Showing that for any complex number $z$, either $ |z + 1| geq 1/sqrt{2} $ or $ |z^2 + 1| geq 1 $












3














I was reading complex Number from A to Z
There was an example which asks to proof that for any complex Number $z$ either
$$ |z + 1| geq 1/sqrt{2} $$
or
$$ |z^2 + 1| geq 1 $$
In the book proof was by contradiction and I had no problem in understanding it, but what rise my problem is how someone can arrive to such a conclusion from nothing.










share|cite|improve this question
























  • please, edit your first equation.
    – Doyun Nam
    yesterday










  • It's a method full of logic (See: en.wikipedia.org/wiki/Proof_by_contradiction). In general, suppose we have to prove a statement $A$ is true implies another statement $B$ is true. Next, for the sake of argument assume $B$ be NOT true then we arrived that $A$ is false. So, "$B$ is false $implies$ $A$ is false"-is true, and so, we have $A$ is true $implies$ $B$ is true. Further, see: math.stackexchange.com/questions/227127/…
    – Sujit Bhattacharyya
    yesterday










  • Thanks for this. But i want to know how can someone without knowing the given inequality , derive them. Not prove them if they are given
    – Keshav Sharma
    yesterday


















3














I was reading complex Number from A to Z
There was an example which asks to proof that for any complex Number $z$ either
$$ |z + 1| geq 1/sqrt{2} $$
or
$$ |z^2 + 1| geq 1 $$
In the book proof was by contradiction and I had no problem in understanding it, but what rise my problem is how someone can arrive to such a conclusion from nothing.










share|cite|improve this question
























  • please, edit your first equation.
    – Doyun Nam
    yesterday










  • It's a method full of logic (See: en.wikipedia.org/wiki/Proof_by_contradiction). In general, suppose we have to prove a statement $A$ is true implies another statement $B$ is true. Next, for the sake of argument assume $B$ be NOT true then we arrived that $A$ is false. So, "$B$ is false $implies$ $A$ is false"-is true, and so, we have $A$ is true $implies$ $B$ is true. Further, see: math.stackexchange.com/questions/227127/…
    – Sujit Bhattacharyya
    yesterday










  • Thanks for this. But i want to know how can someone without knowing the given inequality , derive them. Not prove them if they are given
    – Keshav Sharma
    yesterday
















3












3








3


1





I was reading complex Number from A to Z
There was an example which asks to proof that for any complex Number $z$ either
$$ |z + 1| geq 1/sqrt{2} $$
or
$$ |z^2 + 1| geq 1 $$
In the book proof was by contradiction and I had no problem in understanding it, but what rise my problem is how someone can arrive to such a conclusion from nothing.










share|cite|improve this question















I was reading complex Number from A to Z
There was an example which asks to proof that for any complex Number $z$ either
$$ |z + 1| geq 1/sqrt{2} $$
or
$$ |z^2 + 1| geq 1 $$
In the book proof was by contradiction and I had no problem in understanding it, but what rise my problem is how someone can arrive to such a conclusion from nothing.







complex-numbers






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share|cite|improve this question








edited yesterday









Asaf Karagila

302k32426756




302k32426756










asked yesterday









Keshav Sharma

915




915












  • please, edit your first equation.
    – Doyun Nam
    yesterday










  • It's a method full of logic (See: en.wikipedia.org/wiki/Proof_by_contradiction). In general, suppose we have to prove a statement $A$ is true implies another statement $B$ is true. Next, for the sake of argument assume $B$ be NOT true then we arrived that $A$ is false. So, "$B$ is false $implies$ $A$ is false"-is true, and so, we have $A$ is true $implies$ $B$ is true. Further, see: math.stackexchange.com/questions/227127/…
    – Sujit Bhattacharyya
    yesterday










  • Thanks for this. But i want to know how can someone without knowing the given inequality , derive them. Not prove them if they are given
    – Keshav Sharma
    yesterday




















  • please, edit your first equation.
    – Doyun Nam
    yesterday










  • It's a method full of logic (See: en.wikipedia.org/wiki/Proof_by_contradiction). In general, suppose we have to prove a statement $A$ is true implies another statement $B$ is true. Next, for the sake of argument assume $B$ be NOT true then we arrived that $A$ is false. So, "$B$ is false $implies$ $A$ is false"-is true, and so, we have $A$ is true $implies$ $B$ is true. Further, see: math.stackexchange.com/questions/227127/…
    – Sujit Bhattacharyya
    yesterday










