As it's told on the title I want to find an entire function with real part \begin{equation} x^4-6x^2y^2+y^4 \end{equation} I just begin on the Complex Variable course and I don't even know how to start. I know the definition of entire function is that the function is holomorphic everywhere on the complex plane. I just have to integrate?

Sorry if the question it's easy but I'm stucked.


  • 4
    $\begingroup$ Well, we have $x$ and $y$ to the fourth power, so the first thing I would try, before doing any kind of Cauchy-Riemann equation solving or anything else, is $f(z) = z^4 = (x+iy)^4$. Try that, then come back to us with the result if you're still stuck. $\endgroup$
    – Arthur
    Commented Feb 26, 2018 at 10:17
  • $\begingroup$ Oh, alright. $z^4=(x^4-6x^2y^2+y^4)+ i \cdot (4x^3y-4xy^3)$ $\endgroup$
    – Ginger
    Commented Feb 26, 2018 at 10:21
  • $\begingroup$ And then the answer it's any $z$ with $Im(z)= 4x^3y-4xy^3$ . Supose we have the function $f(z)=z^4$ that's always holomorphic because it's a polynomial? $\endgroup$
    – Ginger
    Commented Feb 26, 2018 at 10:23
  • $\begingroup$ Yes, polynomials are always holomorphic. You can, of course, check that it fulfills the Cauchy-Riemann equations now that you know both the real and the imaginary part of the function, and you should do it at least once so that you know yourself rather than trusting random strangers on the internet. $\endgroup$
    – Arthur
    Commented Feb 26, 2018 at 10:29
  • $\begingroup$ Okey! Much thanks Arthur $\endgroup$
    – Ginger
    Commented Feb 26, 2018 at 10:30

1 Answer 1


Let $f$ be such a function, let $u(x,y)=\operatorname{Re}f(x+yi)=x^4-6x^2y^2+y^4$ and let $v(x,y)=\operatorname{Im}f(x+yi)$. Then, by the Cauchy-Riemann equations,$$v_x=-u_y=12x^2y-4y^3\text{ and }v_y=u_x=4x^3-12xy^2.$$Integrating, it's not hard to see that, for some $k\in\mathbb R$, $v(x,y)=4x^3y-4xy^3+k$. So\begin{align}f(x+yi)&=u(x,y)+v(x,y)\\&=x^4-6x^2y^2+y^4+(4x^3y-4xy^3+k)i\\&=(x+yi)^4+ki.\end{align}


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