Week1Lecture5: Topology in the complex plane

[@probono2876]

Prof Bonfert-Tylor, many thanks for your great lectures.

[@mohamedtlili334]

My first lecture in my Complex Analysis went over the topology of C and it was very confusing and terse. This video helped a ton, thank you!

[@emc2884]

Thanks prof for such great series of lectures .

[@johnq4841]

this is a hard topic to understand, but if u come across this, the world is yours. Thank you, Professor!

[@seanc7472]

thank you so much. Because your great lecture and I love complex analysis.

[@ghostzart]

The information is not exactly in the same order, but your lectures are a great supplement to CA by Gamelin. Thank you for your hard work!

[@evionlast]

Such good videos

[@taegyukang3249]

I doubt the theorem she gave that a subset G of C is connected iff any two points in G can be joined in G by successive line segments. I think the latter statement is equal to the definition of path - connectedness, so I guess that the topologist's sine curve can be a counterexample for that theorem(the graph of topologist's sine curve is a subset of R^2 which is isomorphic to C thus the graph can be regarded as a subset of C). Am I right? or Did I have any serious mistake???

[@danielhader4063]

The topologist's sine curve can be regarded as a subset of C, but it's not open, which is a requirement for the theorem.

[@Rad631]

thank you so much!

[@salumuhamisiramadhani6702]

how can I have this notes I mean this power point

[@MrSazid1]

so do the complex plane contains negative infinity... since R is a subset of C ? Or is it like R is an open set and infinity is its boundary so C contains R but not its boundary Pos. and Neg infinity..:)

[@michel_dutch]

Infinity, be it positive or negative, is not a number, and is therefore not a member of R, nor of C.