# Complex variable...

• Nov 3rd 2010, 10:17 AM
Dinkydoe
Complex variable...
I have problems solving: $\frac{dz}{dt}= \overline{z}e^{it}$

Seperating variables gives: $\frac{dz}{\overline{z}}=e^{it}dt$

But unfortunately, ...I'm not sure if we're able to integrate $\overline{z}$ w.r.t $z$
• Nov 3rd 2010, 10:58 AM
Ackbeet

$\dfrac{dz}{\bar{z}}=\dfrac{z\,dz}{z\bar{z}}=\dfrac {z\,dz}{|z|^{2}}?$
• Nov 3rd 2010, 11:18 AM
Dinkydoe
Thanks for the suggestion...but then again: Integrating $\frac{z}{|z|^2}$ w.r.t z is also a problem I wouldn't know how to solve..

The actual thing I'm trying to show here, is that not all solutions to this equation are bounded..., the question asks to solve the eq. explicitly
But I've no clue how. However, we may be able to use that $|\frac{dz}{dt}|=|z|$ for all t..., then try to show that all solution except the origin are unbounded.

(edit: i meant to show that not all solutions to this problem are bounded)
• Nov 3rd 2010, 10:45 PM
Prove It
$|z|^2$ is a constant.

So you should get

$\displaystyle \frac{1}{|z|^2} \int{z\,dz} = \int{e^{it}\,dt}$
• Nov 4th 2010, 01:25 AM
Dinkydoe
I don't fully understand. I agree if you said: $\int \frac{z}{|z|^2}dz=\int e^{it}dt$

I don't see why you could just treat $|z|^2$ as a constant. It's a real number depending on z. So why is it you can treat it as such?

$\int \frac{z}{|z|^2}dz = \frac{1}{|z|^2}\int zdz$...is this really true?
• Nov 4th 2010, 02:36 AM
Ackbeet
I'm with dinkydoe on this one. I would agree, from the DE, that

$\displaystyle\left|\frac{1}{\bar{z}}\,\frac{dz}{dt }\right|=\left|\frac{z}{|z|^{2}}\,\frac{dz}{dt}\ri ght|=\frac{1}{|z|^{2}}\left|z\,\frac{dz}{dt}\right |$

is a constant. But I don't see how you can get that $|z|^{2}$ is a constant.
• Nov 4th 2010, 05:21 AM
Dinkydoe
Ok, so my professor hinted to something like a substitution $z= r(t)e^{i\phi}$. Then $\overline{z}= r(t)e^{-i\phi}$

So we have $\frac{d}{dt}[r(t)e^{i\phi}] = e^{it}r(t)e^{-i\phi}$

So... $r'(t)= r(t)e^{(t-2\phi)i}\Longrightarrow r(t)= e^{-ie^{(t-2\phi)i}}$

And finally $z= e^{[\phi-e^{(t-2\phi)i}]i}$..

But i'm not even sure if this is a valid way of arguing...Furthermore, doesn't this generate only bounded solutions.
• Nov 4th 2010, 06:00 AM
Ackbeet
Yeah, I was thinking of a substitution somewhat like that, but I ran into difficulties. That one is better, I think. Query: does your solution satisfy the original DE?

I'm not so sure that this generates only bounded solutions. You don't know that the exponent there is purely imaginary, do you?