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I was working on a problem that asked us to find the harmonic conjugate of $u=x^3-3xy^2$.

After determining that $u$ is indeed harmonic by using Laplace's equation $\nabla u^2=\frac{\partial ^2}{\partial x^2}\left(u\right)+\frac{\partial ^2}{\partial y^2}\left(u\right)=0$, I determined its harmonic conjugate to be $v(x,y)=3x^2y-y^3+C$ by using the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$.

Now, the complex function is $f(x,y)=u(x,y)+iv(x,y)=(x^3-3xy^2)+i(3x^2y-y^3)+iC$.

I would like to transform this into $f(z)$ instead of $f(x,y)$. How would I go about doing this?

I have tried to substitute the equations $x=\frac{z+\overline{z}}{2}$ and $y=\frac{z-\overline{z}}{2i}$:

$f(x,y)=(x^3-3xy^2)+i(3x^2y-y^3)+iC$, which becomes:

$f(z)=\left(\frac{z+\overline{z}}{2}\right)^3-3\left(\frac{z+\overline{z}}{2}\right)\left(\frac{z-\overline{z}}{2i}\right)^2+i\left[3\left(\frac{z+\overline{z}}{2}\right)^2\left(\frac{z-\overline{z}}{2i}\right)-\left(\frac{z-\overline{z}}{2i}\right)^3\right]+iC$.

But this seems to simplify with a $\overline{z}$ in the numerator. It is my understanding that an analytic function will not have $\overline{z}$? I am not sure what I am doing wrong.

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1 Answer 1

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Notice the expression

$$\displaystyle z^3 = (x + iy)^3 = x^3 + 3ix^2 y - 3xy^2 - iy^3.$$

From here one sees that

$$\displaystyle f(x,y) = (x^3 - 3xy^2) + i(3x^2 y - y^3) + C = z^3 + C.$$

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