MATH5685 -Assignment 3 Due: Wednesdayday 3 October 2012. The open unit disk is denoted D. Ql. Suppose that an > 0 for all n. Show that Fr i(i + an) converges if and only if EZ1 an converges. [Hint:...

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MATH5685 -Assignment 3 Due: Wednesdayday 3 October 2012.
The open unit disk is denoted D. Ql. Suppose that an > 0 for all n. Show that Fr i(i + an) converges if and only if EZ1 an converges. [Hint: prove that f n 1(1 + an) 5 exp(EnN_i an).]
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Q2. Where does f (z) = z2Th) converge? And what does it converge to? n=0
co Q3. Show that f (z) I e-ttz-1 dt is analytic on the right half plane.
Q4. For n = 1,2,3, .. Step 2 is to prove
kze .Lie-1- -C T —
2z2
n-1 . , find a formula for fl Fn. (Step I is to find a formula; k=1 that it hold for all n.)
(1 + z)2 (a) Find the image of ID under the map f . (b) Find the image of the right half-plane under the map f . Q6. Find a bijective conformal maps which map (a) the square
onto the half-disk
(b) the sector
onto the strip
S = {x iy : 0 Q2 iy : 0



Answered Same DayDec 20, 2021

Answer To: MATH5685 -Assignment 3 Due: Wednesdayday 3 October 2012. The open unit disk is denoted D. Ql....

David answered on Dec 20 2021
124 Votes
1. Note that
ex = 1+
x
1!
+
x2
2!
+ · · · =
∞∑
n=1
xn
n!
≤ 1+ x
Now let sn = a1 + a2 + · · ·+ an, pn = (1+ a1) · (1+ a2) · · · (1+
an). Both sequences {sn} and {pn} are
increasing, and hence to prove the theorem we need to only show that {sn} is bounded if and only if
{pn} is bounded.
First, the inequality pn > sn is obvious. Next, taking x = ak where k = 1,2, · · · ,n and multiplying, we find
pn < e
sn . Hence, {sn} is bounded if and only if {pn} is bounded. Note that {pn} cannot converge to zero
since each pk ≥ 1. Note also that
pk → +∞ if sn → +∞.
2. It converges when |z| < 1. And it converges to the f(z) = 1
1− z2
. To see this note that for n = 3 we have
f3(z) =
3∏
n=1
= (1+ z2) · (1+ z4) · (1+ z8)
= (1+ z2 + z4 + z6) · (1+ z8)
= 1+ z2 + z4 + z6 + z8
So for n → ∞ we have fn(z) = 1+ z2 + z4 + z6 + · · ·+ ad inf =
1
1− z2
3. Convergence follows from |e−xxs−1| = e−xxRe(s)−1. Thus, it suffices to show Γ(s) is holomomorphic in
S�,R = {s ∈ C : � < Re(s) < R} for all 0 < � < R. For all 0 < δ < 1, let
Fδ(s) =
∫ 1/δ
δ
e−x · xsdx
x
Since e−xxs−1 is holomorphic in the half-plane Re(s) > 0 for all fixed x, we know Fδ is holomorphic S�,R.
Write
|Γ(s)− Fδ(s)| ≤
∫ δ
0
e−xxRe(s)
dx
x
The left integral is bounded by δR, so converges uniformly to 0 as δ → 0. We bound the right integral by
A
∫∞
1/δ e
−x/2 dx, which also converges uniformly to 0. Therefore, Fδ → Γ as δ → 0,...
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