# Supplement for Problem 3: About Cauchy-Riemann Equations

The complex plane $\mathbb{C}$ can be viewed as a 2-dimensional real vector space $\mathbb{R}^2$ by writing $z = x + iy$ where $x,y \in \mathbb{R}$ and $i = \sqrt{-1}$. And hence a complex function $f:\mathbb{C} \to \mathbb{C}$ can be written as $$f(x,y) = u(x,y) + i v(x,y) \quad \text{where } u,v:\mathbb{R}^2 \to \mathbb{R}$$

A complex function $f$ is said to be **differentiable** if the following limit exists: $$\lim_{z \to z_{0}}{f(z) - f(z_{0})\over z - z_{0}}$$

We have following theorem:

### Theorem
Let $f:\mathbb{C} \to \mathbb{C}$ be a complex function, and suppose that $f(x+ iy) = u(x,y) + iv(x,y)$, where $x,y, u(x,y),v(x,y)$ are real.
1. If $f$ is differentiable at $c = a+ib$, then the partial derivatives ${\partial u\over \partial x}$,${\partial u \over \partial y}$,${\partial v \over \partial x}$ and ${\partial v\over \partial y}$ exist at $(a,b)$ and satisfy $${\partial u\over \partial x} = {\partial v \over \partial y} \qquad {\partial u\over \partial y} = - {\partial v \over \partial x}$$
2. Suppose $f$ is defined on the disc $D = \{ z\in \mathbb{C}: |z-c|< \epsilon \}$ of $c = a+ ib$ for some $\epsilon >0$, and the partial derivatives ${\partial u\over \partial x}$,${\partial u \over \partial y}$,${\partial v \over \partial x}$ and ${\partial v\over \partial y}$ exist everywhere in $D$. If these first partial derivatives are continuous at $(a,b)$ and satisfy $${\partial u\over \partial x} = {\partial v \over \partial y} \qquad {\partial u\over \partial y} = - {\partial v \over \partial x}$$ at $(a,b)$, then the function is differentiable at $c$.

The equations $${\partial u\over \partial x} = {\partial v \over \partial y} \qquad {\partial u\over \partial y} = - {\partial v \over \partial x}$$ is called **Cauchy-Riemann equations (system)**.

Since $x = {1\over 2}(z + \bar z)$ and $y = {1\over 2i}(z - \bar z)$, where $\bar z$ is the complex conjugate of $z$. One can obtain that $${\partial \over \partial \bar z} = {1\over 2}\left({\partial \over \partial x} + i {\partial \over \partial y}\right)$$ by chain rule. Then it is not hard to see that $${\partial f \over \partial \bar z} = 0 \iff \text{ $f$ satifies Cauchy-Riemann equations. }$$ In short, equivalent to the statement that the function $f$ does not depend on $\bar z$.

In problem 3, all the functions given can be view as a complex function: if $z = x+iy$,
(a) $z^2 =(x + iy)^2 = (x^{2}-y^{2}) +2i xy$.
(b) $e^z = e^{x + iy} = e^{x}e^{iy} = e^x(\cos y + i \sin y)$.
\(c\) $\ln z = \ln (x+iy)$. It is not quite obvious here. Write $x + iy = re^{i\theta}$ where $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan {y\over x}$. Then $$\ln (x+iy) = \ln re^{i\theta} = (\ln r) + i \theta = {1\over 2}\ln (x^2 + y^2) + i \arctan{y\over x}$$
==Note==: $x^2+y^2 = 0$ should not be contained here. In fact, $\ln z$ is not analytic at $z = 0$.
(d) ${1\over z} = {1\over x +i y} = {x - iy\over x^2 + y^2}$.
==Note==: $x^2+y^2 = 0$ should not be contained here. In fact, ${1\over z}$ is not analytic at $z = 0$.

All these functions do not contain $\bar{z}$, hence satisfy Cauchy-Riemann equations.

For the example that a function does not satisfy the Cauchy-Riemann equations, consider $f(x,y) = x^2 + y^2$ which can be viewed as $f(z) = |z|^2 = z\bar z$. It is clear that $f$ depends on $\bar z$, hence it does not satisfy the Cauchy-Riemann equations.

# Problem 9
Since the functions are smooth, you can compute it by $$(\nabla_{\vec u}f)(\vec p) = \vec u \cdot (\nabla f)(\vec p)$$ Alternatively, one can compute it by definition, for example
> \(c\) Find the directional derivative at the point $\vec p$ in the direction $\vec u$: $$f(x,y,z) = xy +yz + zx, \quad \vec p = (1,-1,1)\quad \vec u = {1\over \sqrt{6}}(1,2,1)$$

By definition of directional derivative, we are going to find the first derivative in $t$ at $t = 0$ of following function: $$f(1 +{t\over \sqrt 6},-1 + {2t\over \sqrt6}, 1+ {t\over \sqrt{6}}) = (1 +{t\over \sqrt 6})(-1 + {2t\over \sqrt6}) + (-1 + {2t\over \sqrt6})(1 +{t\over \sqrt 6}) + (1 +{t\over \sqrt 6})^2$$

We can just compute the coefficient of $t$: $${-1\over \sqrt 6} + {2 \over \sqrt 6} + {-1\over \sqrt 6} + {2 \over \sqrt 6} + {2\over \sqrt 6} = {4\over \sqrt 6}$$

This is the answer to the question.