# Trig functions on complex numbers?

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
Homework Helper
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 eix= cos(x)+ i sin(x)?

Then you also know that ex+ iy= ex(cos(y)+ i sin(y))
Of course, then ex- iy= ex(cos(y)- i sin(y))

We can, from those same formulas, derive
cos(x)= (eix+ e-ix)/2 and
sin(x)= (eix- 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+ e4- 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))

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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.

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