }^{2}y},$$

$$\cot (x+iy)=\frac{\sin x\cos x-i\sinh y\cosh y}{{\sin }^{2}x+{\sinh }^{2}y},$$

which may be derived from (1) by substituting  $z:=x+iy$ ($x,y\in ℝ$).

Title	complex tangent and cotangent
Canonical name	ComplexTangentAndCotangent
Date of creation	2013-03-22 16:49:56
Last modified on	2013-03-22 16:49:56
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Definition
Classification	msc 30A99
Classification	msc 30D10
Classification	msc 33B10
Related topic	ExamplesOfInfiniteProducts
Related topic	QuasiPeriodicFunction
Related topic	HyperbolicFunctions
Related topic	QuasiperiodicFunction