After calculating roots of the equation on denominator. It's clear that is the second degree equation and only one of the 2 root are inside the region os -i/2
when denominator is separated (z+i/2) is multiplied by 2 but numerator didnt in the video. Final version of the integral must be 8 /(z+2i)(2z+i). Therefore result of limit must be 8/3i and integral is 16pi/3
Be sure to write the factor of (2z + i) in the denominator as 2(z + i/2). This factor of 2 in the denominator will give the limit as 4/(3i) as shown in the video.
+Catherine Berrouet Hi! If you can get to the step of concluding that the residue at -i/2 is 4/(3i), then remember to multiply that by a full 2(pi)i. The 2 times the 4 creates 8, and the i in fraction's denominator cancels with the i from 2(pi)i, and the pi just sticks around, leaving 8 pi / 3. I'm not sure how you got -4 pi/3, but some things to watch for are making sure your i from the residue is in the *denominator*, and making sure you include the 2 in the 2(pi)i. Good luck!
These videos are saving my academic life
Thanks for turning 5 lectures into 10 minutes. You're a god sent sir!
These videos are a great treatment of complex; ideal for somebody like me who's revisiting the subject after many years.
The steps are very clearly explained, thank you for that.
The steps are very much clearly explained...was really helpful... thankyou
Your explanation is very clear! Thank you!
I like how you say "Hi" in the beginning
Excellent. Thanks Michael for this very clear video.
Integrating 1/(a+cos x)^2 from 0 to 2pi
Thank you so much. You're awesome
thank you so much
This is a helpful video.
I used Cauchy's Integral Formula and got the same answer as yours. Would it have been correct had the pole not been of the first order?
So useful, thanks
What if the limits are not from 0 to 2pi but from 0 to something other than 2pi say 5pi/3 for example.
How do we know that the original and the contour integral are equal?
THANKS A LOT
why did not we find the laurent series then evaluate the a-1 coeff?
How did you take the limit?Didn't understand that part
I love it!
How do you know how to choose other contours and such?
If the question is in integration of sinx^2 form so how can we integrate? Limit from 0 to 2pie
when you say poles above the real axis, does that count for z=0 also?(origin)
morning , where did you get Z-(-i/2) for calculating the limit, thank you very much
After calculating roots of the equation on denominator. It's clear that is the second degree equation and only one of the 2 root are inside the region os -i/2
How we predict that radius is 1?
when denominator is separated (z+i/2) is multiplied by 2 but numerator didnt in the video. Final version of the integral must be 8 /(z+2i)(2z+i). Therefore result of limit must be 8/3i and integral is 16pi/3
I think you are right
+Charlis Johnson I agree bro :)
What if it is integral of dz / ((z^4) +1)
Thank you sir you made me understand this
Thank you
THANKS
Sir why we take contour of unit circle
it's the simplest substitution. we could have chosen z=re^(i*theta) but then rewriting sine would be a pain unless we choose r=1
goat
dont know how you get 4/3i for that limit, i got 4/(3i/2) which give me 8/3i not 4/3i
yes me too
Be sure to write the factor of (2z + i) in the denominator as 2(z + i/2). This factor of 2 in the denominator will give the limit as 4/(3i) as shown in the video.
LACHGAR VISION 2 yes that's there is 2 must be in numerator and then multiply with 4
I agree with you. so the answer is 16pi/3
yes there is a mistake
sir,the given function don't have any singular points so its integral should have been 0.
Sorry, I can't stand to listen to anyone who uses "times" as a verb.
i keep getting -4pie/3 and not 8pie/3.
+Catherine Berrouet Hi! If you can get to the step of concluding that the residue at -i/2 is 4/(3i), then remember to multiply that by a full 2(pi)i. The 2 times the 4 creates 8, and the i in fraction's denominator cancels with the i from 2(pi)i, and the pi just sticks around, leaving 8 pi / 3.
I'm not sure how you got -4 pi/3, but some things to watch for are making sure your i from the residue is in the *denominator*, and making sure you include the 2 in the 2(pi)i. Good luck!
Michael Barrus thank you!!!!!
i figured out my mistake. it's clear now. :)
@@michaelbarrus9935 hey I have a question. How would you evaluate integral of
(cos(x^2) - sin(x^2)) / (1 + x^8) from 0 to infinity?