# What is this equation, and how do I start to tackle it?

1. Jun 29, 2013

### CH3CH2OH

This is my first post. Ive been checking back with this forum for years while studying, but i've got one that I truly cannot solve on my own because I don't know where to start...

1. The problem statement, all variables and given/known data

prove this is true.

L$^{-1}${$\frac{K}{(s+\alpha-i\beta)^{r}}$+$\frac{K^{*}}{(s+\alpha+i\beta)^{r}}$}=$\frac{2|K|}{(r-1)!}$e$^{-\alpha t}$cos(βt+$\vartheta)$u(t)

2. Relevant equations
All I know is that K* has something to do with a modulus of K, and the information that cosine is acting on is the phase angle, possibly an arctan of two constants.

3. The attempt at a solution

I know that there is an inverse laplace involved. I see the complex variables and know that may be where i get the trigonometric functions from. I am unfamiliar with K and Modulus of K, also phase angles. I am not sure where to take that from there. This was a challenge question posted in class. We were given no knowledge of what it relates to or what the name of the equation is, just that we should make an effort to prove it. We are given permission to solicit any resources available to us.
I am not looking for an outright proof, but some hints to move me in the right direction. This is beyond differential equations, but before graduate work level, i think...Also, the course work where this might be found is in either mathematics or engineering

2. Jun 29, 2013

### SteamKing

Staff Emeritus
3. Jun 29, 2013

### Ray Vickson

Assuming that $\alpha$ and $\beta$ are real, the two terms in the Laplace transform are complex conjugates of one another, because $K^*$ is the complex conjugate of $K$ (NOT the modulus of $K$). Therefore, the whole inverse Laplace transform is twice the real part of the inverse transform of the first term. If you write $K = u + i v$, with real $u,v$, and if $r$ is a positive integer, you can write the result in the required form by using some standard trigonometric addition laws.