Open set is complex numbers
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I want to show that each of the inequalities $(1): Re(z)>0$ and $|z+z^2|<1$ defines an open set in $C$. I showed that $|z|<1$ is open. I tried to apply any inequality like triangle inequality but did not work.
I appreciate any help in this regard.
general-topology complex-analysis complex-numbers
asked Jan 17 at 16:41
DavidDavid
1016
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add a comment |
$begingroup$
I want to show that each of the inequalities $(1): Re(z)>0$ and $|z+z^2|<1$ defines an open set in $C$. I showed that $|z|<1$ is open. I tried to apply any inequality like triangle inequality but did not work.
I appreciate any help in this regard.
general-topology complex-analysis complex-numbers
asked Jan 17 at 16:41
DavidDavid
1016
$endgroup$
add a comment |
$begingroup$
I want to show that each of the inequalities $(1): Re(z)>0$ and $|z+z^2|<1$ defines an open set in $C$. I showed that $|z|<1$ is open. I tried to apply any inequality like triangle inequality but did not work.
I appreciate any help in this regard.
general-topology complex-analysis complex-numbers
asked Jan 17 at 16:41
DavidDavid
1016
$endgroup$
I want to show that each of the inequalities $(1): Re(z)>0$ and $|z+z^2|<1$ defines an open set in $C$. I showed that $|z|<1$ is open. I tried to apply any inequality like triangle inequality but did not work.
I appreciate any help in this regard.
general-topology complex-analysis complex-numbers
general-topology complex-analysis complex-numbers
asked Jan 17 at 16:41
DavidDavid
1016
asked Jan 17 at 16:41
DavidDavid
1016
asked Jan 17 at 16:41
DavidDavid
1016
asked Jan 17 at 16:41
DavidDavid
1016
asked Jan 17 at 16:41
DavidDavid
1016
1016
add a comment |
add a comment |
1 Answer
1
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For every continuous function $etacolonmathbb{C}longrightarrowmathbb R$ and every open subset $A$ of $mathbb R$, $eta^{-1}(A)$ is an open subset of $mathbb C$. So, consider the functions $operatorname{Re}$ and $f(z)=lvert z+z^2rvert$.
answered Jan 17 at 16:45
José Carlos SantosJosé Carlos Santos
175k23134243
$endgroup$
$begingroup$
Thank you for your help. It is a well known statement in topology. So, since Re $z > 0$ and $|z+z^2|<1$ are open sets in $R$, the functions define an open set in $C$?
$endgroup$
– David
Jan 17 at 16:58
1
$begingroup$
The sets $(0,infty)$ and $(-infty,1)$ are open sets. Therefore, $operatorname{Re}^{-1}bigl((0,infty)bigr)$ and $f^{-1}bigl((-infty,1)bigr)$ are open subsets of $mathbb C$.
$endgroup$
– José Carlos Santos
Jan 17 at 17:00
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For every continuous function $etacolonmathbb{C}longrightarrowmathbb R$ and every open subset $A$ of $mathbb R$, $eta^{-1}(A)$ is an open subset of $mathbb C$. So, consider the functions $operatorname{Re}$ and $f(z)=lvert z+z^2rvert$.
answered Jan 17 at 16:45
José Carlos SantosJosé Carlos Santos
175k23134243
$endgroup$
$begingroup$
Thank you for your help. It is a well known statement in topology. So, since Re $z > 0$ and $|z+z^2|<1$ are open sets in $R$, the functions define an open set in $C$?
$endgroup$
– David
Jan 17 at 16:58
1
$begingroup$
The sets $(0,infty)$ and $(-infty,1)$ are open sets. Therefore, $operatorname{Re}^{-1}bigl((0,infty)bigr)$ and $f^{-1}bigl((-infty,1)bigr)$ are open subsets of $mathbb C$.
$endgroup$
– José Carlos Santos
Jan 17 at 17:00
add a comment |
$begingroup$
For every continuous function $etacolonmathbb{C}longrightarrowmathbb R$ and every open subset $A$ of $mathbb R$, $eta^{-1}(A)$ is an open subset of $mathbb C$. So, consider the functions $operatorname{Re}$ and $f(z)=lvert z+z^2rvert$.
answered Jan 17 at 16:45
José Carlos SantosJosé Carlos Santos
175k23134243
$endgroup$
$begingroup$
Thank you for your help. It is a well known statement in topology. So, since Re $z > 0$ and $|z+z^2|<1$ are open sets in $R$, the functions define an open set in $C$?
