Prove sin(z1+z2)= sin(z1)cos(z2)+sin(z2)cos(z1), why this way?

  • Thread starter Thread starter nate9228
  • Start date Start date
nate9228
Messages
42
Reaction score
0

Homework Statement


What I had to prove was sin(z1+z2)= sin(z1)cos(z2)+sin(z2)cos(z1). I did this simply using the e function definitions for sin and cos and it turned out fine. I then flipped to the back of my book to double check and they have a completely different method for proving it. The book is Bak and Newman's Complex analysis.


Homework Equations





The Attempt at a Solution


The books method is very non-intuitive, at least to me. It says, "Let z2 be a fixed real number. Then f(z)= sin(z+z2) and g(z)= sin(z)cos(z2)+sin(z2)cos(z) are two entire functions of z which agree for all real values z= z1 and, hence, for all complex values z= z1, as well. Let z=z1 be any such complex number. Then, f(z)= sin(z1+z) and g(z)= sin(z1)cos(z)+ sin(z)cos(z1) agree for all real values z=z2 and, hence, for all complex values z=z2 as well." I do not see how this proves the identity in any way really. Can someone explain, in detail, how this works? And why it would be proved this way when it only takes two minutes to do it using the e function definitions?
 
Physics news on Phys.org
I guess the proof assumes that the identity is true for real numbers, and uses that knowledge to prove it for complex numbers as well.

I agree, it's very silly way of doing it.
 
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