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I'm reading a rather old mechanics paper and the author uses the following to derive certain stresses.

Let x + iy = f(u + iv)

Then, J$e^{i\phi}$ = f'(u+iv)

J is the Jacobian, and $\phi$ is the angle between a tangent to the curve v = constant and the x axis

given f is a conformal mapping.

How can this be derived?

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What's $J$ and $\phi$? – Hans Lundmark Apr 5 '11 at 8:04
oops—sorry. J is the Jacobian, and $\phi$ is the angle between a tangent to the curve v = constant and the x axis – Kunal Bhalla Apr 5 '11 at 8:15
I don't see the question. J is a gradient operator, and it appears the author is stating (not claiming or ignoring a proof for) that the jacobian matrix is $J\{e^{i \phi(u,v)}\} = f'(u + iv)$ -- which is a matrix of 1st derivatives w/respect to each variable. The ambiguity I see is in the definition of $f'(.)$; is it the derivative with respect to the single complex variable $z = u + i v$? – Brian Vandenberg Apr 5 '11 at 8:35
i'm actually asking for a proof of $Je^{iϕ(u,v)}=f′(u+iv)$. f'() is the derivative with respect to a single complex variable z = u+iv. – Kunal Bhalla Apr 5 '11 at 14:00
up vote 1 down vote accepted

Dear Kunal Bhalla, probably the author of the paper is going to introduce just a notation. Given a conformal map $f$ let us denote by $J$ the modulus of $f'$ and by $\phi$ its argument modulo $2\pi$.

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Thanks—based on Brian's answer that's what I was thinking—I misread J to be the Jacobian and considered it to be the complex jacobian and just got stuck on that. – Kunal Bhalla Apr 5 '11 at 17:05

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