the complex exponential function
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- Опубліковано 26 лис 2023
- Advanced MathWear:
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Complex analysis lectures:
• Complex Analysis Lectures
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Your mastery of Complex Analysis is top tier noteworthy!
Not even my Grad Professor taught this good & also this cool! 🤣🤣
Just few days ago I proved that this function is continuous using the limit definition and its Taylor series expansion
First using the ratio test to prove it conv for all z in C
then I got in the deferent of the sum (using the epsilon definition)z^n-a^n
Then I proved a lemma that it's equal to (z-a) times a sum of z^(n-1-k)a^k
k going from 0 to k-1
using mathematical induction
Then I can get z-a out of the absolute value
Now I am very close
I have abs(z-a) × f(a,z)
My professor said in this case we can let z=a in f(a,z)
And divide f(a,a) on both side
(I don't understand why this is okay to do maybe someone can explain it to me)
Then I get a geometric sum in terms of a^(k-1)
So I get (1-a^n)/(1-a)
So I can get 1/(1-a) out of the absolute value
then so simplification and using the fact that e^z
equal to the original sum
So I get the sum equal to e-e^a from 1-a^n that were in the sum
In the end I get that delta equal to epsilon times abs((1-a)/(e-e^a))
And that complete the proof
What is the name of the book and the author please? I couldn’t read your handwriting.
It says Gamelin's Text, which probably means the book "Complex Analysis" by Theodore W. Gamelin
ありがとうございます!
for vertical lines: at x=0 ->unity circle, x0 outside.
So what about kx+b lines? Spirals from (0, 0)?
is it wrong to rewrite (cos theta + i sin theta) as cos theta (1 + i tan theta)? where theta is not equal to Pi/2
(y')!=y , y(x0)=y0 please 🤗🤗
I can't see clearly because of the colour used in writing.
woohooo