Question

Find whether the given complex function is analytic or not.

$f(z) = \dfrac{i}{z^8}$

Use the Polar Form of Cauchy Reimann Equations to check the analyticity. (Formula given below)

$\boxed{\text{Cauchu Reimann Equations}:\\\\u_r = \dfrac{v_\theta}{r} \:\:\&\:\: v_r = \dfrac{-u_\theta}{r}}$

Answer 1

In this question we are given with a complex function , and we have to check that the given function is analytic or not ?

You should remember that in these type of functions , if you use Cartesian system then it will be more difficult . So , we will use Polar form ;

Before doing the question you should know that ;

A complex function let's say be f(z) is said to be analytical iff :-

• f ( z ) should be single valued .

• f ( z ) should have unique derivatives in it's Domain .

• f ( z ) should must follow Cauchy - Riemann Equations .

The function f(z) given isn't an analytical function , Refer to the attachment for explanation :D

Note :-

• I had used the fact that as u & v are sine and cosine functions of $\theta$ & r only so they must have unique derivatives in their domain & must be single valued .

Answer 2

Given :-

A complex function ${\sf{\dfrac{i}{z^8}}}$

To Find :-

Is the given function analytical or not ?

Solution :-

Before starting the answer , we shall aware of some basic formulae & concepts of complex functions & analytical functions ;

• ${\sf{(\cos \theta + i\sin \theta)^{n} = \cos (n\theta)+i \sin (n\theta)}}$

• There are 3 conditions under which a complex function is said to be analytical :-

1. f(z) should be single valued .
2. f(z) should must have unique derivatives in it's domain.
3. f(z) should must follow both Cauchy Riemann Equations .

• In order to check a function analytic , we put f(z) = u + iv & z = x + iy , and then check the other three conditions .

So let's start !!!!

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Here we have ;

${\leadsto \sf f(z)=\dfrac{i}{z^8}}$

Now , put z = x + iy & f(z) = u + iv

${: \implies \quad \sf u+iv=\dfrac{i}{(x+iy)^8}}$

Now , it will be difficult to handle this function in Cartesian form , let's convert it into polar form. So , put ${\sf x = r \cos \theta}$ & ${\sf y = r \sin \theta}$ .

${: \implies \quad \sf u+iv=\dfrac{i}{(r\cos \theta + ri\sin \theta)^8}}$

${: \implies \quad \sf u+iv=\dfrac{i}{r^{8}(\cos \theta +i \sin \theta)^8}}$

${: \implies \quad \sf u+iv=\dfrac{i(\cos \theta +i \sin \theta)^{-8}}{r^8}}$

${: \implies \quad \sf u+iv=\dfrac{i\{\cos(-8\theta)+i\sin (-8\theta)\}}{r^8}}$

${: \implies \quad \sf u+iv=\dfrac{i\{\cos (8\theta)-i\sin (8\theta)\}}{r^8}}$

${: \implies \quad \sf u+iv=\dfrac{i\cos (8\theta) -i^{2}\sin(8\theta)}{r^8}}$

${: \implies \quad \sf u+iv=\dfrac{i\cos (8\theta) +\sin(8\theta)}{r^8}}$

${: \implies \quad \sf u+iv=\dfrac{\sin(8\theta)}{r^8}+\dfrac{i\cos (8\theta)}{r^8}}$

on comparing imaginary and real parts ;

${\quad \boxed{\bf u=\dfrac{\sin(8\theta)}{r^8} \quad and \quad v=\dfrac{\cos(8\theta)}{r^8}}}$

Now consider ;

${: \implies \sf u=\dfrac{\sin(8\theta)}{r^8}}$

Partial Differentiating both sides wr.t.r

${: \implies \sf \dfrac{\partial u}{\partial r}=\sin(8\theta)\dfrac{\partial}{\partial r}(r^{-8})}$

${: \implies \sf \dfrac{\partial u}{\partial r}=\sin(8\theta)\times -8\times r^{-9}}$

${: \implies \sf \dfrac{\partial u}{\partial r}=\dfrac{-8\sin(8\theta)}{r^9} \quad}$

Now consider ;

${: \implies \sf u=\dfrac{\sin(8\theta)}{r^8}}$

Partial Differentiating both sides wr.t.$\theta$

${: \implies \sf \dfrac{\partial u}{\partial \theta}=\dfrac{1}{r^8}\dfrac{\partial}{\partial \theta}\{\sin(8\theta)\}}$

${: \implies \sf \dfrac{\partial u}{\partial \theta}=\dfrac{1}{r^8}\times \cos(8\theta)\times 8}$

${: \implies \sf \dfrac{\partial u}{\partial \theta}=\dfrac{8\cos(8\theta)}{r^8}\quad}$

Now consider ;

${: \implies \sf v=\dfrac{\cos(8\theta)}{r^8}}$

Partial Differentiating both sides w.r.t.r

${: \implies \sf \dfrac{\partial v}{\partial r}=\cos(8\theta)\dfrac{\partial}{\partial r}(r^{-8})}$

${: \implies \sf \dfrac{\partial v}{\partial r}=\cos(8\theta)\times -8\times r^{-9}}$

${: \implies \sf \dfrac{\partial v}{\partial r}=\dfrac{-8\cos(8\theta)}{r^9} \quad}$

Now consider ;

${: \implies \sf v=\dfrac{\cos(8\theta)}{r^8}}$

Partial Differentiating both sides wr.t.$\theta$

${: \implies \sf \dfrac{\partial v}{\partial \theta}=\dfrac{1}{r^8}\dfrac{\partial}{\partial \theta}\{\cos(8\theta)\}}$

${: \implies \sf \dfrac{\partial v}{\partial \theta}=\dfrac{1}{r^8}\times - \sin(8\theta)\times 8}$

${: \implies \sf \dfrac{\partial v}{\partial \theta}=\dfrac{-8\sin(8\theta)}{r^8}\quad }$

Now , as u & v are cosine and sine functions of r & $\theta$ . So , they must follow the two conditions , now checking for CR equations ;

By first equation of Cauchy Riemann :-

• ${\boxed{\bf \dfrac{\partial u}{\partial r}=\dfrac{1}{r}\dfrac{\partial v}{\partial \theta}}}$

Putting the values ;

${:\implies \quad \sf \dfrac{-8\sin(8\theta)}{r^9} =\dfrac{1}{r}\dfrac{-8\sin(8\theta)}{r^8}}$

${:\implies \quad \sf \dfrac{-8 \sin(8\theta)}{r^9} = \dfrac{-8 \sin(8\theta)}{r^9}}$

First CR equation satisfied , now by 2nd equation of Cauchy Riemann ;

• ${\boxed{\bf \dfrac{\partial v}{\partial r}=-\dfrac{1}{r}\dfrac{\partial u}{\partial \theta}}}$

Putting the values ;

${:\implies \quad \sf \dfrac{-8\cos(8\theta)}{r^9} =-\dfrac{1}{r}\dfrac{8\cos(8\theta)}{r^8}}$

${:\implies \quad \sf \dfrac{-8 \cos(8\theta)}{r^9} = \dfrac{-8 \cos(8\theta)}{r^9}}$

Here , both Cauchy Riemann Equations are Satisfied . Hence , the given function is analytical