Complex analysis review 1

Argument

$$Arg \,a=arg\, a +2\pi \mathbb{Z}$$

Stereographic Projection

$$x_1 = \frac{z+\bar{z}}{1+|z|^2}\\ x_2 = \frac{z-\bar{z}}{1+|z|^2}\\ x_3 = \frac{|z|^2-1}{|z|^2+1}\\ $$

Jordan’s Theorem

Every Jordan curve divides the plane into an “interior” region bounded by the curve and an “exterior” region containing all of the nearby and far away exterior points, so that every continuous path connecting a point of one region to a point of the other intersects with that loop somewhere.

Heine-Borel Theorem

Suppose that $A$ is a compact set, $G$ is an open covering of $A$, then there are finite open sets of $G$ which can cover $A$.

Bolzano-Weierstrass Theorem

There is at least an accumulating point in an infinit set.

Cauchy-Riemann Identity

Suppose that $f(z) = u(z)+iv(z)$,

$$\lim_{z\rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0}=f'(z_0)$$
For any path $z\rightarrow z_0$, the limits are equal.

Let $z = x+iy_0,x\rightarrow x_0$,

$$f'(z_0) = u_x+iv_x.$$
Let $z = x_0+iy,y\rightarrow y_0$,
$$f'(z_0) = v_y-iu_y.$$
Alltogether,
$$u_x=v_y,\quad u_y = -v_x.\\ \frac{\partial f}{\partial x}=-i \frac{\partial f}{\partial y}$$

Theorem 1

Complex function $f(z) = u+iv$ is analytic on $D$ if and only if $u,v$ have continuous partial dirivatives and satisfy the Cauchy-Riemann identity.

There is a fact that if $f$ is an analytic function on a domain $D$, then $f'$ is also an analytic function. So $u,v$ have continuous second order derivatives, then

$$\frac{\partial ^2u}{\partial x\partial y} = \frac{\partial ^2u}{\partial y\partial x}.$$
Which means that

$$\frac{\partial ^2u}{\partial^2x}+\frac{\partial ^2u}{\partial^2 y}=0;\\ \frac{\partial ^2v}{\partial^2x}+\frac{\partial ^2v}{\partial^2 y}=0.$$

Two Important Operators

$$z = x+iy,\quad \bar{z} = x-iy$$

$$\frac{\partial f}{\partial z} = \frac{1}{2}(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y})\\ \frac{\partial f}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y})$$

And after a simple calculation

$$df = \frac{\partial f}{\partial z}dz + \frac{\partial f}{\partial \bar{z}}d\bar{z}$$
If $f$ is analytic, then $\frac{\partial f}{\partial \bar{z}}=0$.

Conformality

Suppose that $f(z)$ is analytic on $D$, and $f'(z_0)\neq 0$, $\gamma(t),(0\leq t\leq 1)$ is a smooth curve which pass $z_0$, and $\\gamma(0)=z_0$. Let $\sigma(t) = f(\gamma(t))$, then

$$\sigma'(t) = f'(\gamma(t))\gamma'(t),\quad \sigma'(0) = f'(\gamma(0))\gamma'(0)$$
Therefore,
$$arg \, \sigma'(0) - arg\, \gamma'(0) = arg\, f'(z_0)$$
Now suppose that there are two smooth curves pass through $z_0$, then
$$arg \, \sigma_1'(0) - arg\, \gamma_1'(0) =arg \, \sigma_2'(0) - arg\, \gamma_2'(0)$$
Under the mapping $w = f(z)$, the angles and the directions of rotation between two smooth curves where the derivatives are not zero, are invavirant.

On the other hand, since

$$\lim_{z\rightarrow z_0,z\in \gamma}\frac{|w-w_0|}{|z-z_0|}=|f'(z_0)|.$$
For any smooth curve through $z_0$, the ratio of distance between image points and original points are the same, namely $|f'(z_0)|$.

Integration of Complex Functions

Suppose that $f(t) = u(t)+iv(t)$ defined on $[a,b]$.

• $\int_a^b f(t)dt = \int_a^bu(t)dt+i\int_a^bv(t)dt$.

• $\gamma$ is a rectifiable curve, $f(z) = u(z)+iv(z),dz = dx+idy$, then
  

  $$\int_\gamma f(z)dz = \int_\gamma udx-vdy+i\int_\gamma vdx+udy.$$

$\partial f = \frac{\partial f}{\partial z}dz, \bar{\partial}f = \frac{\partial f}{\partial \bar{z}}d\bar{z}$, then

$$dz \wedge d\bar{z} = -2i dA$$
Define
$$dw = \partial w\wedge dz+\bar{\partial}w \wedge d\bar{z}$$

Theorem 2

For any smooth $(n-1)$ -form with compact support on the oriented n-dimensional manifold $\Omega$,

$$\int _{\partial \Omega}\omega = \iint _\Omega dw.$$