Apply a corollary to show an identity

  • Thread starter Thread starter Shackleford
  • Start date Start date
  • Tags Tags
    Apply Identity
Shackleford
Messages
1,649
Reaction score
2

Homework Statement



Apply corollary to show that 2 sinz*sinw = cos(z-w) - cos(z+w) for any z,w ∈ ℂ

Homework Equations



2 sinz*sinw = cos(z-w) - cos(z+w) for any z,w ∈ ℂ

Corollary: Let f and g be analytic functions defined on a domain D ⊂ ℂ. Let E ⊂ D be a subset that has at least one limit point a in D. If f|E = g|E, then f = g in D.

The Attempt at a Solution



Of course, this trig identity holds in both the real and complex spaces. As I understand it, the notation f|E means that f is restricted to the subset E. I'm not sure if it's as easy as simply plugging in two values from E and then claiming that the functions are equal and then extending that to the larger domain D.
 
Physics news on Phys.org
Two values are not enough, the identity has to hold for every point in E in order to use the corollary - but you have the freedom to choose E as long as it satisfies the condition with the limit point.

I guess you are allowed to use the standard trigonometric identities for real numbers here.
 
mfb said:
Two values are not enough, the identity has to hold for every point in E in order to use the corollary.

I guess you are allowed to use the standard trigonometric identities for real numbers here.

Yeah, I guess I'll simply use the Angle-Sum and -Difference Identities.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top