Enquiry For Math

[shermaine]

Determine the 6 distinct roots for the following equation: Z^6 = [j(1 + j)/(1 - j)]^2, giving answers in the form of a + jb. May i know how to solve this?

 

[shermaine]

Math Enquiry

May i know if sin(z)cos(z) satisfies Cauchy - Riemann?
Is it analticity for all values of Z?
Does 1/[sin(z)cos(z) satisfy similar conditions for z = 0, +/- pi/4, +/- pi/2, +/- 3pi/2?

 

[shermaine]

Math Enquiry 2

Determine the location and the nature of singularities in the finite z plane for (1) f(z) = (z^2 + 1) sin(z)/(z(z + 1)(z - 3)
(2) g(z) = [sinh(z)/z^9]

 

[steveB]

Determine the 6 distinct roots for the following equation: Z^6 = [j(1 + j)/(1 - j)]^2, giving answers in the form of a + jb. May i know how to solve this?

In general, a practical way to solve this is to use a calculator. Just take the 6'th root of the argument ( in this case [latex] (j{{1+j}\over{1-j}})^2[/latex] ). Note that the magnitude of the argument is one, so your roots will be on the unit circle. You need a calculator that can handle complex math (or Matlab, Mathcad etc). Then, once you have at least one root (note calculators usually give only one), the remaining 5 roots will be equally spaced ( in angle ) around the unit circle.

Note that not only is the magnitude of the argument one, but it happens to be equal to one in this case. That is, [latex] (j{{1+j}\over{1-j}})^2=1[/latex] as shown below. So, the problem reduces to finding the sixth root of unity. Obviously, one root is 1 and the remaining 5 are equally spaced around the the unit circle (angle spacing [latex] \pi\over 3 [/latex]). Hopefully, you are familiar with converting complex numbers from polar to rectangular format, and vice versa.

[latex] {{1+j}\over{1-j}}={{(1+j)}\over{(1-j)}}{{(1+j)}\over{(1+j)}} [/latex]

[latex] {{(1+j)}\over{(1-j)}}{{(1+j)}\over{(1+j)}}=j [/latex]

hence,

[latex] (j{{1+j}\over{1-j}})^2=(j j)^2 =-1^2=1[/latex]

 

[picbits]

Is this homework time again ?

 

[steveB]

Determine the location and the nature of singularities in the finite z plane for (1) f(z) = (z^2 + 1) sin(z)/(z(z + 1)(z - 3)
(2) g(z) = [sinh(z)/z^9]

The location of each singularity is simply the value of z that makes the denominator zero. We sometimes call those the "poles". I'm not sure what the question means by "the nature of the singularity". My best guess is that it wants to know if the singularities are degenerate which means that there is more than one pole at that singularity location.

For f(z), the denominator is z(z+1)(z-3). It should be clear what 3 values for z will make the denominator zero. It is also clear that all 3 values are unique, so each pole is not degenerate.

For g(z), the denominator is z^9. It is clear that only z=0 is the pole location. However, the degeneracy is 9.

 

[steveB]

May i know if sin(z)cos(z) satisfies Cauchy - Riemann?
Is it analticity for all values of Z?
Does 1/[sin(z)cos(z) satisfy similar conditions for z = 0, +/- pi/4, +/- pi/2, +/- 3pi/2?

I have a feeling that I may be overlooking a simpler approach, but here is my suggestion.

The way that I would approach this is to use some trig identities to get sin(z)cos(z) into the form u(x,y)+jv(x,y) where z=x+jy. This allows you to test the Cauchy Riemann equations as follows:

du/dx=dv/dy and du/dy = -dv/x where the d indicates partial derivative.

To do this the following identities can be used:

sin(z)cos(z)=sin(2z)/2
sin(x+jy)=sin(x)cos(jy)+cos(x)sin(jy)
cos(jx)=cosh(x)
sin(jx)=j sin(x)

Note there are no singularities in sin(z)cos(z).

The second part of the problem involves the inverse which has singularities. So you need to consider that when looking at the values of z = 0, +/- pi/4, +/- pi/2, +/- 3pi/2. I seem to recall a theorem that says that the inverse of an analytic function is analytic at points that are not singularities, but I'm not sure. Please double check me on that.

