Equivalence relation with complex numbers.












0














In $ mathbb{C} $ we define the binary relations $ R_{j} $ , $ j=1,2,3,4. $



$ z_{1} R_{1} z_{2} Leftrightarrow |z_{1}| = |z_{2}| $



$ z_{1} R_{2} z_{2} Leftrightarrow arg(z_{1}) = arg(z_{2}) $ or $ z_{1} = z_{2} = 0 $



$ z_{1} R_{3} z_{2} Leftrightarrow bar{z_{1}} = z_{2} $



$ z_{1} R_{4} z_{2} Leftrightarrow z_{1} = e^{iphi}z_{2} $ , $ phi in mathbb{R} $



Let X = { z | |z| = 1 }



Find $ R_{j}(mathbb{R}), R_{j}(mathbb{X}), R_{j}(i),R_{j}(mathbb{R}),R_{j}circ R_{k},R_{j}^{-1} $



Find which relation is an equivalence relation and in that case determine it's quotient set.



What is the elegant and right way to find those sets ?
They might seem easy to guess but I am not that familiar with the demonstration.



Some answers : $ R_{1}(mathbb{R}) = mathbb{C} $ , $ R_{2}(mathbb{R}) = mathbb{R} $ , $ R_{3}(mathbb{R}) = mathbb{R} $ , $ R_{4}(mathbb{R}) $ = { $ (a,b) in mathbb{R}^2 | b= atanphi $ } , $ R_{1}(X) = X $, $ R_{2}(X) = mathbb{C}^* $, $ R_{3}(X) = X $, $ R_{4}(X) = X $, $ R_{1}(i) = X $



And the only equivalent relations are the first 2 ( which is pretty obvious because they are Kernels )






complex-numbers relations equivalence-relations







share|cite|improve this question
























  • What’s wrong with the fourth one being an equivalence relation? Or wait is that supposed to be $e^{iphi}$ or $e^{jphi}$?
    – snulty
    Nov 28 '18 at 19:51












  • It's to to i-th power, redactation error it seems :s
    – SADBOYS
    Nov 28 '18 at 19:54










  • So for $R_4$ is it the case that $|z_1|=|z_2|$ if $z_1 R_4 z_2$?
    – snulty
    Nov 28 '18 at 20:03
















0














In $ mathbb{C} $ we define the binary relations $ R_{j} $ , $ j=1,2,3,4. $



$ z_{1} R_{1} z_{2} Leftrightarrow |z_{1}| = |z_{2}| $



$ z_{1} R_{2} z_{2} Leftrightarrow arg(z_{1}) = arg(z_{2}) $ or $ z_{1} = z_{2} = 0 $



$ z_{1} R_{3} z_{2} Leftrightarrow bar{z_{1}} = z_{2} $



$ z_{1} R_{4} z_{2} Leftrightarrow z_{1} = e^{iphi}z_{2} $ , $ phi in mathbb{R} $



Let X = { z | |z| = 1 }



Find $ R_{j}(mathbb{R}), R_{j}(mathbb{X}), R_{j}(i),R_{j}(mathbb{R}),R_{j}circ R_{k},R_{j}^{-1} $



Find which relation is an equivalence relation and in that case determine it's quotient set.



What is the elegant and right way to find those sets ?
They might seem easy to guess but I am not that familiar with the demonstration.



Some answers : $ R_{1}(mathbb{R}) = mathbb{C} $ , $ R_{2}(mathbb{R}) = mathbb{R} $ , $ R_{3}(mathbb{R}) = mathbb{R} $ , $ R_{4}(mathbb{R}) $ = { $ (a,b) in mathbb{R}^2 | b= atanphi $ } , $ R_{1}(X) = X $, $ R_{2}(X) = mathbb{C}^* $, $ R_{3}(X) = X $, $ R_{4}(X) = X $, $ R_{1}(i) = X $



And the only equivalent relations are the first 2 ( which is pretty obvious because they are Kernels )






complex-numbers relations equivalence-relations







share|cite|improve this question
























  • What’s wrong with the fourth one being an equivalence relation? Or wait is that supposed to be $e^{iphi}$ or $e^{jphi}$?
    – snulty
    Nov 28 '18 at 19:51












  • It's to to i-th power, redactation error it seems :s
    – SADBOYS
    Nov 28 '18 at 19:54










  • So for $R_4$ is it the case that $|z_1|=|z_2|$ if $z_1 R_4 z_2$?
    – snulty
    Nov 28 '18 at 20:03














0












0








0


1





In $ mathbb{C} $ we define the binary relations $ R_{j} $ , $ j=1,2,3,4. $



$ z_{1} R_{1} z_{2} Leftrightarrow |z_{1}| = |z_{2}| $



$ z_{1} R_{2} z_{2} Leftrightarrow arg(z_{1}) = arg(z_{2}) $ or $ z_{1} = z_{2} = 0 $



$ z_{1} R_{3} z_{2} Leftrightarrow bar{z_{1}} = z_{2} $



$ z_{1} R_{4} z_{2} Leftrightarrow z_{1} = e^{iphi}z_{2} $ , $ phi in mathbb{R} $



Let X = { z | |z| = 1 }



Find $ R_{j}(mathbb{R}), R_{j}(mathbb{X}), R_{j}(i),R_{j}(mathbb{R}),R_{j}circ R_{k},R_{j}^{-1} $



Find which relation is an equivalence relation and in that case determine it's quotient set.



