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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?

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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
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.

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That makes far more sense. Thank you. – Arthur Skirvin Nov 1 '12 at 0:16

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