 

[shermaine]

In general, a practical way to solve this is to use a calculator. Just take the 6'th root of the argument ( in this case [latex] (j{{1+j}\over{1-j}})^2[/latex] ). Note that the magnitude of the argument is one, so your roots will be on the unit circle. You need a calculator that can handle complex math (or Matlab, Mathcad etc). Then, once you have at least one root (note calculators usually give only one), the remaining 5 roots will be equally spaced ( in angle ) around the unit circle.

Note that not only is the magnitude of the argument one, but it happens to be equal to one in this case. That is, [latex] (j{{1+j}\over{1-j}})^2=1[/latex] as shown below. So, the problem reduces to finding the sixth root of unity. Obviously, one root is 1 and the remaining 5 are equally spaced around the the unit circle (angle spacing [latex] \pi\over 3 [/latex]). Hopefully, you are familiar with converting complex numbers from polar to rectangular format, and vice versa.

[latex] {{1+j}\over{1-j}}={{(1+j)}\over{(1-j)}}{{(1+j)}\over{(1+j)}} [/latex]

[latex] {{(1+j)}\over{(1-j)}}{{(1+j)}\over{(1+j)}}=j [/latex]

hence,

[latex] (j{{1+j}\over{1-j}})^2=(j j)^2 =-1^2=1[/latex]

HI Steve, can i check with u.
For the part 1, how do u get pi/3?
Can i do it this way:
1(cos (pi + 2pi . K)/6 + j sin (pi + 2 pi . K/6))?

 

[shermaine]

I have a feeling that I may be overlooking a simpler approach, but here is my suggestion.

The way that I would approach this is to use some trig identities to get sin(z)cos(z) into the form u(x,y)+jv(x,y) where z=x+jy. This allows you to test the Cauchy Riemann equations as follows:

du/dx=dv/dy and du/dy = -dv/x where the d indicates partial derivative.

To do this the following identities can be used:

sin(z)cos(z)=sin(2z)/2
sin(x+jy)=sin(x)cos(jy)+cos(x)sin(jy)
cos(jx)=cosh(x)
sin(jx)=j sin(x)

Note there are no singularities in sin(z)cos(z).

The second part of the problem involves the inverse which has singularities. So you need to consider that when looking at the values of z = 0, +/- pi/4, +/- pi/2, +/- 3pi/2. I seem to recall a theorem that says that the inverse of an analytic function is analytic at points that are not singularities, but I'm not sure. Please double check me on that.

Hi Steve, to convert to U(x,y) + jV(x,y)...is it as per below:

sin(z)cos(z) = 1/2 sin 2z
= 1/2 [ cosxcos2jy + sinxsin2jy]
= 1/2 [cosxcosh2y - jsinxsinh2y]

Is that correct?

 

[steveB]

HI Steve, can i check with u.
For the part 1, how do u get pi/3?
Can i do it this way:
1(cos (pi + 2pi . K)/6 + j sin (pi + 2 pi . K/6))?

Hello,

Almost, but your expression leads to -1 when raised to the 6th power. Modify it slightly to get ....

[latex] (\cos {{2\pi k}\over {6}} + j \sin {{2\pi k}\over {6}})^6=(e^{{j2\pi k}\over {6}})^6=e^{j2\pi k} = 1 [/latex] for [latex] k=1,2,3,4,5,6 [/latex]

 

[Tesla23]

I have a feeling that I may be overlooking a simpler approach, but here is my suggestion.

The way that I would approach this is to use some trig identities to get sin(z)cos(z) into the form u(x,y)+jv(x,y) where z=x+jy. This allows you to test the Cauchy Riemann equations as follows:

You could use the fact that products of analytic functions are analytic

 

[steveB]

Hi Steve, to convert to U(x,y) + jV(x,y)...is it as per below:

sin(z)cos(z) = 1/2 sin 2z
= 1/2 [ cosxcos2jy + sinxsin2jy]
= 1/2 [cosxcosh2y - jsinxsinh2y]

Is that correct?

No, that does not look correct to me. There are a few mistakes. I think you should have 2x and not x for the arguments of the functions. Also, the final form should be sin()cosh() + j cos()sinh().

I noticed that I had a typo in what I wrote before:
I should have wrote sin(jx)=j sinh(x) and not sin(jx)=j sin(x)

 

[steveB]

You could use the fact that products of analytic functions are analytic

Yes, very good! Working with functions individually is easier, generally. Although in this particular case cos(z)sin(z)=sin(2z)/2, so it is about the same amount of work either way.

