# partial derivative of a complex function.

Let $f(z)=\sin\left(z+\mathrm{e}^{3z}\right)$. Find $\frac{\partial f}{\partial \bar{z}}(z)$.

I tried to start with the well known result $$\frac{\partial^2}{\partial z\partial\bar{z}}=4\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right).$$ Am I right to use this? Please help me to solve this problem.

$\newcommand{\i}{i\,} \newcommand{\R}{\mathcal R} \newcommand{\I}{\mathcal I} \newcommand{\l}{\left(} \newcommand{\r}{\right)} \newcommand{\z}{\overline{z}} \newcommand{\xy}{x + \i y} \newcommand{\y}{\i y} \newcommand{\x}{x} \newcommand{\e}[1]{\,\mathrm{e}^{\,#1}} \newcommand{\u}{\Big(\!\sin x \cos\l\e{3z}\r+\cos x \sin\l\e{3z}\r\!\Big)\cosh y} \newcommand{\v}{\Big(\!\cos x\cosh\l\e{3z}\r-\sin x\sinh\l\e{3z}\r\!\Big)\sinh y}$ HINT: Represent your complex-value function $f$ as in terms of real and imaginary components.

Recall that complex derivatives with respect to $z = x + \i y$ and $\z = x - \i y$ are defined as: $$\frac{\partial }{\partial z} = \frac{1}{2}\left( \frac{\partial }{\partial x}- \i\frac{\partial}{\partial y} \right), \quad \frac{\partial }{\partial \z} = \frac{1}{2}\left( \frac{\partial }{\partial x} + \i\frac{\partial}{\partial y} \right).$$

Indeed,
$$\begin{cases} z = x + \i y, \\ \z = x - \i y \end{cases} \implies \begin{cases} x = \frac{1}{2}\l z + \z\r , \\ y = \frac{1}{2i}\l z - \z\r \end{cases} \implies \begin{cases} \frac{\partial{x}}{\partial{z}} = \frac{1}{2}, & \frac{\partial{x}}{\partial{\z}} = \frac{1}{2}, \\ \frac{\partial{y}}{\partial{z}} = \frac{1}{2i}, & \frac{\partial{y}}{\partial{\z}} = -\frac{1}{2i}. \\ \end{cases}$$ Therefore $$\frac{\partial{f}}{\partial{\z}} = \frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{\z}} + \frac{\partial{f}}{\partial{y}}\frac{\partial{y}}{\partial{\z}} = \frac{1}{2}\bigg( \frac{\partial{f}}{\partial{x}} + \i\frac{\partial{f}}{\partial{y}} \bigg)$$

Any complex function $\ f:\Bbb Z \to \Bbb Z$ can be written as $$f\left(z\right) = f\left(x,y\right) = u\left(x,y\right) + \i v\left(x,y\right),$$ where $u = \Re\left(f\right)$ and $v = \Im\left(f\right)$ are real functions which are real and imaginary components of $f$ respectively. For your specific case $\, f\left(z\right)=\sin\left(z+\mathrm{e}^{3z}\right)\,$ the real and imaginary components $u$ and $v$ can be computed as following: \begin{aligned} \sin z & = \sin\l\xy\r = \sin\l x\r \cos\l \y\r + \cos\l x\r\sin\l \y\r = \\ & = \sin x \cosh y + \i \cos x \sinh y, \\ \e{z} & = \e{\xy} = \e x \big(\cos y +\i \sin y \big), \\ z + \e{3z} & = \xy + \e{3x} \big(\cos y +\i \sin y \big) = x \e{3x} \cos y + \i \l y+\e{3x}\sin y \r. \end{aligned} Denoting $\ \R := x \e{3x} \cos y$ and $\I := y+\e{3x}\sin y, \,$ we write \begin{aligned} f\l z\r & = f \big(\R+\i\I\big) = \sin \big(\R+\i\I\big) = \underbrace{\sin \R \cosh\I}_{:=u} + \i \underbrace{\cos\R\sinh\I}_{:=v} \end{aligned} Therefore $$f\l z \r = f\l \xy\r = u\l x, y \r + \i v\l x, y \r,\ \ \text{where} \ \ \ \begin{cases} u \l x, y \r = \sin \l x \e{3x} \cos y\r \cosh \l y+\e{3x}\sin y\r \\ v \l x, y \r = \cos\l x \e{3x} \cos y\r \sinh \l y+\e{3x}\sin y\r \end{cases}$$ Finally, we write $$\bbox[5pt, border:2pt solid #FF0000]{\ f\l x,y\r = \sin \l x \e{3x}\! \cos y\r \cosh \l y+\!\e{3x}\!\sin y\r + \i \cos\l x \e{3x}\! \cos y\r \sinh \l y+\!\e{3x}\!\sin y\r\ }$$

I hope you can pick it from here and compute the derivative $$\frac{\partial{f}}{\partial{\z}} = \frac{1}{2}\bigg( \frac{\partial{f}}{\partial{x}} + \i\frac{\partial{f}}{\partial{y}} \bigg)$$

• There are (unsurprisingly) typos in your formula for $f$. Would you honestly suggest expansion in real and imaginary parts is a good way to solve this? It seems like time not very well-spent. – mrf Aug 17 '15 at 11:50
• The point of Vlad's answer is to show what's going on "under the hood" when people take partial derivatives with respect to the complex numbers $z$ and $z^*$. After all, $z$ and $z^*$ are not independent of each other in the naive sense; the value of one completely determines the value of the other. – Jess Riedel Nov 6 '17 at 23:09
• Once the student learns the restricted sense in which they can treat them as independent number (i.e., because $\partial_z z^* = 0=\partial_{z^*} z$ and $\partial_z z = 1 =\partial_{z^*} z^*$, where $\partial_z$ and $\partial_{z^*}$ are defined as above), the student can perform the computation quickly by forgetting about the real and imaginary parts $x$ and $y$. – Jess Riedel Nov 6 '17 at 23:09

Your $f$ is a composition of holomorphic functions, so is itself holomorphic. Hence $$\frac{\partial f}{\partial \bar z} = 0.$$ (This is just Cauchy-Riemanns' equations in disguise.)

• mrf ..thanks a lot.. – Nitin Uniyal Aug 17 '15 at 13:31

I think you are misunderstanding the question (but I might misunderstand it as well). You are probably asked to compute the Wirtinger derivative $$\frac{\partial}{\partial \bar{z}} = \frac12 \left( \frac{\partial}{\partial x} + \mathrm{i} \frac{\partial}{\partial y} \right). \tag{1}$$ Hence you should write $f(z)$ as $f(x,y)$ where $x = \Re z$ and $y = \Im z$. Then you should differentiate with respect to $x$ and $y$ and insert them into (1).