I am trying to prove that $$\frac{-z^2}{z+1}$$ is holomorphic.

I know that to do this, I want to express $z=x+iy=u(x,y)+iv(x,y)$ and then confirm the CR equations .

My problem is with getting it into that form where I can have my functions u and v.

This is what I tried.

$$\frac{z^2}{z+1}=\frac{z(z'+1)}{zz'+z+z'+1}$$ where $z'$ denotes the complex conjugate.

I am also confused on other basic topics such as modulus.

I know that $|z|^{2}=x^2+y^2=zz'$, but at one point I have $|z+1|^{2}$ on the bottom, so would this be $x^2+y^2+1$ or would it be $(x+1)^2+y^2$ ? This may be what is causing me some problems above.

Any and all help is appreciated.

Thank you

(PS: I ignore the initial negative sign for now as I thought that this alone wont change holomorphic, is that correct or is this potentially a major mistake in my understanding?)

  • $\begingroup$ I don't think your function is holomorphic on $\mathbb{C}$ since it has a simple pole at $z=-1$. $\endgroup$ Commented Sep 20, 2016 at 2:07

1 Answer 1


Suppose you want to show your $f(z)$ is holomorphic on $\mathbb{C}\backslash\{0\}$ using Cauchy-Riemann equation. Let us begin by rewriting $f(z) = u(x, y) + iv(x, y)$. First, let us observe \begin{align} \frac{z^2}{1+z}= \frac{z^2(1+\bar z)}{|1+z|^2} = \frac{z^2}{|1+z|^2} + \frac{z|z|^2}{|1+z|^2}. \end{align} Using the facts \begin{align} |1+z|^2=|1+x+iy|^2= (1+x)^2+ y^2 \ \ \text{ and } \ \ z^2= (x+iy)^2= x^2-y^2+2ixy \end{align} we get that \begin{align} \frac{z^2}{1+z} =&\ \frac{(x^2-y^2)+2ixy}{(1+x)^2+y^2} + \frac{x(x^2+y^2) + i y(x^2+y^2)}{(1+x)^2+y^2}\\=&\ \frac{x^2-y^2+x^3+xy^2}{(1+x)^2+y^2} + i \frac{2xy+yx^2+y^3}{(1+x)^2+y^2}. \end{align}

Now you can check your CR-equations.

Edit: It's probably easier to check holomorphicity of $f(z)$ on $\mathbb{C}\backslash\{0\}$ using the complex differentiation definition.

  • $\begingroup$ what do you mean by using the complex diff definition? isn't that what I have to do CR anyway? $\endgroup$
    – PersonaA
    Commented Sep 21, 2016 at 19:21

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