M4.1 Complex Analysis-Introdution

Review of basic definitions & Key points

• The numbers are of the form $x+iy$ are called as complex numbers, where $x$ & $y$ are Real numbers and $i=\sqrt{-1}$.
• $i=\sqrt{-1}$ is obtained from the equation $x^2=-1$.
• $1,i,i^2,i^3$ are the fourth roots of the unity and $i^2=-1, i^3=-i, i^4=1$.
• Geometrically a complex number $x+iy$ is a point in Cartesian plane as shown in the following figure
• From the figure we can say that $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}{y/x}$ are called as modulus and amplitude of the complex number $z$ denoted by $|z|$ and $amp(z)$ or $arg(z)$ respectively
• $z\bar=x-iy$ is the complex conjugate of the complex number $z=x+iy$

Euler's Formula

We know the following series 

• $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$
• $\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-+...$
• $\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-+...$

If we multiply $i$ to the above equation we get

• $i \sin(x)=ix-\frac{ix^3}{3!}+\frac{ix^5}{5!}-+...$ Now
• $\cos(x)+i \sin(x)=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-+... \,\,\,\,\,\,\,\,\,\, (1)$
• Now 
• $e^{ix}=1+ix-\frac{x^2}{2!}-\frac{ix^3}{3!}+\frac{x^4}{4!}+\frac{ix^5}{5!}-+... \,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$

If we compare (1) and (2) we can conclude that

$cos(x) + i sin(x) = e^{ix}$

• $e^{-ix} = \cos(x)-i \sin(x)$ Hence,
• $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$  & $\sin(x)=\frac{e^{ix}-e^{-ix}}{2}$
• De-Moivre Theorem: $(\cos(\theta)+i \sin(\theta))^n=\cos(n\theta)+i \sin(n\theta)$

Complex Functions

Complex functions are mapping from complex numbers to complex numbers. Ex: $f(z)=\frac{1}{z}=\frac{1}{x+iy}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}=u(x,y)+iv(x,y)$

Limit of a Complex Function

Let $f(z)$ be a complex valued function, limit $l$ as $z$ tends to some point $z_0$, there exists $\epsilon, \delta$, when $|z\to z_0|<\epsilon, |f(z)−l|<\delta$. i.e., $lim_{z\to z_0}f(z)=l$

Continuity of a complex function

If $lim_{z\to z_0} f(z)=f(z_0)$ then $f(z)$ is continuous at $z=z_0$
Example

Differentiablility of a complex function

Let $\delta z=z−z_0$, as $z\to z_0$ or $\delta z\to 0, lim_{z\to z_0}\frac{f(z+\delta z)−f(z)}{\delta z}$ is the derivative of $f$ at $z_0$

Analytic Functions

A complex function $w=f(z)$ is said to be analytic at some point $z_0$ if $\frac{dw}{dz}=f′(z)=lim_{\delta z \to z_0} \frac{f(z+\delta z)−f(z)}{\delta z}$ exists and unique.