Yes, C is isomorphic to R2 which really means they are the same vectorapace, like they are really the same space, only C has a multiplication which R2 doesn't have. The only problem that you have is that you cannot perform easy changes with inequalities, as an example: Let z be complex and ε reel and >0 then |z|
Very thorough. Nicely done. Good details.
You're a beast bro. I liek the way you prove shit.
Good work
But why did not use sandwitch theoreme after you had the double inequality
⏳🧙♀️
By using the sandwich theorem you’re using the proof he wrote by assuming it’s true
@ no you’re not, when you have cosx
@@rudeusdoto but then the limit for cos(x) still has to be proven with epsilon-delta
@filipeoliveira7001no it doesn't as he already proved cos is continuous
What good textbooks use this approach (like properties of goniometric functions from definitions like these)?
Happy holidays!
Did you prove that lim{(e^z-1)/z=1 as z->0} for complex z?
Nice proof . Can this epsilon delta proof be generalized to complex number z approaches 0+0i?
Yes, C is isomorphic to R2 which really means they are the same vectorapace, like they are really the same space, only C has a multiplication which R2 doesn't have. The only problem that you have is that you cannot perform easy changes with inequalities, as an example: Let z be complex and ε reel and >0 then |z|
bro you spelled "like" wrong on your channel 😉
うおw