Query on Cauchy Riemann Condition question

[NJunJie]

Dear Friends and Colleagues!

I have this practise question:-

Show that z(sin(z))(cos(z)) statisfies the Cauchy-Riemann Conditions for analyticity for all values of z.
Does 1/[z(sin(z))(cos(z))] statisify simiar conditions?
Calculate the derivative of 1/[z(sin(z))(cos(z))] at z=0, + pie/4, + pie/2, + 3pie/2.
Comment on each answer.


I'm stuck!
Pls help and advise! :)
Thanks in advanced.

 

[NJunJie]

Aim:
Want to prove analyticity.
So using Cauchy Riemann – need to put nicely in u(x,y) = j v(x,y) format.

(z)sin(z)cos(z)
= (z)[ ½ 2(sin(z)cos(z))]
= (z) [½ sin 2z ]
= ½ (x + jy) [ sin (2x + j 2y) ]
= ½ (x + jy) [ sin 2x cos j 2y + cos 2x sin j 2y ]


{We know that cos jy = cosh y; sin jy = sinh y}


= ½ (x + jy) [ sin 2x cosh 2y + j cos 2x sinh 2y ]
= ½{ (x) [ sin 2x cosh 2y + j cos 2x sinh 2y ] + (jy) [ sin 2x cosh 2y + j cos 2x sinh 2y ]}

= ½ {x sin 2x cosh 2y + j x cos 2x sin h 2y + jy sin 2x cosh 2y + j2ycos 2x sinh 2y }

Rearranging:-

= ½{ x sin 2x cosh 2y – y cos 2x sinh 2y + jy sin 2x cosh 2y + j x cos 2x sin h 2y }

= ½ { x sin 2x cosh 2y – y cos 2x sinh 2y
+ j (y sin 2x cosh 2y + x cos 2x sin h 2y)}

-- stuck from here... pse help :D

 

[Dick]

Ouch. You did a very good job on finding the u(x,y) and v(x,y). Now you want to show du/dx=dv/dy and du/dy=(-dv/dx). But are you sure you really want to? It's a painfully complicated exercise to no real purpose. z is analytic and sin(2z) is analytic. Therefore the product is analytic. In my opinion, this problem is way too complicated to make it worth verifying Cauchy-Riemann directly. You CAN do it. But why? Similarly, 1/(z*sin(2z)) is also analytic except where z*sin(2z)=0. Are you sure you want to suffer through verifying CR directly rather than citing theorems about analyticity that will let you say that CR is true without directly computing u and v and the derivatives?

 

[NJunJie]

But any better short cuts?? Pse advise.

Here i goes on:-

(z)sin(z)cos(z)
= (z)[ ½ 2(sin(z)cos(z))]
= (z) [½ sin 2z ]
= ½ (x + jy) [ sin (2x + j 2y) ]
= ½ (x + jy) [ sin 2x cos j 2y + cos 2x sin j 2y ]


{We know that cos jy = cosh y; sin jy = sinh y}


= ½ (x + jy) [ sin 2x cosh 2y + j cos 2x sinh 2y ]
= ½{ (x) [ sin 2x cosh 2y + j cos 2x sinh 2y ] + (jy) [ sin 2x cosh 2y + j cos 2x sinh 2y ]}

= ½ {x sin 2x cosh 2y + j x cos 2x sin h 2y + jy sin 2x cosh 2y + j2ycos 2x sinh 2y }

Rearranging:-

= ½{ x sin 2x cosh 2y – y cos 2x sinh 2y + jy sin 2x cosh 2y + j x cos 2x sin h 2y }

= ½ { x sin 2x cosh 2y – y cos 2x sinh 2y
+ j (y sin 2x cosh 2y + x cos 2x sinh 2y)}

Re part:
u(x,y) = x sin 2x cosh 2y – y cos 2x sinh 2y

(treat ‘d’ as partial differentiation here)

du/dx=
cosh 2y [ d(x sin 2x) ] – ysinh2y [ d(cos 2x) ]

du/dy=
x sin 2x [ d(cosh 2y) ] – cos 2x [ d(ysinh2y) ]


Img part:
v(x,y)= y sin 2x cosh 2y + x cos 2x sinh 2y

(treat ‘d’ as partial differentiation here)

dv/dx=
ycosh 2y [ d (sin 2x) ] + sinh 2y [d(xcos 2x)]

dv/dy=
sin 2x [ d (ycosh 2y) ] + xcos 2x [d(sinh 2y)]

--really gets complicated--

Theorems:-
1. The sum or product of analytic functions is analytic.
2. All polynomials are analytic.
3. A rational function (the quotient of two polynomials) is analytic, except
at zeroes of the denominator.
4. An analytic function of an analytic function is analytic.
5. Functions e z , s i n z , c o s z , s i n h z , c o s h z are analytic everywhere.

 

[Dick]

Forget the complicated part. Use the theorems. You know the product of analytic functions is analytic. z*sin(z)*cos(z) is a product of analytic functions. Analytic functions satisfy Cauchy-Riemann. That should be one of your theorems or maybe a definition depending on how you define 'analytic'.

 

[NJunJie]

Sure. Thanks!

For the first part, i will prove each is analytic and use theorem to define the whole z sin(z) cos (z) as analytic.

by the way the second part i didn't appreciated it yet.
Any hint? Thanks.

 

[Dick]

The quotient of two analytic functions is also analytic as long as the denominator doesn't equal zero. That's not actually in your list of theorems, but you can get it by putting 2, 3 and 4 together. f(z)=1/z is analytic for z not equal 0 and g(z)=z*sin(z)*cos(z) is analytic. Now use 4. What's f(g(z))? Where is it not analytic?

 

[NJunJie]

Hi,

Should be (3):
3. A rational function (the quotient of two polynomials) is analytic, except
at zeroes of the denominator.

Calculate the derivative of 1/[z(sin(z))(cos(z))] at z=0, + pie/4, + pie/2, + 3pie/2.
-- i think its related to the points at which the function of sine and cosine will cut the x-axis which makes the denominator zero?

pie = 3.142... that 'pie'

 

[Dick]

Sort of. You do need to use 4 as well, since z*sin(z)*cos(z) is not a polynomial. Yes, I think the values given are related to zeros of z*sin(z)*cos(z). But pi/4 isn't a zero, is it?

 

[NJunJie]

Thanks alot.

sort of proved the Cauchy Riemann complicated steps already after some careful derivation.
And i have also proved through the theorems indirectly from this exercise.

For the second part; i attempted to take the limit of [1/[z(sin(z))(cos(z))] as (change in Z)->0.

Heres the result - what can i say from this mathematically?

1 / [ (change in y)*(sin j(change in y)* cos j(change in y)] --- (1)

1 / [ (change in x)*(sin (change in x)* cos(change in x)] --- (2)


note (1): set change in x = 0; and (change in y) tends to zero
note (2): set change in y = 0; and (change in x) tends to zero


Generally ; the limit will fail when the change in y or x tends to zero for 'z' or 'sin z' part?

 

[Dick]

Now I'm really not sure what you are doing. Define g(z)=z*sin(z)*cos(z). If g(z) is not equal to zero, then 1/g(z) is analytic at z. If g(z) is equal to zero then 1/g(z) is not analytic at z. It's not even defined at z. Why do you have to do limits?

 

[NJunJie]

was thinking to proof the second part from first principles.
seems like your method clearer.

:)

Thanks.

 

[MSLeong]

Hi NJunJie,

i believe you are a student in my class. Is your mat no. u0800640 ? Pls see me after class on monday. I will review your marks for the assigment.


Rgds
MSLeong