Trig functions on complex numbers?

[davee123]

Out of curiosity, what happens when you try to perform a trig function on a complex number? So, say, sin(4i+3)? Is this undefined since angles are only capable of being real numbers, or is there an agreed behavior for complex numbers?

DaveE

 

[HallsofIvy]

Even in terms of real numbers, the arguments for the functions sin(x) and cos(x) are NOT assumed to be angles!

Do you know that e^ix= cos(x)+ i sin(x)?

Then you also know that e^{x+ iy}= e^x(cos(y)+ i sin(y))
Of course, then e^{x- iy}= e^x(cos(y)- i sin(y))

We can, from those same formulas, derive
cos(x)= (e^ix+ e^-ix)/2 and
sin(x)= (e^ix- e^-ix)/(2i)
While those are derived, originally, with x real, we can easily extend them as definitions for functions of complex x.

In particular, if x= 4i+ 3, then
cos(4i+3)= (e^{-4+ 3i}+ e^{4- 3i})/2
= (e^-4(cos(3)+ i sin(3))+ e⁴(cos(3)- i sin(3))
= (e^-4+ e⁴)cos(3)/2 + i(e
^{and
sin(4i+ 3)= (e-4i+3- e-4+ 3i)/(2i)
= (e-4(cos(3)+ i sin(3))/2i+ i(e4cos(3)+ i sin(3))}

 

[belliott4488]

You might also want to look up the definitions of the hyperbolic trig functions, cosh and sinh. Their definitions look a lot like the ones HallsofIvy gave for cos and sin, but without the factors of i. Thus, putting an i in the argument of a sin or cos gives you a cosh or sinh, and vice versa.

 

[Kummer]

$$\sin (x+iy) = \sin x \cos iy + \cos x \sin iy$$
$$=\sin x \cosh y + i\cos x\sinh y$$