Line Integral of real linear function Let $L: \mathbb C \rightarrow \mathbb C $ be a real linear function and $\gamma $ a closed $C^1$ curve in $\mathbb C$.
Then $\int_\gamma L(z) dz = 0 \Leftrightarrow \quad $L is complex linear
Why is that so? I cant find to find a reasonable solution.
 A: Viewing $L$ as a real operator, if we choose a basis for $\mathbb{C}$, say $\{1,i\}$, we may write it as a 2x2 matrix:
$$L(x+iy)=\begin{pmatrix}a & b\\c &d\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}=(ax+by)+i(cx+dy)$$
Then $L$ is $\mathbb{C}$-linear if $L(i)=iL(1)$ which amounts to $b+di=ia-c$ so $a=d$ and $b=-c.$
We may also view $L$ as a function of $z=x+iy$ and $\bar z=x-iy$. After this change of variables we have 
$$L(z,\bar z)= \frac{z}{2}((a+d) + i(c-b))+\frac{\bar z}{2}((a-d)+i(b+c))$$
Thus the line integral becomes 
$$\oint L(z,\bar z)\,dz = \frac{1}{2}((a+d) + i(c-b))\oint z\,dz+\frac{1}{2}((a-d)+i(b+c))\oint\bar z\,dz$$
But $\oint z\,dz=0$ by Cauchy residue. And $\oint \bar z\,dz=\int_0^{2\pi}e^{-i\theta}ie^{i\theta}\,d\theta=2\pi i$. So the integral vanishes only if $a=d$ and $b=-c$.
Or more succinctly if you know the relevant theorems, the contour integral vanishes iff the function is holomorphic iff the antiholomorphic part vanishes and for linear maps homolorphicity is equivalent to $\mathbb{C}$-linearity.
