Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The only way I know how to verify something as an exact differential is when it is in the form $P(x,y) dx + Q(x,y) dy$. The book's definition states that such a differential is exact if there exists a function $g(x,y)$ such that $\frac{\partial g}{\partial x} = P$ and $\frac{\partial g}{\partial y} = Q$.

That's all well and good, but I'm not sure how to express $\cos{z}$ $dz$ in the form $P(x,y) dx + Q(x,y) dy$.

Considering $z=x+iy$ and $cos{z} = \frac{1}{2}(e^{iz}+e^{-iz})$, I know that that $cos{z} = \frac{1}{2}(e^{-y+ix}+e^{y-ix})$, but I'm not sure how $dx$ and $dy$ fit in here or what the rules are for getting them from $dz$. Or even if that question makes sense. Should I even be trying to express $cos{z}$ $dz$ in the form $P(x,y) dx + Q(x,y) dy$?

Please help?

share|improve this question
    
I think you might be misunderstanding a concept and it is hard to interpret your question. Please elaborate because $cos(z)dz$ is not in the form $P(x,y)dx + Q(x,y)dy$. To be exact we must have ${P_y} = {Q_x}$. –  glebovg Oct 31 '12 at 23:01
add comment

1 Answer 1

up vote 3 down vote accepted

The complex differential $\cos z \, dz$ is exact if there is a function $f$ such that $df = \cos z \, dz$, i.e., an analytic function $f(z)$ with $f'(z) = \cos z$. Now you know (I hope) that $f(z) = \sin z$ does the job, so it is exact. You can actually spell out everything in terms of real and imaginary parts, and use $dz = dx + i \, dy$ to make it fit into the template for real forms, but I would not do it unless you have to.

share|improve this answer
    
That makes far more sense. Thank you. –  Arthur Skirvin Nov 1 '12 at 0:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.