  • Thanks for this. But i want to know how can someone without knowing the given inequality , derive them. Not prove them if they are given
    – Keshav Sharma
    yesterday


















please, edit your first equation.
– Doyun Nam
yesterday




please, edit your first equation.
– Doyun Nam
yesterday












It's a method full of logic (See: en.wikipedia.org/wiki/Proof_by_contradiction). In general, suppose we have to prove a statement $A$ is true implies another statement $B$ is true. Next, for the sake of argument assume $B$ be NOT true then we arrived that $A$ is false. So, "$B$ is false $implies$ $A$ is false"-is true, and so, we have $A$ is true $implies$ $B$ is true. Further, see: math.stackexchange.com/questions/227127/…
– Sujit Bhattacharyya
yesterday




It's a method full of logic (See: en.wikipedia.org/wiki/Proof_by_contradiction). In general, suppose we have to prove a statement $A$ is true implies another statement $B$ is true. Next, for the sake of argument assume $B$ be NOT true then we arrived that $A$ is false. So, "$B$ is false $implies$ $A$ is false"-is true, and so, we have $A$ is true $implies$ $B$ is true. Further, see: math.stackexchange.com/questions/227127/…
– Sujit Bhattacharyya
yesterday












Thanks for this. But i want to know how can someone without knowing the given inequality , derive them. Not prove them if they are given
– Keshav Sharma
yesterday






Thanks for this. But i want to know how can someone without knowing the given inequality , derive them. Not prove them if they are given
– Keshav Sharma
yesterday












2 Answers
2






active

oldest

votes


















4














Let us think of complex plane. I used the program, GeoGebra to draw the figure below.



Let $A = 0$, $B = -0.5+0.5i$, $C = -0.5 -0.5 i$, and $D=-1$.



Your first inequality is true if and only if $z$ is on the circle or outside the circle.





Thus it is enough to show that if $z$ is in the circle, then that $z$ satisfies the second inequality.



The circle is between the half-line $overrightarrow{AB}$ and $overrightarrow{AC}$.



The angle of $overrightarrow{AB}$ is $frac{3pi}{4}$, and the angle of
$overrightarrow{AC}$ is $frac{5pi}{4}$.



Generally, assume that $z$ is a complex number between these two half-line.



We know that the angle of $z^2$ is twice of the angle of $z$.



Thus the angle of $z^2$ is between $frac{3pi}{2}$ and $frac{5pi}{2}$.



It means that the real part of $z^2$ is non-negative.



Thus, the real part of $z^2 +1$ is greater than or equal to $1$.



Because $|z^2 + 1| geq |Re(z^2 +1)|$, it follows that $|z^2 + 1| geq 1$. Q.E.D.



figure in complex plane



I think that this geometric proof let us be able to make similar problems.



For example, let the second inequality be the same. And let the first inequality change as follows:



$$ |z+r| leq frac{r}{sqrt{2}} , $$



for any fixed $r>0$.



Then we can say that every complex number $z$ satisfies the (changed) first or second inequality.






share|cite|improve this answer































    3














    Geometrically, looking at the equality case, the first equation corresponds to a circle centered at $-1$ with radius $1/sqrt{2}$, whereas the second is the locus of points whose product of the distances from $i$ and $-i$ is equal to $1$. The second locus of points is called the Lemiscate of Bernoulli. The thesis is equivalent to saying that the Bernoulli Lemniscate and the circle do not intersect.



    Rotating the frame of reference the lemniscate is given by the equation https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli:



    $(x^2+y^2)^2=2(x^2-y^2) [1]$



    and the rotated circle:



    $x^2+(y-1)^2=r^2 [2]$



    The optimal $r$ is given when the circle and the lemniscate are tangent, i.e. the point of intersections are only 2. Substituting $x^2$ from [2] into [1] and putting the determinant to zero I get $sqrt{2-sqrt{2}} sim 0.76$ as optimal r, which is greater than $1/sqrt{2}sim 0.70$ the one proposed and proves that the circle and the leminscate do not intersect.






    share|cite|improve this answer























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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

      oldest

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      4














      Let us think of complex plane. I used the program, GeoGebra to draw the figure below.



      Let $A = 0$, $B = -0.5+0.5i$, $C = -0.5 -0.5 i$, and $D=-1$.



      Your first inequality is true if and only if $z$ is on the circle or outside the circle.





      Thus it is enough to show that if $z$ is in the circle, then that $z$ satisfies the second inequality.



      The circle is between the half-line $overrightarrow{AB}$ and $overrightarrow{AC}$.



      The angle of $overrightarrow{AB}$ is $frac{3pi}{4}$, and the angle of
      $overrightarrow{AC}$ is $frac{5pi}{4}$.



      Generally, assume that $z$ is a complex number between these two half-line.



      We know that the angle of $z^2$ is twice of the angle of $z$.



      Thus the angle of $z^2$ is between $frac{3pi}{2}$ and $frac{5pi}{2}$.



