# Proof of Taylor's Theorem with Wirtinger Derivatives (Complex coordinates)

Suppose that $f$, defined in $D_1(0)$, is infinitely differentiable. Show that for each $n \in \mathbb{N}$ we have \begin{equation*} f(z,\bar{z}) = \sum\limits_{0 \leq j + k \leq n} \frac{\partial_z^j\partial_{\bar{z}}^kf(0,0) }{j!k!}z^j\bar{z}^k + \mathcal{O}(|z|^{n+1}). \end{equation*}

I've tried to expand Taylor's theorem for reals to get this result, but everything I've tried has worked out badly. I'm sure there's an elegant way to do this that I'm just not seeing. This IS a homework problem, so feel free to give partial solutions/hints if you prefer.

Edit: My starting point was that we know: \begin{equation*} f(x,y) = \sum\limits_{0 \leq j + k \leq n} \frac{\partial_x^j\partial_{y}^kf(0,0) }{j!k!}x^jy^k + \mathcal{O}(\sqrt{x^2 + y^2}^{n+1}). \end{equation*} from Taylor's theorem for two variables. It's likely possible to get one from the other from the extremely ugly, brute-force method of substituting in $z = x + iy,~\bar{z} = x - iy$ and $\partial_z = \frac{1}{2}(\partial_x - i\partial_y),~\partial_{\bar{z}} = \frac{1}{2}(\partial_x + i\partial_y)$. My gut feeling tells me that there must be a better way to solve this problem then that. I just can't figure it out. Any help would be extremely appreciated.

Starting from the Taylor formula for functions of a real variable,

$$g(x) = \sum_{k=0}^n \frac{g^{(k)}(0)}{k!}x^k + \frac{1}{n!}\int_0^x (x-t)^n \cdot g^{(n+1)}(t)\,dt,$$

we can obtain the result by considering $g_\varphi(r) = f(re^{i\varphi},re^{-i\varphi})$ and expressing the derivatives of $g_\varphi$ in terms of the Wirtinger derivatives of $f$.

Inductively, we have

\begin{align} g_\varphi^{(k+1)}(t) &= \frac{\partial}{\partial t} g_\varphi^{(k)}(t)\\ &= \frac{\partial}{\partial t} \sum_{m=0}^k \binom{k}{m} \partial_z^m\partial_{\overline{z}}^{k-m}f(te^{i\varphi},te^{-i\varphi})e^{im\varphi}e^{-i(k-m)\varphi}\\ &= \sum_{m=0}^k \binom{k}{m} \partial_z^{m+1}\partial_{\overline{z}}^{k-m}f(te^{i\varphi},te^{-i\varphi})e^{i(m+1)\varphi}e^{-i(k-m)\varphi}\\ &\quad + \sum_{m=0}^k \binom{k}{m} \partial_z^m \partial_{\overline{z}}^{k+1-m}f(te^{i\varphi},te^{-i\varphi})e^{im\varphi}e^{-i(k+1-m)\varphi}\\ &= \sum_{m=0}^{k+1} \binom{k}{m-1} \partial_z^m\partial_{\overline{z}}^{k+1-m}f(te^{i\varphi},te^{-i\varphi})e^{im\varphi}e^{-i(k+1-m)\varphi}\\ &\quad + \sum_{m=0}^{k+1} \binom{k}{m} \partial_z^m\partial_{\overline{z}}^{k+1-m}f(te^{i\varphi},te^{-i\varphi})e^{im\varphi}e^{-i(k+1-m)\varphi}\\ &= \sum_{m=0}^{k+1}\binom{k+1}{m} \partial_z^m\partial_{\overline{z}}^{k+1-m}f(te^{i\varphi},te^{-i\varphi})e^{im\varphi}e^{-i(k+1-m)\varphi}\\ \end{align}

by the chain rule just like for the real partial derivatives $\partial_x,\,\partial_y$, and so for $z = \lvert z\rvert e^{i\varphi}$ we obtain

\begin{align} f(z,\overline{z}) &= g_\varphi(\lvert z\rvert)\\ &= \sum_{k=0}^n \frac{g_\varphi^{(k)}(0)}{k!}\lvert z\rvert^k + \underbrace{\frac{1}{n!}\int_0^{\lvert z\rvert} (\lvert z\rvert-t)^n g_\varphi^{(n+1)}(t)\,dt}_{R_n(z,\overline{z})}\\ &= \sum_{j+m\leqslant n} \frac{\partial_z^j\partial_{\overline{z}}^m f(0,0)}{j!m!} e^{ij\varphi}e^{-im\varphi}\lvert z\rvert^{j+m} + R_n(z,\overline{z})\\ &= \sum_{j+m\leqslant n} \frac{\partial_z^j\partial_{\overline{z}}^m f(0,0)}{j!m!}z^j\overline{z}^m + R_n(z,\overline{z}), \end{align}