$endgroup$
– David
Jan 17 at 16:58
1
$begingroup$
The sets $(0,infty)$ and $(-infty,1)$ are open sets. Therefore, $operatorname{Re}^{-1}bigl((0,infty)bigr)$ and $f^{-1}bigl((-infty,1)bigr)$ are open subsets of $mathbb C$.
$endgroup$
– José Carlos Santos
Jan 17 at 17:00
add a comment |
$begingroup$
For every continuous function $etacolonmathbb{C}longrightarrowmathbb R$ and every open subset $A$ of $mathbb R$, $eta^{-1}(A)$ is an open subset of $mathbb C$. So, consider the functions $operatorname{Re}$ and $f(z)=lvert z+z^2rvert$.
answered Jan 17 at 16:45
José Carlos SantosJosé Carlos Santos
175k23134243
$endgroup$
For every continuous function $etacolonmathbb{C}longrightarrowmathbb R$ and every open subset $A$ of $mathbb R$, $eta^{-1}(A)$ is an open subset of $mathbb C$. So, consider the functions $operatorname{Re}$ and $f(z)=lvert z+z^2rvert$.
answered Jan 17 at 16:45
José Carlos SantosJosé Carlos Santos
175k23134243
answered Jan 17 at 16:45
José Carlos SantosJosé Carlos Santos
175k23134243
answered Jan 17 at 16:45
José Carlos SantosJosé Carlos Santos
175k23134243
answered Jan 17 at 16:45
José Carlos SantosJosé Carlos Santos
175k23134243
175k23134243
$begingroup$
Thank you for your help. It is a well known statement in topology. So, since Re $z > 0$ and $|z+z^2|<1$ are open sets in $R$, the functions define an open set in $C$?
$endgroup$
– David
Jan 17 at 16:58
1
$begingroup$
The sets $(0,infty)$ and $(-infty,1)$ are open sets. Therefore, $operatorname{Re}^{-1}bigl((0,infty)bigr)$ and $f^{-1}bigl((-infty,1)bigr)$ are open subsets of $mathbb C$.
$endgroup$
– José Carlos Santos
Jan 17 at 17:00
add a comment |
$begingroup$
Thank you for your help. It is a well known statement in topology. So, since Re $z > 0$ and $|z+z^2|<1$ are open sets in $R$, the functions define an open set in $C$?
$endgroup$
– David
Jan 17 at 16:58
1
$begingroup$
The sets $(0,infty)$ and $(-infty,1)$ are open sets. Therefore, $operatorname{Re}^{-1}bigl((0,infty)bigr)$ and $f^{-1}bigl((-infty,1)bigr)$ are open subsets of $mathbb C$.
$endgroup$
– José Carlos Santos
Jan 17 at 17:00
$begingroup$
Thank you for your help. It is a well known statement in topology. So, since Re $z > 0$ and $|z+z^2|<1$ are open sets in $R$, the functions define an open set in $C$?
$endgroup$
– David
Jan 17 at 16:58
$begingroup$
Thank you for your help. It is a well known statement in topology. So, since Re $z > 0$ and $|z+z^2|<1$ are open sets in $R$, the functions define an open set in $C$?
$endgroup$
– David
Jan 17 at 16:58
1
1
$begingroup$
The sets $(0,infty)$ and $(-infty,1)$ are open sets. Therefore, $operatorname{Re}^{-1}bigl((0,infty)bigr)$ and $f^{-1}bigl((-infty,1)bigr)$ are open subsets of $mathbb C$.
$endgroup$
– José Carlos Santos
Jan 17 at 17:00
$begingroup$
The sets $(0,infty)$ and $(-infty,1)$ are open sets. Therefore, $operatorname{Re}^{-1}bigl((0,infty)bigr)$ and $f^{-1}bigl((-infty,1)bigr)$ are open subsets of $mathbb C$.
$endgroup$
– José Carlos Santos
Jan 17 at 17:00
add a comment |
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