      It means that the real part of $z^2$ is non-negative.



      Thus, the real part of $z^2 +1$ is greater than or equal to $1$.



      Because $|z^2 + 1| geq |Re(z^2 +1)|$, it follows that $|z^2 + 1| geq 1$. Q.E.D.



      figure in complex plane



      I think that this geometric proof let us be able to make similar problems.



      For example, let the second inequality be the same. And let the first inequality change as follows:



      $$ |z+r| leq frac{r}{sqrt{2}} , $$



      for any fixed $r>0$.



      Then we can say that every complex number $z$ satisfies the (changed) first or second inequality.






      share|cite|improve this answer




























        4














        Let us think of complex plane. I used the program, GeoGebra to draw the figure below.



        Let $A = 0$, $B = -0.5+0.5i$, $C = -0.5 -0.5 i$, and $D=-1$.



        Your first inequality is true if and only if $z$ is on the circle or outside the circle.





        Thus it is enough to show that if $z$ is in the circle, then that $z$ satisfies the second inequality.



        The circle is between the half-line $overrightarrow{AB}$ and $overrightarrow{AC}$.



        The angle of $overrightarrow{AB}$ is $frac{3pi}{4}$, and the angle of
        $overrightarrow{AC}$ is $frac{5pi}{4}$.



        Generally, assume that $z$ is a complex number between these two half-line.



        We know that the angle of $z^2$ is twice of the angle of $z$.



        Thus the angle of $z^2$ is between $frac{3pi}{2}$ and $frac{5pi}{2}$.



        It means that the real part of $z^2$ is non-negative.



        Thus, the real part of $z^2 +1$ is greater than or equal to $1$.



        Because $|z^2 + 1| geq |Re(z^2 +1)|$, it follows that $|z^2 + 1| geq 1$. Q.E.D.



        figure in complex plane



        I think that this geometric proof let us be able to make similar problems.



        For example, let the second inequality be the same. And let the first inequality change as follows:



        $$ |z+r| leq frac{r}{sqrt{2}} , $$



        for any fixed $r>0$.



        Then we can say that every complex number $z$ satisfies the (changed) first or second inequality.






        share|cite|improve this answer


























          4












          4








          4






          Let us think of complex plane. I used the program, GeoGebra to draw the figure below.



          Let $A = 0$, $B = -0.5+0.5i$, $C = -0.5 -0.5 i$, and $D=-1$.



          Your first inequality is true if and only if $z$ is on the circle or outside the circle.





          Thus it is enough to show that if $z$ is in the circle, then that $z$ satisfies the second inequality.



          The circle is between the half-line $overrightarrow{AB}$ and $overrightarrow{AC}$.



          The angle of $overrightarrow{AB}$ is $frac{3pi}{4}$, and the angle of
          $overrightarrow{AC}$ is $frac{5pi}{4}$.



          Generally, assume that $z$ is a complex number between these two half-line.



          We know that the angle of $z^2$ is twice of the angle of $z$.



          Thus the angle of $z^2$ is between $frac{3pi}{2}$ and $frac{5pi}{2}$.



          It means that the real part of $z^2$ is non-negative.



          Thus, the real part of $z^2 +1$ is greater than or equal to $1$.



          Because $|z^2 + 1| geq |Re(z^2 +1)|$, it follows that $|z^2 + 1| geq 1$. Q.E.D.



          figure in complex plane



          I think that this geometric proof let us be able to make similar problems.



          For example, let the second inequality be the same. And let the first inequality change as follows:



          $$ |z+r| leq frac{r}{sqrt{2}} , $$



          for any fixed $r>0$.



          Then we can say that every complex number $z$ satisfies the (changed) first or second inequality.






          share|cite|improve this answer














          Let us think of complex plane. I used the program, GeoGebra to draw the figure below.



          Let $A = 0$, $B = -0.5+0.5i$, $C = -0.5 -0.5 i$, and $D=-1$.



          Your first inequality is true if and only if $z$ is on the circle or outside the circle.





          Thus it is enough to show that if $z$ is in the circle, then that $z$ satisfies the second inequality.



          The circle is between the half-line $overrightarrow{AB}$ and $overrightarrow{AC}$.



          The angle of $overrightarrow{AB}$ is $frac{3pi}{4}$, and the angle of
          $overrightarrow{AC}$ is $frac{5pi}{4}$.



          Generally, assume that $z$ is a complex number between these two half-line.



          We know that the angle of $z^2$ is twice of the angle of $z$.



          Thus the angle of $z^2$ is between $frac{3pi}{2}$ and $frac{5pi}{2}$.



          It means that the real part of $z^2$ is non-negative.



          Thus, the real part of $z^2 +1$ is greater than or equal to $1$.