What is the elegant and right way to find those sets ?
They might seem easy to guess but I am not that familiar with the demonstration.



Some answers : $ R_{1}(mathbb{R}) = mathbb{C} $ , $ R_{2}(mathbb{R}) = mathbb{R} $ , $ R_{3}(mathbb{R}) = mathbb{R} $ , $ R_{4}(mathbb{R}) $ = { $ (a,b) in mathbb{R}^2 | b= atanphi $ } , $ R_{1}(X) = X $, $ R_{2}(X) = mathbb{C}^* $, $ R_{3}(X) = X $, $ R_{4}(X) = X $, $ R_{1}(i) = X $



And the only equivalent relations are the first 2 ( which is pretty obvious because they are Kernels )






complex-numbers relations equivalence-relations







share|cite|improve this question















In $ mathbb{C} $ we define the binary relations $ R_{j} $ , $ j=1,2,3,4. $



$ z_{1} R_{1} z_{2} Leftrightarrow |z_{1}| = |z_{2}| $



$ z_{1} R_{2} z_{2} Leftrightarrow arg(z_{1}) = arg(z_{2}) $ or $ z_{1} = z_{2} = 0 $



$ z_{1} R_{3} z_{2} Leftrightarrow bar{z_{1}} = z_{2} $



$ z_{1} R_{4} z_{2} Leftrightarrow z_{1} = e^{iphi}z_{2} $ , $ phi in mathbb{R} $



Let X = { z | |z| = 1 }



Find $ R_{j}(mathbb{R}), R_{j}(mathbb{X}), R_{j}(i),R_{j}(mathbb{R}),R_{j}circ R_{k},R_{j}^{-1} $



Find which relation is an equivalence relation and in that case determine it's quotient set.



What is the elegant and right way to find those sets ?
They might seem easy to guess but I am not that familiar with the demonstration.



Some answers : $ R_{1}(mathbb{R}) = mathbb{C} $ , $ R_{2}(mathbb{R}) = mathbb{R} $ , $ R_{3}(mathbb{R}) = mathbb{R} $ , $ R_{4}(mathbb{R}) $ = { $ (a,b) in mathbb{R}^2 | b= atanphi $ } , $ R_{1}(X) = X $, $ R_{2}(X) = mathbb{C}^* $, $ R_{3}(X) = X $, $ R_{4}(X) = X $, $ R_{1}(i) = X $



And the only equivalent relations are the first 2 ( which is pretty obvious because they are Kernels )






complex-numbers relations equivalence-relations




complex-numbers relations equivalence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 19:54

























asked Nov 28 '18 at 19:46









SADBOYS

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4288












  • What’s wrong with the fourth one being an equivalence relation? Or wait is that supposed to be $e^{iphi}$ or $e^{jphi}$?
    – snulty
    Nov 28 '18 at 19:51












  • It's to to i-th power, redactation error it seems :s
    – SADBOYS
    Nov 28 '18 at 19:54










  • So for $R_4$ is it the case that $|z_1|=|z_2|$ if $z_1 R_4 z_2$?
    – snulty
    Nov 28 '18 at 20:03


















  • What’s wrong with the fourth one being an equivalence relation? Or wait is that supposed to be $e^{iphi}$ or $e^{jphi}$?
    – snulty
    Nov 28 '18 at 19:51












  • It's to to i-th power, redactation error it seems :s
    – SADBOYS
    Nov 28 '18 at 19:54










  • So for $R_4$ is it the case that $|z_1|=|z_2|$ if $z_1 R_4 z_2$?
    – snulty
    Nov 28 '18 at 20:03
















What’s wrong with the fourth one being an equivalence relation? Or wait is that supposed to be $e^{iphi}$ or $e^{jphi}$?
– snulty
Nov 28 '18 at 19:51






What’s wrong with the fourth one being an equivalence relation? Or wait is that supposed to be $e^{iphi}$ or $e^{jphi}$?
– snulty
Nov 28 '18 at 19:51














It's to to i-th power, redactation error it seems :s
– SADBOYS
Nov 28 '18 at 19:54




It's to to i-th power, redactation error it seems :s
– SADBOYS
Nov 28 '18 at 19:54












So for $R_4$ is it the case that $|z_1|=|z_2|$ if $z_1 R_4 z_2$?
– snulty
Nov 28 '18 at 20:03




So for $R_4$ is it the case that $|z_1|=|z_2|$ if $z_1 R_4 z_2$?
– snulty
Nov 28 '18 at 20:03










