Functions and limits · Definition

Complex trigonometric functions

complex analysiscomplex functions

Before you start you need to know

They let sine and cosine take complex arguments and show that real trigonometry and the exponential are two faces of the same function.

Definition

The sine \(\sin z\) and cosine \(\cos z\) of a complex number \(z\) are the functions that extend the real ones through the complex exponential \(e^{iz}\).

\[\sin z=\frac{e^{iz}-e^{-iz}}{2i},\qquad \cos z=\frac{e^{iz}+e^{-iz}}{2}\]

Euler's formula \(e^{iy}=\cos y+i\sin y\) and its conjugate \(e^{-iy}=\cos y-i\sin y\) can be solved for sine and cosine: adding and subtracting gives exactly the expressions in the box. They are taken as the definition for any \(z\), and agree with the real functions when \(z\) is real. The other functions are defined as usual: \(\tan z=\sin z/\cos z\), and so on.

Almost every real identity survives: \(\sin^2z+\cos^2z=1\), \(\sin(z+w)=\sin z\cos w+\cos z\sin w\), \(\sin(z+2\pi)=\sin z\) and \(\sin(z+\pi)=-\sin z\). They are checked by substituting the exponentials, so knowing the properties of \(e^{z}\) is enough. The zeros are also the familiar ones: \(\sin z=0\) only at \(z=k\pi\), and \(\cos z=0\) only at \(z=\frac{\pi}{2}+k\pi\).

What changes is the size. Writing \(z=x+iy\) gives \(\sin z=\sin x\cosh y+i\cos x\sinh y\) and \(\cos z=\cos x\cosh y-i\sin x\sinh y\). Since \(\cosh y\) and \(\sinh y\) grow without bound, neither sine nor cosine is bounded in the plane, and for instance \(\cos(iy)=\cosh y\) takes the value 2. The inequality \(|\sin z|\le1\) is a property of the real line only.

A common mistake is to treat the hyperbolic functions as a separate topic: in fact \(\cos(iy)=\cosh y\) and \(\sin(iy)=i\sinh y\), so trigonometric and hyperbolic functions are one family seen along different axes. This will be used when studying entire functions and when solving equations such as \(\cos z=w\).

Think about it

Why does \(\cos z=2\) have a solution in \(\mathbb{C}\) even though \(|\cos x|\le 1\) for real \(x\)?

Because the complex cosine is unbounded: \(\cos(iy)=\cosh y\) takes every value \(\ge1\), so \(z=i\ln(2+\sqrt3)\) works.

Use the definitions to prove that \(\sin(-z)=-\sin z\).

Replacing \(z\) by \(-z\) swaps \(e^{iz}\) and \(e^{-iz}\), so the numerator changes sign while the denominator \(2i\) does not.

Exercises 10

References

  • Elias M. Stein, Rami Shakarchi, Complex Analysis, 1st edition, Princeton University Press, 2003. Cap. 1, §2.3
  • James Ward Brown, Ruel V. Churchill, Complex Variables and Applications, 9th edition, McGraw-Hill, 2014. §34–35