am i doing this prove by liouville's theorem?
can i do it like this:
suppose f = 1/e^f(z)=1/e^ Im(f) < 1/e^M
so it is bounded
by liouville's theorem, it is constant
1. Evaluate ∫C (z)/z2+9 dz , where C is the circle │z-2i│=4.
what i have done so far is :
z(t) = 2i + 4eit
z'(t) = 4ieit
f(z(t)) = 4ieit/(4ieit)2+9
∫ (4ieit/(4ieit)2+9) (4ieit) dt
intergrate from 0->2pi
but i don't know how to solve this intergral, can anyone help?
2. ∫c...
Q:Let f be entire and suppose that I am f(z) ≥ M for all z. Prove that f must be a constant function.
A: i suppose M is a constant. So I am f(z) is a constant which means the function is a constant.
Am i doing this right ?
but i don't think there will be such a stupid question in my...
compute the integral ∫Cr (z - z0)n dz,
with an integer and Cr the circle │z - z0│= r traversed once in the counterclockwise direction
Solution:
A suitable parametrization for Cr is give by z(t)= z0 + reit 0≤t≤2π
...
...
My question is , how to find that suitable z(t)?
i have no idea...
thanks, i know eπi=-1
but the problem is , when i was in exam i was looking at the -1 , how can i know e?, what about the question is -2 , do i need to try it one by one like π/2,π/3...
Prove that cos z = 0 if and only if z = π/2 + kπ, where k is an integer.
The solution is :
cos z = (eiz+e-iz)/2 = 0
<=>eiz+e-iz)=0
<=>e2iz=-1
<=>2iz=πi +2kπi
<=>z=π/2 + kπ
But i don't understand how does these 2 steps work
<=>e2iz=-1
<=>2iz=πi +2kπi
how can a exp function...
In my complex analysis textbook, there is a definition about log function, which is:
log z := Log |z| + i arg z
= Log |z| + i Arg z + 2kπ where k = 0,±1,±2...
my question is , is there any different between log and Log, and arg and Arg?
If yes, what's the different between them