I'm having trouble understanding some of your notation. I would be easier to help you if you can either scan a hand written document, or if you learn to use the latex equation notation as follows
[latex] x+y=z [/latex]
[latex]e^{j \omega t} [/latex]
[latex]\sin x [/latex] etc.
However, I notice a number of mistakes above. When you ask about the zeros and poles, you are not looking at the correct value of z, for example you need to ask about z=-1 and not z=1 in the first question. Also, that Laurent expansion does not look correct.
Well it was a little clearer, but you should use the equation editor in Word to make it better. Or, the latex notation is good too.
Anyway, I've made some suggestions in the attached document. I went through quickly, so I could have made some mistakes, but I did correct a few of your mistakes and you should be able to figure it out from there.
Thanks so much for your advise.
I got one question for (b) and (c)
For (b), how come there is a minus sign for -3j/z2 + 5z + 1)?
For (c), Yes, i mean For sin z = 0, z = 0, z = +/- , z = +/- 3 , z = +/- 4 (you mean 0, ±p, ±2p, ±3p )
I had amended a bit for (c), but i still blurred with finding residue for it.
Maybe if you can assist me abit further?
Thanks so much for your advise.
I got one question for (b) and (c)
For (b), how come there is a minus sign for -3j/z2 + 5z + 1)?
For (c), Yes, i mean For sin z = 0, z = 0, z = +/- , z = +/- 3 , z = +/- 4 (you mean 0, ±p, ±2p, ±3p )
I had amended a bit for (c), but i still blurred with finding residue for it.
Maybe if you can assist me abit further?
I got a negative sign from the j being in the denominator. When you move the j to the numerator it becomes -j. However, like I said, I went quickly, so I could have made a mistake. Go through the calculation slowly and you should get it right.
Also, why are you looking at the point +/-4pi? Isn't 4pi greater than 12 and outside the circle of integration?
Also, why to you need to take the limit of the exponential function. You have already shown that the pole at zero is canceled by a zero at zero. Just use the residue formula at +/-pi, +/-2pi and +/-3pi.
Note that is is very hard to interpret your document. The notation is very confusing. I'm doing my best to interpret it, but I could be misunderstanding something. Anyway, I'm just trying to give you hints that will allow you to get it to the right answer.
I got a negative sign from the j being in the denominator. When you move the j to the numerator it becomes -j. However, like I said, I went quickly, so I could have made a mistake. Go through the calculation slowly and you should get it right.
Also, why are you looking at the point +/-4pi? Isn't 4pi greater than 12 and outside the circle of integration?
Also, why to you need to take the limit of the exponential function. You have already shown that the pole at zero is canceled by a zero at zero. Just use the residue formula at +/-pi, +/-2pi and +/-3pi.
Note that is is very hard to interpret your document. The notation is very confusing. I'm doing my best to interpret it, but I could be misunderstanding something. Anyway, I'm just trying to give you hints that will allow you to get it to the right answer.
Thanks so much! From my understanding,since after taking lim [z.exp z ]/[sin z] = 0/0
z -> 0
then i need to take L'hospital rule, which is lim [z.exp z + exp z]/[sin z] = 1/1 = 1
z -> 0
Is that correct method?
As for the Residue, i am not sure how to express it out for the numerator portion.
Is it for -3pi,
Its as followed:
Res(f(z),-3pi) = lim (z + 3pi) [(z.exp z)/sin z]
z - > -3pi
May i know what is exp (- 3pi)?
Do i need to differentiate again in residue?
May i know what should be the correct answer for this part?
How do i solve the numberator part?
Plesase advise me a bit more. Thnks.
Thanks so much! From my understanding,since after taking lim [z.exp z ]/[sin z] = 0/0
z -> 0
then i need to take L'hospital rule, which is lim [z.exp z + exp z]/[sin z] = 1/1 = 1
z -> 0
You really need to be careful when your write mathematical equations. Everything you are writing has mistakes in them. I don't know if these are just typos, or if you just don't understand. Anyway the above is not making sense.
Res(f(z),-3pi) = lim as z-> -3pi ( (z + 3pi).z.exp(z)/sin(z)] )
Note that (z+3pi) goes to zero and sin(z) goes to zero, but in the limit as z goes to - 3pi, (z+3pi)/sin(z) goes to 1.
Thanks for your reply.
But i was thinking for Res(f(z),-3pi) = lim as z-> -3pi ( (z + 3pi).z.exp(z)/sin(z)] )
(z + 3pi) and sin (z) cant cancelled off, so it remains there.
And thus the equation becomes
( ([-3pi] + 3pi).[-3pi].exp(-3pi)/sin(-3pi)] )
sin(-3pi)] is 0 and ([-3pi] + 3pi) = 0 , so left [-3pi].exp(-3pi)
so can i left the answer in -3pi.exp(-3pi) = -540 exp (-540)
But i got one last question: exp (-540) when keyed into calculator, its a math error.
what answer should i left that be?
Steve, really thanks for all your help
so can i left the answer in -3pi.exp(-3pi) = -540 exp (-540)
But i got one last question: exp (-540) when keyed into calculator, its a math error.
what answer should i left that be?
Thanks so much for your help.
I was working on z = -3pi,
Res(f(z),-3pi) = lim z ---> -3pi ( z + 3pi) . [z.exp z / sin z]
since z = -3pi , thus -3pi + 3pi = 0 so 0 multiply anything = 0
I'm not sure if i'm correct and it seem that all the equation for -3pi, 3pi,-2pi is equal to zero.
Could you advise me? Thanks!
Thanks so much for your help.
I was working on z = -3pi,
Res(f(z),-3pi) = lim z ---> -3pi ( z + 3pi) . [z.exp z / sin z]
since z = -3pi , thus -3pi + 3pi = 0 so 0 multiply anything = 0
I'm not sure if i'm correct and it seem that all the equation for -3pi, 3pi,-2pi is equal to zero.
Could you advise me? Thanks!
I believe you can take the limit as z goes to -3pi. Both the numerator and denominator approach zero, but the ratio is finite when you take the limit. I think if you get a nonzero value for the answer, you will have the right answer. See section 2.3.(4) on page 5 of the attached document.