# Homework Help: Coordinates of a particle (coupled differential equations)

1. Dec 14, 2011

### Linday12

1. The problem statement, all variables and given/known data
I'm given two equations for coordinates of a certain particle in the xy plane, $\dot{x}$+ωy=0
and $\dot{y}$-ωx=0.
Then using the complex variable z=x+iy, find the differential for z, and solve it. Hence give x and y as functions of time.

2. Relevant equations

3. The attempt at a solution
I'm not sure how this is gotten:
$\dot{z}$-iωz=0

Any help would be highly appreciated. Thank you!

2. Dec 14, 2011

### I like Serena

Hi Linday12!

Can you write $\dot z$ in terms of x and y?
And then substitute the equations you have?

Can you also write iωz in terms of x and y?

3. Dec 14, 2011

### Linday12

Hi. I'm not really sure what to do. As far as I know, taking $\dot{z}$=$\dot{x}$+i$\dot{y}$, and $\dot{y}$=ωx and $\dot{x}$=-ωy, which doesn't seem to get me anywhere. I have no idea what I am doing.

4. Dec 14, 2011

### I like Serena

So substitute your 2nd and 3rd equation in the first?

5. Dec 14, 2011

### Linday12

Yes. I then get $\dot{z}$=-ωy+iωx=ω(-y+ix). So I can sort of see the relation here now, except my variables seem to be the wrong way, and I have no idea where the $\dot{z}$-iωz=o comes from still, because i would be gone if the variables were the proper way and I subbed in z for the x-iy.

6. Dec 14, 2011

### I like Serena

What is iz in terms of x and y?

7. Dec 14, 2011

### Linday12

Wow, interesting. So $\dot{z}$=iωz → $\dot{z}$-iωz=0 because iz=ix-y. Thank you! Sorry, one more question. Now that I have that, when solving it I get z=$z_{0}$$e^{ωt}$, but the answer has an additional phase angle $\phi$ in it, so I was just wondering if there was a quick explanation for that.

It's been a while since I did any differentials (I haven't had a class in them yet), I think perhaps I'm used to the general formula that includes the phase angle in the constant z_0, but if you were to solve it you would get ln(z)=iωt+c, which is where the phase angle came from, but then where does the z_0 come from? Because the answer is z=z_0$e^{ωt+\phi}$ Sorry, I hope that makes sense.

8. Dec 15, 2011

### I like Serena

Your z0 is the integration constant that is itself a complex number.
Write $z_0=r_0e^{i\phi}$ and you get:
$$z=z_0e^{i\omega t}=r_0e^{i\phi}e^{i\omega t}=r_0e^{i(\omega t + \phi)}$$

Btw, note the extra $i$ and the use of the non-negative real number r0 instead of z0.