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finding the number of complex number


Simplified way of finding a complex number raised to another complex numberNumber of Infinities in complex numbersComplex number polar form equationFinding numbers $a$ and $b$ for a complex numberFinding a Trigonometric Form of Complex NumberFinding the root of a complex numberFind the polar form of the complex numberInequation complex number with fractionsThe Geometry of Complex Number and Its Multiplicative InverseFinding a complex number













2












$begingroup$



How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$?




My attempt :



I got $2$, that is $ z=-2, z= +2$ , $|z| = {sqrt{ 2^2+1}}$, $|z| = {sqrt{(-2^2) +1}}$



Is it true ?






geometry proof-verification complex-numbers euclidean-geometry analytic-geometry







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$?




    My attempt :



    I got $2$, that is $ z=-2, z= +2$ , $|z| = {sqrt{ 2^2+1}}$, $|z| = {sqrt{(-2^2) +1}}$



    Is it true ?






    geometry proof-verification complex-numbers euclidean-geometry analytic-geometry







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$?




      My attempt :



      I got $2$, that is $ z=-2, z= +2$ , $|z| = {sqrt{ 2^2+1}}$, $|z| = {sqrt{(-2^2) +1}}$



      Is it true ?






      geometry proof-verification complex-numbers euclidean-geometry analytic-geometry







      share|cite|improve this question











      $endgroup$





      How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$?




      My attempt :



      I got $2$, that is $ z=-2, z= +2$ , $|z| = {sqrt{ 2^2+1}}$, $|z| = {sqrt{(-2^2) +1}}$



      Is it true ?






      geometry proof-verification complex-numbers euclidean-geometry analytic-geometry




      geometry proof-verification complex-numbers euclidean-geometry analytic-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 28 mins ago









      Anirban Niloy

      599118




      599118










      asked 2 hours ago









      jasminejasmine

      1,806418




      1,806418






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            (+1) for explaining it geometrically
            $endgroup$
            – José Carlos Santos
            2 hours ago



















          3












          $begingroup$

          Let $A(-1,0)$ and $B(0,-1)$.



          Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$



          with the circle $x^2+y^2=25.$



          Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7












            $begingroup$

            From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              (+1) for explaining it geometrically
              $endgroup$
              – José Carlos Santos
              2 hours ago
















            7












            $begingroup$

            From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              (+1) for explaining it geometrically
              $endgroup$
              – José Carlos Santos
              2 hours ago














            7












            7








            7





            $begingroup$

            From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$






            share|cite|improve this answer











            $endgroup$



            From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago

























            answered 2 hours ago









            greedoidgreedoid

            45k1156111




            45k1156111








            • 2




              $begingroup$
              (+1) for explaining it geometrically
              $endgroup$
              – José Carlos Santos
              2 hours ago














            • 2




              $begingroup$
              (+1) for explaining it geometrically
              $endgroup$
              – José Carlos Santos
              2 hours ago








            2




            2




            $begingroup$
            (+1) for explaining it geometrically
            $endgroup$
            – José Carlos Santos
            2 hours ago




            $begingroup$
            (+1) for explaining it geometrically
            $endgroup$
            – José Carlos Santos
            2 hours ago











            3












            $begingroup$

            Let $A(-1,0)$ and $B(0,-1)$.



            Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$



            with the circle $x^2+y^2=25.$



            Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Let $A(-1,0)$ and $B(0,-1)$.



              Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$



              with the circle $x^2+y^2=25.$



              Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Let $A(-1,0)$ and $B(0,-1)$.



                Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$



                with the circle $x^2+y^2=25.$



                Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.






                share|cite|improve this answer









                $endgroup$



                Let $A(-1,0)$ and $B(0,-1)$.



                Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$



                with the circle $x^2+y^2=25.$



                Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Michael RozenbergMichael Rozenberg

                105k1893198




                105k1893198






























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