          Because $|z^2 + 1| geq |Re(z^2 +1)|$, it follows that $|z^2 + 1| geq 1$. Q.E.D.



          figure in complex plane



          I think that this geometric proof let us be able to make similar problems.



          For example, let the second inequality be the same. And let the first inequality change as follows:



          $$ |z+r| leq frac{r}{sqrt{2}} , $$



          for any fixed $r>0$.



          Then we can say that every complex number $z$ satisfies the (changed) first or second inequality.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          Doyun Nam

          4169




          4169























              3














              Geometrically, looking at the equality case, the first equation corresponds to a circle centered at $-1$ with radius $1/sqrt{2}$, whereas the second is the locus of points whose product of the distances from $i$ and $-i$ is equal to $1$. The second locus of points is called the Lemiscate of Bernoulli. The thesis is equivalent to saying that the Bernoulli Lemniscate and the circle do not intersect.



              Rotating the frame of reference the lemniscate is given by the equation https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli:



              $(x^2+y^2)^2=2(x^2-y^2) [1]$



              and the rotated circle:



              $x^2+(y-1)^2=r^2 [2]$



              The optimal $r$ is given when the circle and the lemniscate are tangent, i.e. the point of intersections are only 2. Substituting $x^2$ from [2] into [1] and putting the determinant to zero I get $sqrt{2-sqrt{2}} sim 0.76$ as optimal r, which is greater than $1/sqrt{2}sim 0.70$ the one proposed and proves that the circle and the leminscate do not intersect.






              share|cite|improve this answer




























                3














                Geometrically, looking at the equality case, the first equation corresponds to a circle centered at $-1$ with radius $1/sqrt{2}$, whereas the second is the locus of points whose product of the distances from $i$ and $-i$ is equal to $1$. The second locus of points is called the Lemiscate of Bernoulli. The thesis is equivalent to saying that the Bernoulli Lemniscate and the circle do not intersect.



                Rotating the frame of reference the lemniscate is given by the equation https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli:



                $(x^2+y^2)^2=2(x^2-y^2) [1]$



                and the rotated circle:



                $x^2+(y-1)^2=r^2 [2]$



                The optimal $r$ is given when the circle and the lemniscate are tangent, i.e. the point of intersections are only 2. Substituting $x^2$ from [2] into [1] and putting the determinant to zero I get $sqrt{2-sqrt{2}} sim 0.76$ as optimal r, which is greater than $1/sqrt{2}sim 0.70$ the one proposed and proves that the circle and the leminscate do not intersect.






                share|cite|improve this answer


























                  3












                  3








                  3






                  Geometrically, looking at the equality case, the first equation corresponds to a circle centered at $-1$ with radius $1/sqrt{2}$, whereas the second is the locus of points whose product of the distances from $i$ and $-i$ is equal to $1$. The second locus of points is called the Lemiscate of Bernoulli. The thesis is equivalent to saying that the Bernoulli Lemniscate and the circle do not intersect.



                  Rotating the frame of reference the lemniscate is given by the equation https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli:



                  $(x^2+y^2)^2=2(x^2-y^2) [1]$



                  and the rotated circle:



                  $x^2+(y-1)^2=r^2 [2]$



                  The optimal $r$ is given when the circle and the lemniscate are tangent, i.e. the point of intersections are only 2. Substituting $x^2$ from [2] into [1] and putting the determinant to zero I get $sqrt{2-sqrt{2}} sim 0.76$ as optimal r, which is greater than $1/sqrt{2}sim 0.70$ the one proposed and proves that the circle and the leminscate do not intersect.






                  share|cite|improve this answer














                  Geometrically, looking at the equality case, the first equation corresponds to a circle centered at $-1$ with radius $1/sqrt{2}$, whereas the second is the locus of points whose product of the distances from $i$ and $-i$ is equal to $1$. The second locus of points is called the Lemiscate of Bernoulli. The thesis is equivalent to saying that the Bernoulli Lemniscate and the circle do not intersect.



                  Rotating the frame of reference the lemniscate is given by the equation https://en.wikipedia.org/wiki/Lemniscate_of_Bernoulli:



                  $(x^2+y^2)^2=2(x^2-y^2) [1]$



                  and the rotated circle:



                  $x^2+(y-1)^2=r^2 [2]$



                  The optimal $r$ is given when the circle and the lemniscate are tangent, i.e. the point of intersections are only 2. Substituting $x^2$ from [2] into [1] and putting the determinant to zero I get $sqrt{2-sqrt{2}} sim 0.76$ as optimal r, which is greater than $1/sqrt{2}sim 0.70$ the one proposed and proves that the circle and the leminscate do not intersect.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited yesterday

























                  answered yesterday









                  Thomas

                  16318




                  16318






























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