# Simplifying a solution such that the imaginary number goes away

1. Sep 25, 2007

### jwang34

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

Given a general solution to a system of differential equations:

Y(t)= C1(1;2i)e^(2it)+C2(1;-2i)e^(-2it)

side note: i is sqrt of -1, and the (1;2i) is a 2 by 1 matrix. The idea is to simplify the solution such that the imaginary components go away.

2. Relevant equations

Euler's formula.

3. The attempt at a solution
I have subbed the e^(2it) and e^(-2it) with their appropriate sin and cos counterparts. This is what I have have:

Y(t)= C1(1;2i)[cos(2t)+isin(2t)]+C2(1;-2i)[cos(2t)-isin(2t)]

and so from here, I lumped the cosines and sines together:

Y(t)=[C1(1;2i)+C2(1;-2i)]cos(2t)+i[C1(1;2i)+C2(1;-2i)]sin(2t)

I know I need to introduce two new free parameters to replace C1 and C2 but I'm not sure if I do that now or is there another way? I'm especially stumped as how to rid of the i from the (1;2i) and (1;-2i). Any suggestions are welcome. thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 25, 2007

### Dick

This is really just Y=C1*z+C2*conjugate(z) where z is a complex matrix. The condition for Y to be real is Y=conjugate(Y). Putting it all together, don't you just get C1=conjugate(C2)?

3. Sep 25, 2007

### jwang34

I'm not sure I totally understand. So you're saying I set C1=the conjugate of C2? Or should I do set C1=C2*conjugate(z)?

4. Sep 25, 2007

### Dick

C1 and C2 are constants, right? I'm just saying if C1=(a+bi) then C2=(a-bi). a and b also constants. That will cancel all of the imaginary parts. You need to check this yourself. Shouldn't this:

Y(t)=[C1(1;2i)+C2(1;-2i)]cos(2t)+i[C1(1;2i)+C2(1;-2i)]sin(2t)

be this:

Y(t)=[C1(1;2i)+C2(1;-2i)]cos(2t)+i[C1(1;2i)-C2(1;-2i)]sin(2t)

5. Sep 26, 2007

### jwang34

Yes, you are right. I think I understand it now. So I should set C1=(a+bi) and C2=(a-bi) and the complex matrices will be resolved. Thank you so much!