3
$\begingroup$

I would like to calculate the Jacobian transformation of the measure $dx dy$ $z = e^{x+iy}, \bar{z}=e^{x-iy}$ which should give $-2i|z|^2$. I am not sure how to do this in Mathematica. Further, I need it to generalize it to more variables. So, how could I write this transformation in "1-step"?

$\endgroup$
4
  • 2
    $\begingroup$ What's the definition of the Jacobian that you use? $\endgroup$
    – yohbs
    Commented Jun 19, 2017 at 19:05
  • $\begingroup$ The standard one, the matrix $h_{ij} = \partial f_{i}/\partial u^j$. $\endgroup$
    – Gorbz
    Commented Jun 20, 2017 at 9:51
  • 1
    $\begingroup$ Funny. Neither h, f or u appear in your question. Also, this is a matrix and in the question you said it should be a scalar. It would help us to help you if you gave a more detailed explanation of what you are looking for, or at least a derivation of the expected result (I got a different answer) $\endgroup$
    – yohbs
    Commented Jun 20, 2017 at 12:32
  • $\begingroup$ Related, possible duplicate: How to make Jacobian automatically in Mathematica $\endgroup$
    – Jens
    Commented Jun 22, 2017 at 4:26

2 Answers 2

5
$\begingroup$
z = Exp[x + I y];

zconj = Exp[x - I y];


vars = {x, y};

funcs = {z, zconj};

matJacobi = Outer[D, funcs, vars]
{{E^(x + I y), I E^(x + I y)}, {E^(x - I y), -I E^(x - I y)}}
detJacobi = Det[matJacobi]
-2 I E^(2 x)

but I do not get -2i Abs[z]^2, but -2i Exp[2x]

$\endgroup$
4
$\begingroup$

You can just use D. The Jacobian is:

j = D[{Exp[x + I y], Exp[x - I y]}, {{x, y}}]

and the Determinant of this is:

Det[j]

-2 I E^(2 x)
$\endgroup$

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