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I have $$F(z) = \phi + i\psi$$

Also, it is given that $$u = \frac{\delta \phi}{\delta x}, \ \ v = \frac{\delta\phi}{\delta y}$$ and $$u = \frac{\delta \psi}{\delta x}, \ \ v = -\frac{\delta\psi}{\delta y}$$

Now I have to prove that $$\frac{dF}{dz} = u - iv$$

But I am getting terms like $\dfrac{dx}{dz}$ and $\dfrac{dy}{dz}$ in the differentiated term. How do I proceed?

EDIT:

As pointed out the function $\psi$ was wrongly defined by me. It's actually

$$u = \frac{\delta \psi}{\delta y}, \ \ v = -\frac{\delta\psi}{\delta x}$$

Ravi Upadhyay
  • 791

1 Answers1

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Let's get some formulas straight in the first place. You have $$F(z) = \phi + i\psi $$ Also, it is given that $$u = \frac{\partial \phi}{\partial x}, \ \ v = \frac{\partial \phi}{\partial y} $$ And no, what you have instead is this: $$u = \frac{\partial \psi}{\partial y}, \ \ v = -\frac{\partial \psi}{\partial x} $$ Differentiate in the $x$-direction or in the $y$-direction - it doesn't matter (why?): $$ \frac{dF}{dz} = \frac{\partial F}{\partial x}\frac{\partial x}{\partial z} \qquad ; \qquad \frac{dF}{dz} = \frac{\partial F}{\partial y}\frac{\partial y}{\partial z} $$ With $$ \frac{\partial z}{\partial x} = 1 \quad \Longrightarrow \quad \frac{\partial x}{\partial z} = 1/1 = 1 \qquad ; \qquad \frac{\partial z}{\partial y} = i \quad \Longrightarrow \quad \frac{\partial y}{\partial z} = 1/i = -i $$ Hence in the $x$- or $y$-direction, take your favorite: $$ \frac{dF}{dz} = 1 \left(\frac{\partial \phi}{\partial x} + i \frac{\partial \psi}{\partial x} \right) = -i \left( \frac{\partial \phi}{\partial y} + i \frac{\partial \psi}{\partial y}\right) = u - i v $$ More about this subject in Cauchy-Riemann Equations (PDF).

Han de Bruijn
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