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I have no problems doing this question but failed to understand how chain rule works again. For example, i have $\overline{f(\overline{z})} = A(x,y) + iB(x,y)$

Which also means that $A(x,y) = u(x,-y)$ and $B(x,y) = -v(x,-y)$

Although i got the answer that $A_x(x,y) = u_x(x,-y), A_y(x,y) = (-1) \cdot u_y(x,-y)$, i failed to see how the answer $A_y(x,y) = (-1) \cdot u_y(x,-y)$ is achieved via chain rule. Maybe someone can enlighten me on the exact steps and formulas used to achieved the chain rules result.

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    This result is known as the Schwartz reflexion principle see \url{documents.lamacs.fr/cours/macs2/ana_app/holomorphe.pdf} – 2017-02-07

2 Answers 2

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We know that $u_x(x,y)=v_y(x,y)$ and $u_y(x,y)=-v_x(x,y)$.

Now, let $s(x,y)=u(x,-y)$ and $t(x,y)=-v(x,-y)$. Then, we can write

$$\begin{align} \frac{\partial s(x,y)}{\partial x}&=\frac{\partial u(x,-y)}{\partial x}\\\\ &=\frac{\partial v(x,-y)}{\partial y}\\\\ &=-\frac{\partial t(x,y)}{\partial y} \end{align}$$

We can also write

$$\begin{align} \frac{\partial s(x,y)}{\partial y}&=\frac{\partial u(x,-y)}{\partial y}\\\\ &=-\frac{\partial v(x,-y)}{\partial x}\\\\ &=\frac{\partial t(x,y)}{\partial x} \end{align}$$

Hence, $s(x,y)-it(x,y)$ satisfies the CREs. And we are done!

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    yes , but i still dont understand how to apply the chain rule in this scenario, there is some variation to the question i posted previosly.. – 2017-02-07
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    Which step is posing a challenge? $$\frac{\partial u(x,-y)}{\partial y}=\frac{\partial (-y)}{\partial y}\frac{\partial u(x,-y)}{\partial (-y)}=-\frac{\partial u(x,-y)}{\partial (-y)}=-\left. u_y(x,y)\right|_{y=-y}$$ – 2017-02-07
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    For example, in your notations, i can not find a "right" formula or way to partial differentiate $s(x,y) = u(x,-y)$ with respect to $y$. One reason is because now the inside of $u$ has a $-y$. – 2017-02-07
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    Please see my previous comment. – 2017-02-07
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    ok, so i finally see how the first few steps are done, but in the last step, i did not quite get it, according to your steps, shouldnt $-\dfrac{du(x,-y)}{d(-y)} = -u_{-y}(x,-y)$? – 2017-02-07
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    It is indeed. And that is equal to $-u_y(x,y)$ evaluated at $y=-y$. – 2017-02-07
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First, let's write $u=u(x_1,x_2)$ for a moment. Suppose $A(x,y) = u(x,g(y))$. Doesn't the chain rule tell you that $A_y(x,y) = u_2(x,g(y))g'(y)$ [where $u_2$ denotes the partial derivative with respect to the second variable]?