Finding minimum value of the equation given.











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Question - Find the minimum value of $|1 + z| + |1-z|$.



I'm trying to solve the question by thinking of them as points in the Argand plane.
The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane.
But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.










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  • $|1-z| = |-(z-1)|=|z-1|$
    – krirkrirk
    Nov 16 at 12:42










  • @krirkrirk Thanks! It didn't click at the moment. Got the answer!
    – Kaustuv Sawarn
    Nov 16 at 12:47















up vote
0
down vote

favorite












Question - Find the minimum value of $|1 + z| + |1-z|$.



I'm trying to solve the question by thinking of them as points in the Argand plane.
The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane.
But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.










share|cite|improve this question
























  • $|1-z| = |-(z-1)|=|z-1|$
    – krirkrirk
    Nov 16 at 12:42










  • @krirkrirk Thanks! It didn't click at the moment. Got the answer!
    – Kaustuv Sawarn
    Nov 16 at 12:47













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Question - Find the minimum value of $|1 + z| + |1-z|$.



I'm trying to solve the question by thinking of them as points in the Argand plane.
The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane.
But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.










share|cite|improve this question















Question - Find the minimum value of $|1 + z| + |1-z|$.



I'm trying to solve the question by thinking of them as points in the Argand plane.
The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane.
But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.







complex-analysis complex-numbers






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edited Nov 17 at 17:45









Ted Shifrin

62.1k44489




62.1k44489










asked Nov 16 at 12:32









Kaustuv Sawarn

515




515












  • $|1-z| = |-(z-1)|=|z-1|$
    – krirkrirk
    Nov 16 at 12:42










  • @krirkrirk Thanks! It didn't click at the moment. Got the answer!
    – Kaustuv Sawarn
    Nov 16 at 12:47


















  • $|1-z| = |-(z-1)|=|z-1|$
    – krirkrirk
    Nov 16 at 12:42










  • @krirkrirk Thanks! It didn't click at the moment. Got the answer!
    – Kaustuv Sawarn
    Nov 16 at 12:47
















$|1-z| = |-(z-1)|=|z-1|$
– krirkrirk
Nov 16 at 12:42




$|1-z| = |-(z-1)|=|z-1|$
– krirkrirk
Nov 16 at 12:42












@krirkrirk Thanks! It didn't click at the moment. Got the answer!
– Kaustuv Sawarn
Nov 16 at 12:47




@krirkrirk Thanks! It didn't click at the moment. Got the answer!
– Kaustuv Sawarn
Nov 16 at 12:47










3 Answers
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1
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The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.






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    down vote













    $|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.



    BEcause the axis must be longer than the distance between focus, $cge 2$.






    share|cite|improve this answer




























      up vote
      1
      down vote













      A bit of geometry.



      In the complex plane:



      $A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.



      $triangle ABC$ has sides of lengths:



      $|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.



      The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:



      $|z+1|+|z-1| > 2.$



      $2$ is a lower bound. Is there a minimum?



      If yes , $z=?$






      share|cite|improve this answer























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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        1
        down vote



        accepted










        The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.






        share|cite|improve this answer

























          up vote
          1
          down vote



          accepted










          The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.






          share|cite|improve this answer























            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.






            share|cite|improve this answer












            The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 16 at 14:03









            i. m. soloveichik

            3,68011125




            3,68011125






















                up vote
                1
                down vote













                $|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.



                BEcause the axis must be longer than the distance between focus, $cge 2$.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  $|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.



                  BEcause the axis must be longer than the distance between focus, $cge 2$.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    $|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.



                    BEcause the axis must be longer than the distance between focus, $cge 2$.






                    share|cite|improve this answer












                    $|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.



                    BEcause the axis must be longer than the distance between focus, $cge 2$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 16 at 13:24









                    Tito Eliatron

                    1,012621




                    1,012621






















                        up vote
                        1
                        down vote













                        A bit of geometry.



                        In the complex plane:



                        $A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.



                        $triangle ABC$ has sides of lengths:



                        $|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.



                        The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:



                        $|z+1|+|z-1| > 2.$



                        $2$ is a lower bound. Is there a minimum?



                        If yes , $z=?$






                        share|cite|improve this answer



























                          up vote
                          1
                          down vote













                          A bit of geometry.



                          In the complex plane:



                          $A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.



                          $triangle ABC$ has sides of lengths:



                          $|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.



                          The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:



                          $|z+1|+|z-1| > 2.$



                          $2$ is a lower bound. Is there a minimum?



                          If yes , $z=?$






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            A bit of geometry.



                            In the complex plane:



                            $A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.



                            $triangle ABC$ has sides of lengths:



                            $|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.



                            The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:



                            $|z+1|+|z-1| > 2.$



                            $2$ is a lower bound. Is there a minimum?



                            If yes , $z=?$






                            share|cite|improve this answer














                            A bit of geometry.



                            In the complex plane:



                            $A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.



                            $triangle ABC$ has sides of lengths:



                            $|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.



                            The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:



                            $|z+1|+|z-1| > 2.$



                            $2$ is a lower bound. Is there a minimum?



                            If yes , $z=?$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 16 at 13:38

























                            answered Nov 16 at 13:29









                            Peter Szilas

                            10.1k2720




                            10.1k2720






























                                 

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