Complex Analysis L10: Cauchy Integral Formula

This video explores the Cauchy Integral Formula (CIF), which is one of the most important theorems for complex contour integrals.

@eigensteve on Twitter

This video was produced at the University of Washington

[shahabm.]

The charisma and humour and the way you explained the subject makes it not only fun and interesting but also clear to absorb and understand the concept. I am happy l discovered your lectures. Thanks

[GoinHamm-yd]

"Don't drink and derive" gave me a chuckle! Dry humour is best humour.

[wes]

Been searching for a video like this for a while. Great job

[Pauly-ulob]

Many thanks for these videos, they are truly entertaining and brilliant.

[cowgomoo]

ah i used a similar but less rigorous argument for I_2. i just taylor expanded f(z) around a (because f(z) is analytic there) and said that for a sufficiently small circle f(z) \approx f(a). then the numerator is 0 and the whole integral is 0. it's pretty much exactly what you did but not as rigorous as the epsilon delta stuff. i like what you did better. thanks for the videos they have been great for helping me prepare for my exam.

[bSharpHacker]

Oh I remember this topic when taking Mathematical Methods for Physicists, by Arfken back in the mid 90s. I wish I had YouTube videos like this back then! So jealous 😁

[ingeniandy]

Thank you so much! I can understand complex numbers thanks to you!

[rafvermeer]

You can also write f as a Taylor series to prove the theorem.

[kashifislam]

Dear sir, as you have established the inequality I2 is less than or equal to 2pi*iota*epsilon. But we know that "iota" does not obey order relations <, =, >.

[ralphhebgen]

2:59! That’s why! I never knew that but have always been wondering. Cool.

[Alan-zftt]

These are great! Thank you for sharing and... is this okay?
integral over closed curve of [f(z) / (z-a)]dz = I = I₁+ I₂ as shown about 7:09
By linearity of integral operator on I₂ the numerator f(z) - f(a) over common denominator (z-a)
can be split into two parts wlog f(z)/(z-a) and -f(a)/(z-a) and this is negative result to I₁ so sum to zero
And so integral I = I₁+ I₂₍₁₎ - I₂₍₂₎ becomes integral over closed curve of [f(z)/(z-a)]dz =f(a)2πi from Cauchy Integral Formula since second part of integral in I₂ is negative value in I₁

Equivalently I₂ = I₂₍₁₎ + I₂₍₂₎ and I₂₍₂₎ = -I₁

I cannot help but wonder at some Newtonian things happening here such as the center of gravity of a sphere equals sum of all its 'parts'.
When these are summed up the value is zero.

[mumina_b]

Thank you for this video
Remember people, "Don't drink and derive"😂

[jayCal]

"Analytic functions are continuous on steroids" man😂🔥

[JgNt]

Whats this board used to write called?

[ralphhebgen]

“Don’t drink and derive”. Seriously? 😂😂

[hoseinzahedifar]

The link you solved the exercise on Matlab (I didn't find any code in python):

[final_gamesy]

For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???

[hoseinzahedifar]

In time 5:08: Don't drink in drive 😂😂😂.

[final_gamesy]

For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???

[final_gamesy]

For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???