Solutions manual for complex analysis quiz 2024 2nd edition update/

IV 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
1 X X X X X X X X X X X
2 X X X X X X X X
3 X X X X X X X
4 X X X X
5 X X X X
6 X X X X
7 X X X X X X X X X X X
8 X X X X X X X X




1

,I.1.1
Identifynandnsketchnthensetnofnpointsnsatisfying.
(a) |xn—n 1n—n i|n =n 1 (f) 0n <n Fmnxn <n v
(b)n 1n <n|2xn—n 6|n <n2 (g) —vn <nRenxn <nv
(c) |xn—n 1|2n +n|xn+n1|2n <n8n (h)n |Renx|n <n|x|
(d)n |xn—n 1|n +n|xn+n1|n ≤n 2 (i) Ren(ixn+n2)n >n0
(e) |xn—n 1|n <n|x| (j) |xn—n i|2n +n|xn+ni|2n <n2
Solution
Letn xn =n xn+niy,n wheren x,ny ∈ R.
(a) Circle,ncentren1n+ni,nradiusn1.


|xn—n 1n—n i|n =n1nen |(xn—n 1)n+nin(yn—n 1)|n =n1nen (xn—n 1)n +2n(yn—n 1)n =2n1 2


(b) Annulusnwithncentren3,ninnernradiusn1/2,nouternradiusn1.


1n<n|2xn—n 6|n <n2nen 1n<n2n|xn—n 3|n <n2ne
2 2 2 2
en 1/2n<n|xn—n 3|n <n1nen (1/2) <n(xn—n 3)n +nyn <n1
√
(c) Disk,ncentren0,nradiusn 3.

2
2
|xn+niyn—n 1|n +n|xn+niyn+n1|n <n8ne
2 2 2 2 2 √nnnn 2
2
en(xn—n1)nn+ny +n(xn+n1)n +ny <n8nenxn +nyn < 3

(d) Intervaln [—1,n1].

q q
|xn—n 1|n +n|xn+n1|n ≤n 2nen (xn—n 1)2n+ny2n ≤n 2n—n (xn+n1)2n+ny2n e
2 2
nn q nnnnnn q nnnnn


enn (xn—n1)2n+ny2 ≤ 2n—n (xn+n1) n+ny
2 2 e
2
q nn q nnnnn 2

en (xn+n1)2n+ny2n ≤nxn+n1nenn (xn+n1)2n+ny2 ≤n (xn+n1)n enyn=n0
Now,ntakenyn=n0n inntheninequality,nandncomputenthenthreenintervals

2

, xn<n—1, thenn |xn—n 1|n +n|xn+n1|n =n —n(xn—n 1)n—n (xn+n1)n =n —2xn≥n 2,
—1n≤n xn≤n 1n thenn |xn—n 1|n +n|xn+n1|n n =n —n(xn—n 1)n+n(xn+n1) =n 2n≤n2
xn>n1, thenn |xn—n 1|n +n|xn+n1|n n =n (xn—n 1)n+n(xn+n1) =n 2xn≥n 2.

(e) Half–planenxn>n1/2.

2
2
|xn—n 1|n <n|x|n en |xn—n 1|n <n|x| en |xn+niyn—n 1| 2 <n|xn+niy|2n e
en(xn—n1) 2 +ny2n <nx2n+ny2 enxn>n1/2

(f) Horizontaln strip,n 0n <nyn <nv.
(g) Verticalnstrip,n—vn<nxn<nv.
(h) s\R.

|Renx|n <n|x|n en |Ren(xn+niy)|n 2<n|xn+niy|n 2en xn <2nxn +2nyn e2n |y|n >n0
(i) Halfnplanenyn<n2.

Ren(ixn+n2)n>n0nen Ren(in(xn+niy)n+n2)n>n0nen —yn+n2n>n0nen yn<n2
(j) Emptynset.


|xn—n i| 2 +n|xn+ni|2 <n2nen |xn+niyn—n i|2 +n|xn+niyn+ni|2n <n2ne
en xn 2+n(yn—n 1)2 +nx2n+n(yn+n1)2 <n2nenx 2 +ny2n <n0




3

, I.1.4
Verifynfromnthendefinitionsneachnofnthenidentities 2
(a)nnn xn+nun =n x̄ n+nū (b)nnn xun =n x̄ū (c)nnn |x̄|n =n |x| (d)nnn |x|nnn=n xx̄
Drawnsketchesntonillustraten(a)nandn(c).


Solution
Substituten xn =n xn+niyn andn un =n un+niv,n andn usen then definitions.n(a)


xn+nun=n(xn+niy)n+n(un+niv)n=n(xn+nu)n+n(yn+nv)nin=
=n (xn+nu)n—n(yn+nv)nin =n (xn—niy)n+n(un—niv)n =n x̄ n+nū.

(b)


xun =n (xn+niy)n(un+niv)n =n (xun—n yv)n+n(xvn+nyu)nin =
=n (xun—nyv)n—n(xvn+nyu)nin =n (xn—niy)n(un—niv)n =n x̄ū.

(c)

q √
. .
|x¯|n =n xn+niyn =n|xn—n iy|n =n x2n +n(—y)2n= x2n +ny2n =n |xn+niy|n =n |x|n.

(d)

n √ n
2
|x|2 =n|xn+niy|2 = x2 +ny2 =nx2n+ny2n=
=n x2n—ni2 y 2n =n (xn+niy)n(xn—niy)n =n xx̄.




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