5) Ex 9.5 from Lecture Notes determine which of the Using following functions are entire. Justify your answer: i) cos 1214 iv) Rez (ii) z³²_-4iz ³³ + (1+8i) z + (17 + i√z) iii) z Rez v) Imz vi) i Rez

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Chapter2: Second-order Linear Odes
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Questions: 5(ii) 5(v) 5(vi) Written out on paper please
(61
Pf EXERCISE
Ex 9.5: Prove: if f(z) is analytic and real on a domain ✓
then I is a constant function..
Pf: & is analytic and real on √✓ => f(x, y) = u(x, y ) +i. 0,
namely v(x₂ y) = 0₁ CR equations hold for ZEN
every
I from analyticity on №}:
=0 and 24 = -3x
DV=0
04 OV
dy
24
• Du = 0 => M (x₁y) = CER =) f(x,y) = c, namely f is a
constant function.
四
Prop 7.2: If f is analytic on a domain I and f'(Z) =0 for
all ZEN, then & is constant on 2.
Rmk: Prop 7.2 is false if I is not connected- & can
take different constants (constant values) on each connected.
component of R.
exists
Pf. The version of this proposition for real functions is proved.
using the Mean Value Theorem, that asserts that under
appropriate continuity assumption for f: (a, b) → IR there
ce(a, b) sit. f(b)-f(a) = f'(c) (b-a). So, if f'(x) = 0
for all XE (2₂6), then f(b) = f(a). The problem in the complex case
is that there is no analogous idea
in I, since there are many routes from a to b.
However, we can use the Mean Value Theorem for
Ref and I'm & as followS!
and b
b
Since f'lz)=0 for all ZEN by CR equations
0= f'(z) = 0x tiax
24
for all ZEN
Du
RV
DV
by ² ox = by = 0
Y
DU
-DX
Claim:
u is
H: We have
= 0 for all (x, y) = √.
u: R → R,
Pick (xo, yo), (x₁, y₁) ER. I is connected, thus by Def 4.2
any I points in I can be connected by a path that is
entirely in №. We join (Xo, yo) and (x₁, y₁) by a path
comprised of pieces parallel to x-axis or y-axis [One needs.
to show that such path exists!] Since ou =0 by the Mean Value
Theorem, u is constant on the horizontal pieces. In the same way
of "between a
4
for all ZEN.
a constant function.
04
DX
=
Transcribed Image Text:(61 Pf EXERCISE Ex 9.5: Prove: if f(z) is analytic and real on a domain ✓ then I is a constant function.. Pf: & is analytic and real on √✓ => f(x, y) = u(x, y ) +i. 0, namely v(x₂ y) = 0₁ CR equations hold for ZEN every I from analyticity on №}: =0 and 24 = -3x DV=0 04 OV dy 24 • Du = 0 => M (x₁y) = CER =) f(x,y) = c, namely f is a constant function. 四 Prop 7.2: If f is analytic on a domain I and f'(Z) =0 for all ZEN, then & is constant on 2. Rmk: Prop 7.2 is false if I is not connected- & can take different constants (constant values) on each connected. component of R. exists Pf. The version of this proposition for real functions is proved. using the Mean Value Theorem, that asserts that under appropriate continuity assumption for f: (a, b) → IR there ce(a, b) sit. f(b)-f(a) = f'(c) (b-a). So, if f'(x) = 0 for all XE (2₂6), then f(b) = f(a). The problem in the complex case is that there is no analogous idea in I, since there are many routes from a to b. However, we can use the Mean Value Theorem for Ref and I'm & as followS! and b b Since f'lz)=0 for all ZEN by CR equations 0= f'(z) = 0x tiax 24 for all ZEN Du RV DV by ² ox = by = 0 Y DU -DX Claim: u is H: We have = 0 for all (x, y) = √. u: R → R, Pick (xo, yo), (x₁, y₁) ER. I is connected, thus by Def 4.2 any I points in I can be connected by a path that is entirely in №. We join (Xo, yo) and (x₁, y₁) by a path comprised of pieces parallel to x-axis or y-axis [One needs. to show that such path exists!] Since ou =0 by the Mean Value Theorem, u is constant on the horizontal pieces. In the same way of "between a 4 for all ZEN. a constant function. 04 DX =
5
Using
Ex 9.5 from Lecture Notes determine which of the
following functions are entire. Justify your answer:
i) cos 1214
iv) Rez
ii) z²_4₁7³ + (1+8i) z + (17 + i√z)
3
v) Imz
iii) z Rez
vi) i Rez
Transcribed Image Text:5 Using Ex 9.5 from Lecture Notes determine which of the following functions are entire. Justify your answer: i) cos 1214 iv) Rez ii) z²_4₁7³ + (1+8i) z + (17 + i√z) 3 v) Imz iii) z Rez vi) i Rez
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