with

\begin{align} \lvert R_n(z,\overline{z})\rvert &= \frac{1}{n!} \left\lvert \int_0^{\lvert z\rvert} (\lvert z\rvert-t)^n g_\varphi^{(n+1)}(t)\,dt\right\rvert\\ &\leqslant \frac{1}{n!}\sum_{j=0}^{n+1}\binom{n+1}{j}\int_0^{\lvert z\rvert} (\lvert z\rvert-t)^n \left\lvert \partial_z^j\partial_{\overline{z}}^{n+1-j}f(te^{i\varphi},te^{-i\varphi})\right\rvert\,dt\\ &\leqslant \left(\sum_{j=0}^{n+1} \frac{\lVert \partial_z^j\partial_{\overline{z}}^{n+1-j} f\rVert_{R}}{j!(n+1-j)!}\right)\lvert z\rvert^{n+1} \end{align}

where $R$ is arbitrary between $\lvert z\rvert$ and $1$, and $\lVert h\rVert_R = \sup \{ \lvert h(z,\overline{z})\rvert : \lvert z\rvert \leqslant R\}$.

Note: we cannot have a bound $C\cdot \lvert z\rvert^{n+1}$ for the remainder term uniformly on all of $D_1(0)$, since $f$ could be unbounded on the disk, but a polynomial always is bounded on bounded subsets of $\mathbb{C}$. We can only expect to have for every compact $K\subset D_1(0)$ a constant $C_K$ such that $\lvert R_n(z,\overline{z})\rvert \leqslant C_K\cdot \lvert z\rvert^{n+1}$ holds for all $z\in K$. The expression with the $\lVert\cdot\rVert_R$ gives exactly that.

You may have noticed that the proof is exactly like the/a standard proof of the Taylor formula for a function of several (in this case two) real variables. The point is the formula for the higher derivatives of $g_\varphi$, which matches exactly the formula for the derivatives expressed in terms of the real partial derivatives. That they behave just like true partial derivatives in many ways (chain rule, product rule, ...) makes the Wirtinger derivatives useful.

So, you know there is a polynomial $P$ of degree $\le n$ such that $$f(x,y) = P(x,y) + \mathcal{O}((x^2 + y^2)^{(n+1)/2}) \tag{1}$$ Note that the error term has all derivatives of orders $\le n$ vanishing at the origin.

Plug $x=(z+\bar z)/2$ and $y=(z-\bar z)/(2i)$ in (1). Treating $z$ and $\bar z$ as abstract variables for the moment, observe that this is a linear invertible change of variables: a polynomial becomes another polynomial $Q$ of same degree. So, $$f(z) = Q(z,\bar z) + \mathcal{O}(|z|^{n+1}) \tag{2}$$ As before, the error term has all derivatives of orders $\le n$ vanishing at the origin. Assuming as known that $$\frac{\partial }{\partial z}(z^m \bar z^n)=mz^{m-1} \bar z^n,\qquad \frac{\partial }{\partial \bar z}(z^m \bar z^n)=nz^{m} \bar z^{n-1} \tag{3}$$ we find that the coefficients of $Q$ are what is claimed by taking derivatives on both sides and evaluating them at $0$.

One way to prove (3) is to

• check that Wirtinger derivatives satisfy the product rule (easy, since they are just the sum of two things that satisfy it)
• check that $\frac{\partial }{\partial z} z =1$, $\frac{\partial }{\partial z} \bar z =0$, $\frac{\partial }{\partial \bar z} z =0$, $\frac{\partial }{\partial \bar z} \bar z =1$. (Something that should be done to motivate said derivatives, anyway.)
• Thanks for the comment, but I'm having a bit of trouble understanding what you mean by "we find that the coefficients of $Q$ are what is claimed by taking derivatives on both sides and evaluating them at $0$." What exactly do you mean by that? – user165388 Sep 12 '14 at 20:03
• Say, you differentiated both sides of (2) twice in $z$ and three times in $\bar z$. Then on the right you have a polynomial where every monomial lost two factors of $z$ and three factors of $\bar z$ (and gained some coefficient). Now when we plug $0$ in there, the only monomial that survives is the one that came from $z^2\bar z^3$. This gives a relation between the derivatives of $f$ and coefficients of $Q$. – user147263 Sep 12 '14 at 21:41