1 Answer
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oldest

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2














You should think of $R_j(S)$ (where $S$ is some subset of $Bbb{C}$) as the image of the set $S$ under the relation $R_j$. What it means is as follows:
$$R_j(S)={z in Bbb{C}, | , s R_j z ,,text{ for some } s in S}$$



For example, if $S=Bbb{R}$ and we are using relation $R_1$, then
begin{align*}
R_1(Bbb{R}) & ={z in Bbb{C}, | , s R_j z ,,text{ for some } s in Bbb{R}}\
& ={z in Bbb{C}, | , |s|=|z| ,,text{ for some } s in Bbb{R}}\
end{align*}

Since every complex number $z=a+ib$ has magnitude $|z|=sqrt{a^2+b^2}$, so we can say that the real number $sqrt{a^2+b^2}$ is related to $z$ via the relation $R_1$. Thus
$$R_1(Bbb{R})=Bbb{C}.$$



Hope you can take it from here.






share|cite|improve this answer























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    1 Answer
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    active

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    You should think of $R_j(S)$ (where $S$ is some subset of $Bbb{C}$) as the image of the set $S$ under the relation $R_j$. What it means is as follows:
    $$R_j(S)={z in Bbb{C}, | , s R_j z ,,text{ for some } s in S}$$



    For example, if $S=Bbb{R}$ and we are using relation $R_1$, then
    begin{align*}
    R_1(Bbb{R}) & ={z in Bbb{C}, | , s R_j z ,,text{ for some } s in Bbb{R}}\
    & ={z in Bbb{C}, | , |s|=|z| ,,text{ for some } s in Bbb{R}}\
    end{align*}

    Since every complex number $z=a+ib$ has magnitude $|z|=sqrt{a^2+b^2}$, so we can say that the real number $sqrt{a^2+b^2}$ is related to $z$ via the relation $R_1$. Thus
    $$R_1(Bbb{R})=Bbb{C}.$$



    Hope you can take it from here.






    share|cite|improve this answer




























      2














      You should think of $R_j(S)$ (where $S$ is some subset of $Bbb{C}$) as the image of the set $S$ under the relation $R_j$. What it means is as follows:
      $$R_j(S)={z in Bbb{C}, | , s R_j z ,,text{ for some } s in S}$$



      For example, if $S=Bbb{R}$ and we are using relation $R_1$, then
      begin{align*}
      R_1(Bbb{R}) & ={z in Bbb{C}, | , s R_j z ,,text{ for some } s in Bbb{R}}\
      & ={z in Bbb{C}, | , |s|=|z| ,,text{ for some } s in Bbb{R}}\
      end{align*}

      Since every complex number $z=a+ib$ has magnitude $|z|=sqrt{a^2+b^2}$, so we can say that the real number $sqrt{a^2+b^2}$ is related to $z$ via the relation $R_1$. Thus
      $$R_1(Bbb{R})=Bbb{C}.$$



      Hope you can take it from here.






      share|cite|improve this answer


























        2












        2








        2






        You should think of $R_j(S)$ (where $S$ is some subset of $Bbb{C}$) as the image of the set $S$ under the relation $R_j$. What it means is as follows:
        $$R_j(S)={z in Bbb{C}, | , s R_j z ,,text{ for some } s in S}$$



        For example, if $S=Bbb{R}$ and we are using relation $R_1$, then
        begin{align*}
        R_1(Bbb{R}) & ={z in Bbb{C}, | , s R_j z ,,text{ for some } s in Bbb{R}}\
        & ={z in Bbb{C}, | , |s|=|z| ,,text{ for some } s in Bbb{R}}\
        end{align*}

        Since every complex number $z=a+ib$ has magnitude $|z|=sqrt{a^2+b^2}$, so we can say that the real number $sqrt{a^2+b^2}$ is related to $z$ via the relation $R_1$. Thus
        $$R_1(Bbb{R})=Bbb{C}.$$



        Hope you can take it from here.






        share|cite|improve this answer














        You should think of $R_j(S)$ (where $S$ is some subset of $Bbb{C}$) as the image of the set $S$ under the relation $R_j$. What it means is as follows:
        $$R_j(S)={z in Bbb{C}, | , s R_j z ,,text{ for some } s in S}$$



        For example, if $S=Bbb{R}$ and we are using relation $R_1$, then
        begin{align*}
        R_1(Bbb{R}) & ={z in Bbb{C}, | , s R_j z ,,text{ for some } s in Bbb{R}}\
        & ={z in Bbb{C}, | , |s|=|z| ,,text{ for some } s in Bbb{R}}\
        end{align*}

        Since every complex number $z=a+ib$ has magnitude $|z|=sqrt{a^2+b^2}$, so we can say that the real number $sqrt{a^2+b^2}$ is related to $z$ via the relation $R_1$. Thus
        $$R_1(Bbb{R})=Bbb{C}.$$



        Hope you can take it from here.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 '18 at 3:56

























        answered Nov 28 '18 at 20:12









        Anurag A

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        25.6k12249






























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