Proving a function is constant, if a sum of the real and imaginary parts is bounded
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Let $g$ be an entire function such that there exist three
real numbers $a$, $b$, $c$ such that $a$ and $b$ are not both $0$ and such that
$$a operatorname{Re}(g(z)) + b operatorname{Im}(g(z)) ≤ c$$
for all $ z ∈ mathbb{C}$.
Show that $g$ is constant.
Solving simpler versions of this problem involves defining a new function, say $h(z)$, that is some (often exponential) function we can find to be bounded. What do we do in our case?
complex-analysis
edited Nov 22 at 18:42
egreg
176k1484198
asked Nov 22 at 18:31
Dino
836
add a comment |
up vote
0
down vote
favorite
Let $g$ be an entire function such that there exist three
real numbers $a$, $b$, $c$ such that $a$ and $b$ are not both $0$ and such that
$$a operatorname{Re}(g(z)) + b operatorname{Im}(g(z)) ≤ c$$
for all $ z ∈ mathbb{C}$.
Show that $g$ is constant.
Solving simpler versions of this problem involves defining a new function, say $h(z)$, that is some (often exponential) function we can find to be bounded. What do we do in our case?
complex-analysis
edited Nov 22 at 18:42
egreg
176k1484198
asked Nov 22 at 18:31
Dino
836
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $g$ be an entire function such that there exist three
real numbers $a$, $b$, $c$ such that $a$ and $b$ are not both $0$ and such that
$$a operatorname{Re}(g(z)) + b operatorname{Im}(g(z)) ≤ c$$
for all $ z ∈ mathbb{C}$.
Show that $g$ is constant.
Solving simpler versions of this problem involves defining a new function, say $h(z)$, that is some (often exponential) function we can find to be bounded. What do we do in our case?
complex-analysis
edited Nov 22 at 18:42
egreg
176k1484198
asked Nov 22 at 18:31
Dino
836
Let $g$ be an entire function such that there exist three
real numbers $a$, $b$, $c$ such that $a$ and $b$ are not both $0$ and such that
$$a operatorname{Re}(g(z)) + b operatorname{Im}(g(z)) ≤ c$$
for all $ z ∈ mathbb{C}$.
Show that $g$ is constant.
Solving simpler versions of this problem involves defining a new function, say $h(z)$, that is some (often exponential) function we can find to be bounded. What do we do in our case?
complex-analysis
complex-analysis
edited Nov 22 at 18:42
egreg
176k1484198
asked Nov 22 at 18:31
Dino
836
edited Nov 22 at 18:42
egreg
176k1484198
asked Nov 22 at 18:31
Dino
836
edited Nov 22 at 18:42
egreg
176k1484198
edited Nov 22 at 18:42
egreg
176k1484198
edited Nov 22 at 18:42
egreg
176k1484198
176k1484198
asked Nov 22 at 18:31
Dino
836
asked Nov 22 at 18:31
Dino
836
asked Nov 22 at 18:31
Dino
836
836
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?
answered Nov 22 at 18:39
José Carlos Santos
147k22117217
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?
answered Nov 22 at 18:39
José Carlos Santos
147k22117217
add a comment |
up vote
2
down vote
accepted
Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?
answered Nov 22 at 18:39
José Carlos Santos
147k22117217
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?
answered Nov 22 at 18:39
José Carlos Santos
147k22117217
Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?
answered Nov 22 at 18:39
José Carlos Santos
147k22117217
answered Nov 22 at 18:39
José Carlos Santos
147k22117217
answered Nov 22 at 18:39
José Carlos Santos
147k22117217
answered Nov 22 at 18:39
José Carlos Santos
147k22117217
147k22117217
add a comment |
add a comment |
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