Functions of a Complex Variable : Cauchy Riemann Equation

Complex Variable :

z = x + i y is a complex variable where x is real part & y is imaginary part .

If z = x + i y , then the real part of z is denoted by Re (z) and the imaginary part by Im (z)

If w be the Function of a Complex Variable then w = f(z)

Limits , Continuity & Differentiability of a Function of Complex Variable

Let f(z) be a single valued function .

(i) The limit of f(z) as z approaches z0 is wo .

$\displaystyle lim_{z \rightarrow z_0} f(z) = w_0$

(ii) f(z) is said to be continuous at z = z0 if

$\displaystyle lim_{z \rightarrow z_0} f(z) = f(z_0)$

(iii) If f(z) be a single valued function of variable z then,

$\displaystyle f'(z) = lim_{\delta z \rightarrow 0} \frac{f(z + \delta z) – f(z)}{\delta z}$

Analytic Function

Suppose f(z) be a single valued function which is differentiable at z = z0 then it is said to be Analytic at z = z0 .

The Point at which the function is not differentiable is called a Singular Point of the function .

The Necessary Condition for function to be Analytic

The Necessary Condition for function f(z) = u + i v to be Analytic at all points in a region R are :

(i) $\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} $

(ii) $\displaystyle \frac{\partial u}{\partial y} = – \frac{\partial v}{\partial x} $ ; Provided $\frac{\partial u}{\partial x} , \frac{\partial u}{\partial y} , \frac{\partial v}{\partial x} , \frac{\partial u}{\partial y} \; exist $

This is Known as Cauchy Riemann Equation

Sufficient Condition for function to be Analytic

The Sufficient Condition for function f(z) = u + i v to be Analytic at all points in a region R are :

(i) $\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \; , \frac{\partial u}{\partial y} = – \frac{\partial v}{\partial x}$

(ii) $\displaystyle \frac{\partial u}{\partial x} , \frac{\partial u}{\partial y} , \frac{\partial v}{\partial x} , \frac{\partial u}{\partial y} $ are Continuous Function of x and y in Region R .

Example : Show that sinz is an Analytic Function .

Solution : Let w = sinz = sin(x + i y)

w = sinx cos iy + cosx sin iy

w = sinx cos hy + i cosx sin hy

(Since , cos ix = cos hx , sin ix = i sin hx)

u = sinx cos hy ,

v = cosx sin hy

$\displaystyle \frac{\partial u}{\partial x} = cosx \; cos hy $

$\displaystyle \frac{\partial u}{\partial y} = sinx \; sin hy $

$\displaystyle \frac{\partial v}{\partial x} = – sinx \; sin hy $

$\displaystyle \frac{\partial v}{\partial y} = cosx \; cos hy $

Since C-R equation is satisfied , hence sinz is an Analytic Function .

Example : Show that the function z|z| is not analytic anywhere .

Solution : Let w = z |z|

w = u + iv and z = x + i y

$\displaystyle u + iv = (x + iy) \sqrt{x^2 + y^2}$

$u = x \sqrt{x^2 + y^2} $

$ v = y \sqrt{x^2 + y^2} $

$\displaystyle \frac{\partial u}{\partial x} = \sqrt{x^2 + y^2} + \frac{2 x^2}{2 \sqrt{x^2 + y^2}} $

$\displaystyle \frac{\partial u}{\partial x} = \frac{x^2 + y^2 + x^2}{\sqrt{x^2 + y^2}} $

$\displaystyle \frac{\partial v}{\partial y} = \sqrt{x^2 + y^2} + \frac{2 y^2}{2 \sqrt{x^2 + y^2}} $

$\displaystyle \frac{\partial v}{\partial y} = \frac{x^2 + y^2 + y^2}{\sqrt{x^2 + y^2}} $

$\displaystyle \frac{\partial u}{\partial y} = \frac{2 x y}{2 \sqrt{x^2 + y^2}} $

$\displaystyle \frac{\partial v}{\partial x} = \frac{2 x y}{2 \sqrt{x^2 + y^2}} $

$\displaystyle \frac{\partial v}{\partial x} = \frac{x y}{\sqrt{x^2 + y^2}} $

Since C-R Equation is not satisfied , hence function z|z| is not Analytical .

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