 

[shermaine]

Hello,

Almost, but your expression leads to -1 when raised to the 6th power. Modify it slightly to get ....

[latex] (\cos {{2\pi k}\over {6}} + j \sin {{2\pi k}\over {6}})^6=(e^{{j2\pi k}\over {6}})^6=e^{j2\pi k} = 1 [/latex] for [latex] k=1,2,3,4,5,6 [/latex]

steve, should that be e^j(pi/3)k?


i got one doubt, hmm how cm i cant use pi + 2pi?

btw, can i also write it in Z = 1^1/6 . ej^(pi/6 + 2pi.k/6)?

 

[shermaine]

No, that does not look correct to me. There are a few mistakes. I think you should have 2x and not x for the arguments of the functions. Also, the final form should be sin()cosh() + j cos()sinh().

I noticed that I had a typo in what I wrote before:
I should have wrote sin(jx)=j sinh(x) and not sin(jx)=j sin(x)

I got another question. How do i go about searching for analytic and singularites?
Can i said if it is analytic, it will be non singularities?

 

[steveB]

steve, should that be e^j(pi/3)k?

Should which part be e^j(pi/3)k? Do you mean e^j(pi/3)k=e^j(2pi/6)k? If so, yes, you can write it either way.

i got one doubt, hmm how cm i cant use pi + 2pi?

If you do that and raise it to the 6'th power, you get -1 instead of 1.

btw, can i also write it in Z = 1^1/6 . ej^(pi/6 + 2pi.k/6)?

I'm not sure I understand your notation ej^(pi/6 + 2pi.k/6), but I don't think it's right, no matter how it's interpreted.

 

[shermaine]

thanks!! so much

 

[steveB]

I got another question. How do i go about searching for analytic and singularites?
Can i said if it is analytic, it will be non singularities?

I'm a little hesitant to answer this, because I've seen people disagree over what the definition of analytic is. Some people refer to points being analytic and others refer to functions being analytic.

I was taught that an analytic function has at least one regular point. A regular point is one that is not singular, and is formally defined as follows:
"If a function possesses a derivative at some point Z0 in the z-plane, and if we can draw a small circle around the point such that the derivative exists at all points in this circle, then Z0 is a regular point. "

From "Complex Variables and the Laplace Transform for Engineers" by Wilbur R. LePage, section 2-5:

So basically, to answer your question, you need to look at points and decide if they are regular. If you can find one regular point, the function is analytic. If you find a point that is not regular, then that point is singular.

 

[shermaine]

Determine the location and the nature of singularities in the finite z plane for (1) f(z) = (z^2 + 1) sin(z)/(z(z + 1)(z - 3)
(2) g(z) = [sinh(z)/z^9]

using Cauchy's integral formula , evaluate

f(z) dz with C: |z + j| = 2 (Counter clock wise)
and g(z) dz with C: | z -1| = 5 (counter clock wise)

how do i go about finding the test removable and test for poles?
And how to i plot the contour?

 

[steveB]

using Cauchy's integral formula , evaluate

f(z) dz with C: |z + j| = 2 (Counter clock wise)
and g(z) dz with C: | z -1| = 5 (counter clock wise)

how do i go about finding the test removable and test for poles?
And how to i plot the contour?

I'm not sure what you mean by "test removable", but I think I understand what you are trying to do. It's difficult to explain the whole procedure here, and usually people learn this with a good book for reference. Try the following web-link for some details.

Cauchy's integral formula - Wikipedia, the free encyclopedia

The basic approach involves plotting (either mentally or physically) the circular contours. Contours of the form | z - z0| = r are just circles in the complex plane with center at z0 and radius r. Remember your formula for a circle? (x-x0)^2+(y-y0)^2=r^2 It's the same thing.

Next you want to see if the functions are analytic and if any poles are within the circular contour. If no poles are in there, the integral is zero. If there are poles, then you can use a theorem to express the integral as a sum of integrals around each pole, which is easy and given in the article above.

If your poles are degenerate (which is the case for g(z)), you need to be more careful, but I recommend you get through f(z) before thinking about that issue. It's better to go one step at a time, otherwise it will be overwhelming.

EDIT: After thinking further, I see you probably are referring to testing for removable poles with "test removable". You need to look at that too. I'll try to find a reference about that.