complex primitive function of 1/z^2

Question

It is known that the function of complex variable $f:\mathbb{C}\setminus\{0\}\to\mathbb{C}$, $f(z)=\frac{1}{z}$ does not have a primitive function since the integral of $f$ along closed circle containing zero is equal to $2\pi i$.

Now, define $g:\mathbb{C}\setminus\{0\}\to\mathbb{C}$, $g(z)=\frac{1}{z^2}$. The integral of $g$ along closed circle containing zero is equal to zero. From this, we deduce that the integral of $g$ along any curve (not neccessarily closed) not passing through zero depends only on starting and ending point of the curve.

My question is: Can we then say that the function $-f$ is primitive to $g$ on $\mathbb{C}\setminus\{0\}$ ?

Yes, of course $-\frac 1 z$ is a primitive of $\frac 1 {z^{2}}$ if primitive means anti-derivative. — Kavi Rama Murthy, May 14 '19 at 09:11
Primitive function? I think the correct term describing that is 'anti-derivative' — RMWGNE96, May 14 '19 at 09:11
Primitive and anti-derivative are synonymous . https://en.wikipedia.org/wiki/Antiderivative — 19aksh, May 14 '19 at 09:13
ok, i found the answer. https://math.stackexchange.com/questions/1916365/why-the-integral-of-frac1z-over-a-closed-curve-is-not-0 — elliptic, May 14 '19 at 09:52

score 0 · Answer 1 · answered May 14 '19 at 09:28

Yes, we can say that, simply because, for each $z_0\in\mathbb C\setminus\{0\}$,\begin{align}(-f)'(z_0)&=\lim_{z\to z_0}\frac{-\frac1z+\frac1{z_0}}{z-z_0}\\&=\lim_{z\to z_0}\frac{z-z_0}{zz_0(z-z_0)}\\&=\lim_{z\to z_0}\frac1{zz_0}\\&=\frac1{{z_0}^2}\\&=g(z_0).\end{align}

complex primitive function of 1/z^2

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