- #1

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Can you please help me how to do it ?

I am really struggle with this question.

Thank you in advance

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In summary, the conversation discusses how a harmonic conjugate function is related to the concept of holomorphic functions and complex differentiability. The next step in the process is to integrate the partial derivative of the harmonic conjugate function with respect to one of the variables, treating the other variable as a constant. This will result in a function that satisfies the necessary conditions for being a harmonic conjugate. f

- #1

- 58

- 0

Can you please help me how to do it ?

I am really struggle with this question.

Thank you in advance

- #2

Homework Helper

MHB

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- #3

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Can you tell what is the next step that I should do please ?Hint:Aharmonic conjugate function$v$ must have that $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$, so that the function given by $f(x+iy)=u(x,y)+iv(x,y)$ isholomorphic(complex differentiable).

- #4

Homework Helper

MHB

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It's aCan you tell what is the next step that I should do please ?

More concisely:

$$\frac{\partial}{\partial x}(y)=0$$

So you can simplify your expression for $\frac{\partial u}{\partial x}$ a bit.

Next step is to integrate $\frac{\partial u}{\partial x}$ with respect to $y$ to find $v(x,y)$.

When we do so, we treat $x$ as a constant.

Let me give an example.

Suppose we have $u(x,y)=xy$. Then we have $\frac{\partial u}{\partial x}=y$.

And $v(x,y)=\int \frac{\partial u}{\partial x} \,dy = \int y\,dy = \frac 12y^2 + C(x)$, where $C(x)$ is some function of $x$.

We can verify by evaluating $\frac{\partial v}{\partial y} = \frac{\partial}{\partial y}\Big(\frac 12y^2 + C(x)\Big) = y$.

As we can see, we get indeed that $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$.

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