Recent content by sweetvirgogirl

so i know what it is (i think lol) ... but what are its applications?

f (z) = lamdba (a - z)/(1 - abar*z) |lambda| = 1; |a| <1 where f is any one-to-one analytic function mapping delta = {z:|z| < 1} onto itself set g(z) = (a - f(z))/(1-abar*f(z)) now they say that g is one-to-one analytic function mapping delta onto delta (why?) and g(0) = 0 and then they...

so the way book states it, for the sc. lemma to work, |f(z)| has to be less than or equal to 1 and and z has to be less than 1. However, the book seems to use the lemma in some problems even if one of the conditions is not satisfied ... any help with gretaly appreciated also ...if you could...

thats what i thought ... i wrote down "open" as my answer and the prof circles it and I don't think I got any points for it ... yes, i didnt write connected, but I should at least get half the points or something. oh well maybe he didnt gimme any credit, because I didnt explain why I think it's...

in that case ... wouldn't it be an open set? and it will be above real axis? (meaning the boundary is upper plane or lower plane? getting confused with terminology a little)

how exactly do you determine that? coz the way i saw it ... -1 was on the line ... maybe i need some rest

Let C be a simple closed curve. Show that the area enclosed by C is given by 1/2i * integral of conjugate of z over the curve C with respect to z. the hint says: use polar coordinates i can prove it for a circle, but i am not sure how to extend it to prove it for any given closed curve

{z^2: z = x+iy, x>0, y>0} i am a lil confused about the notation to represent the set ... i'm used to seeing {z: z = x+iy, x>0, y>0} but what effect does squaring z have? i thought the set was open simply because x>0 and y>0 ... but apprently i was wrong ... (or maybe not?) ... i...

no we haven't covered cross ratio in class yet. so i *sorta* get how to send a circle to a line ... ... the circle |z| = 1 and the line Re((1+i)w) = 0 ... so i sent 1 to infinity, -1 to 0 and i = -1 and i came up with i*(z+1)/(z-1) ... the book has the answer as (1-i)*(z+1)/(z-1) i...

woopsie ... let's restart then ... I send i to infinity, -i to zero ... what about the third point? it has to do something with fixing orientation, right? how does that work? (like i know the third point can not be totally arbitrary like the first two points were, right?)

oops typo ... i meant we could send i to infinity and -i to - infinity and then the third point? (-infinity and infinity are different, right?)

if i apply max/min modulus to both f and 1/f, then it means both f and 1/f don't have a local max/min ... so what does it tell me? i'm sorry ... i don't know if it's lack of confidence or what .. but i still don't see it

thats what i thought ... but i still have no insight about f having any zero inside lambda ... like i kinda see it visually ... like i know it makes sense ... but i got no clue how to "prove" it

the way prof showed it in class ... he said pick three points on the circle and send it to three points on the line (or vice versa if we are trying to map a line to a cirle) ... but i am not sure how do i know what those three points are going to map to ... on a circle |z| = 1, we could send...

what do i do with cases infinity/infinity and/or 0/0?