The mean value theorem for vector-valued function in the real domain $f: \mathcal{R}^n \rightarrow \mathcal{R}^d$ can be expressed as
\begin{equation}
f(x)-f(y)=\int\_{0}^{1}\nabla f(x(\tau))d \tau \cdot (x -y),
\end{equation}
where $x(\tau) = y + \tau (x - y)$. 


I wonder if there exists similar results for vector-valued functions in the complex domain. Suppose $f:\mathcal{C}^n \rightarrow \mathcal{C}^d$, do we have
\begin{equation}
f(z\_1)-f(z\_2)=\int\_{0}^{1}\nabla f(z(\tau))d \tau \cdot \left(\begin{array}{c}
z\_{1}-z\_{2}\\
\overline{z\_{1}}-\overline{z\_{2}}
\end{array}\right),
\end{equation}
where $z(\tau) = z\_2 + \tau (z\_1 - z\_2)$, $\overline{z\_1} - \overline{z\_2}$ denotes the conjugate of $z\_1 -z\_2$ and $\nabla f(z(\tau))$ is the Wirtinger Jacobian.



Mean value theorem in terms of Wirtinger calculus?
The desired relation can be written in components as
$$f\_n(\mathbf{z}^{(1)},\mathbf{\bar z}^{(1)})-f\_n(\mathbf{z}^{(2)},\mathbf{\bar z}^{(2)})=\int\_0^1 d\tau\,\sum\_{m}\left(\frac{\partial f\_n}{\partial z\_m}(z\_m^{(1)}-z\_m^{(2)})+\frac{\partial f\_n}{\partial \bar{z}\_m}(\bar{z}\_m^{(1)}-\bar{z}\_m^{(2)})\right),\;\;[1]$$
with the prescribed $\tau$ dependence:
$$\mathbf{z}(\tau)=\mathbf{z}^{(2)}+\tau(\mathbf{z}^{(1)}-\mathbf{z}^{(2)}),\;\;
\mathbf{\bar z}(\tau)=\mathbf{\bar z}^{(2)}+\tau(\mathbf{\bar z}^{(1)}-\mathbf{\bar z}^{(2)}).\;\;[2]$$
We start from the equation
$$f\_n(\mathbf{z}^{(1)},\mathbf{\bar z}^{(1)})-f\_n(\mathbf{z}^{(2)},\mathbf{\bar z}^{(2)})=\int\_0^1 d\tau\,\frac{d}{d\tau}f\_n(\mathbf{z},\mathbf{\bar z}).\;\;[3]$$
Equation [3] implies equation [1] if
$$df\_n=\sum\_m\left(\frac{\partial f\_n}{\partial z\_m}dz\_m+\frac{\partial f\_n}{\partial \bar{z}\_m}d\bar{z}\_m\right).\;\;[4]$$
This is indeed a property of the Wirtinger differential.$^\ast$




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$^\ast$ **Proof**  

Denote $z=x+iy$, $\bar{z}=x-iy$ and insert the definitions $\partial/\partial x=\partial/\partial z+\partial/\partial \bar{z}$, $\partial/\partial y=i\partial/\partial z-i\partial/\partial \bar{z}$ of the Wirtinger derivatives:


$$df\_n=\sum\_m\left(\frac{\partial f\_n}{\partial x\_m}dx\_m+\frac{\partial f\_n}{\partial y\_m}dy\_m\right)$$
$$\qquad=\sum\_m\left(\frac{\partial f\_n}{\partial z\_m}dx\_m+\frac{\partial f\_n}{\partial \bar{z}\_m}dx\_m+i\frac{\partial f\_n}{\partial z\_m}dy\_m-i\frac{\partial f\_n}{\partial \bar{z}\_m}dy\_m\right)$$
$$\qquad=\sum\_m\left(\frac{\partial f\_n}{\partial z\_m}dz\_m+\frac{\partial f\_n}{\partial \bar{z}\_m}d\bar{z}\_m\right).$$


