Trig functions on complex numbers?

  • Thread starter davee123
  • Start date
  • #1
664
3
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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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))
 
Last edited by a moderator:
  • #3
662
0
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.
 
  • #4
291
0
[tex]\sin (x+iy) = \sin x \cos iy + \cos x \sin iy[/tex]
[tex]=\sin x \cosh y + i\cos x\sinh y[/tex]
 

Related Threads on Trig functions on complex numbers?

Replies
4
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
4
Views
3K
  • Last Post
Replies
22
Views
3K
Replies
5
Views
3K
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
1
Views
2K
Top