$d(f(z)dz)=0$ iff $f$ satisfies Cauchy Riemann equations

Question

Define $$d(\omega+i\nu)=d\omega + id\nu $$ $$(\omega+i\nu)\land(\theta+i\lambda)=\omega \land\theta-\nu\land\lambda+i(\nu\land\theta +\omega\land\lambda) $$ $$dz=dx+idy$$ Proof $d(f(z)dz)=0$ iff $f$ satisfies the Cauchy Riemann equations
Idk where can i find information, please some help for bibliography or related questions

Answer 1

Hint:

Say $f = u + iv$. Then

$$f \ dz = (u + iv)\ (dx + i \ dy) = (u \ dx - v \ dy) + i (v \ dx + u \ dy)$$

Now we can compute $d(f\ dz)$ using the usual computation rules for (real) differential forms. What do we get? What happens when we set the real and imaginary parts to be $0$ (since we're assuming $d(f \ dz) = 0$)?

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I hope this helps ^_^