How do I prove the multiplicative inverse in complex arithmetic?












1












$begingroup$


I am working with Sheldon Axler's Linear Algebra Done Right and I could use some help from you guys. I have the following property of complex arithmetic:




for every $alpha in mathbb{C}$ with $alpha neq$ 0, there exist a >unique $beta in mathbb{C}$ such that $alphabeta = 1$




I want to prove this. I start by using de definition of multiplication of complex number.




$(a + bi)(c + di) = (ac - bd)+(ad+bc)i$




And the fact that:



$1 = (1 + 0i)$



Aaaand I run out of ideas. I really have no idea on how to continue so a clue would be greatly appreciated.






complex-numbers







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am working with Sheldon Axler's Linear Algebra Done Right and I could use some help from you guys. I have the following property of complex arithmetic:




    for every $alpha in mathbb{C}$ with $alpha neq$ 0, there exist a >unique $beta in mathbb{C}$ such that $alphabeta = 1$




    I want to prove this. I start by using de definition of multiplication of complex number.




    $(a + bi)(c + di) = (ac - bd)+(ad+bc)i$




    And the fact that:



    $1 = (1 + 0i)$



    Aaaand I run out of ideas. I really have no idea on how to continue so a clue would be greatly appreciated.






    complex-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am working with Sheldon Axler's Linear Algebra Done Right and I could use some help from you guys. I have the following property of complex arithmetic:




      for every $alpha in mathbb{C}$ with $alpha neq$ 0, there exist a >unique $beta in mathbb{C}$ such that $alphabeta = 1$




      I want to prove this. I start by using de definition of multiplication of complex number.




      $(a + bi)(c + di) = (ac - bd)+(ad+bc)i$




      And the fact that:



      $1 = (1 + 0i)$



      Aaaand I run out of ideas. I really have no idea on how to continue so a clue would be greatly appreciated.






      complex-numbers







      share|cite|improve this question











      $endgroup$




      I am working with Sheldon Axler's Linear Algebra Done Right and I could use some help from you guys. I have the following property of complex arithmetic:




      for every $alpha in mathbb{C}$ with $alpha neq$ 0, there exist a >unique $beta in mathbb{C}$ such that $alphabeta = 1$




      I want to prove this. I start by using de definition of multiplication of complex number.




      $(a + bi)(c + di) = (ac - bd)+(ad+bc)i$




      And the fact that:



      $1 = (1 + 0i)$



      Aaaand I run out of ideas. I really have no idea on how to continue so a clue would be greatly appreciated.






      complex-numbers




      complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 28 '18 at 5:30









      Eric Wofsey

      189k14216347




      189k14216347










      asked Apr 14 '18 at 9:18









      A39-A20A39-A20

      206




      206






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          Basic operations: $;alpha:=x+iyneq0iff x^2+y^2neq0;$ ( since $;x,yinBbb R;$) , so:



          $$alphaalpha^{-1}=1impliesalpha^{-1}=frac1alpha=frac1{x+iy}cdotfrac{x-iy}{x-iy}=frac x{x^2+y^2}-frac y{x^2+y^2}i$$



          and there you have a cartesian expression for $;alpha^{-1};$ whenever $;0neqalphainBbb C;$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you for your help! What you did in the beginning by saying that $alpha^-1 = beta$ made it a lot easier for me to understand. I just want to be sure that I understand the proof: $alphaalpha^-1 = 1 Rightarrow alpha^1 = frac{1}{alpha} = frac{1}{x+iy}cdot1 Rightarrow frac{1}{x+iy}cdotfrac{x-iy}{x-iy}$ Since $1 = frac{x-iy}{x-iy}$
            $endgroup$
            – A39-A20
            Apr 15 '18 at 8:00





















          1












          $begingroup$

          You start from $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$. Then impose the conditions $ac-bd = 1$ and $ad+bc=0$. From the second you get $d = -lambda b$ and $c = lambda a$ for some $lambda$. Plugging these in the first equation you end up with $lambda = frac{1}{a^2+b^2}$ and so $c = frac{a}{a^2+b^2}, d = -frac{b}{a^2+b^2}$. So your inverse is
          $$beta = frac{a}{a^2+b^2}-ifrac{b}{a^2+b^2}.$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You have the following:



            $ac-bd=1$, Eq. 1



            $bc+ad=0$, Eq. 2



            Recognize this as a linear system of two equations in two unknowns $c, d$ and solve using linear combinations. But before you do so check the matrix of known coefficients:



            $left[begin{matrix}a&-b\b&aend{matrix}right]$



            Observe that this guarantees a unique solution if and only if the determinant $a^2+b^2$ is nonzero, but for real numbers $a$ and $b$ this requirement is met unless $a=b=0$; the square of one nonzero number is (strictly) positive and the square of the other real number is never negative. Thereby, a unique solution is guaranteed for any nonzero complex number input $a+bi$.



            Now for a linear combination solution. Setting the multipliers to make the terms containing $d$ cancel gives:



            $a^2c-abd=a$, Eq. 1 × $a$



            $b^2c+abd=0$, Eq. 2 × $b$



            And then the sum contains only one unknown:



            $(a^2+b^2)c=a, c=frac{a}{a^2+b^2}$



            Substitute this for $c$ in either Eq. 1 or Eq. 2 and you can then solve for $d=-frac{b}{a^2+b^2}$.






            share|cite|improve this answer











            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Basic operations: $;alpha:=x+iyneq0iff x^2+y^2neq0;$ ( since $;x,yinBbb R;$) , so:



              $$alphaalpha^{-1}=1impliesalpha^{-1}=frac1alpha=frac1{x+iy}cdotfrac{x-iy}{x-iy}=frac x{x^2+y^2}-frac y{x^2+y^2}i$$



              and there you have a cartesian expression for $;alpha^{-1};$ whenever $;0neqalphainBbb C;$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Thank you for your help! What you did in the beginning by saying that $alpha^-1 = beta$ made it a lot easier for me to understand. I just want to be sure that I understand the proof: $alphaalpha^-1 = 1 Rightarrow alpha^1 = frac{1}{alpha} = frac{1}{x+iy}cdot1 Rightarrow frac{1}{x+iy}cdotfrac{x-iy}{x-iy}$ Since $1 = frac{x-iy}{x-iy}$
                $endgroup$
                – A39-A20
                Apr 15 '18 at 8:00


















              1












              $begingroup$

              Basic operations: $;alpha:=x+iyneq0iff x^2+y^2neq0;$ ( since $;x,yinBbb R;$) , so:



              $$alphaalpha^{-1}=1impliesalpha^{-1}=frac1alpha=frac1{x+iy}cdotfrac{x-iy}{x-iy}=frac x{x^2+y^2}-frac y{x^2+y^2}i$$



              and there you have a cartesian expression for $;alpha^{-1};$ whenever $;0neqalphainBbb C;$






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Thank you for your help! What you did in the beginning by saying that $alpha^-1 = beta$ made it a lot easier for me to understand. I just want to be sure that I understand the proof: $alphaalpha^-1 = 1 Rightarrow alpha^1 = frac{1}{alpha} = frac{1}{x+iy}cdot1 Rightarrow frac{1}{x+iy}cdotfrac{x-iy}{x-iy}$ Since $1 = frac{x-iy}{x-iy}$
                $endgroup$
                – A39-A20
                Apr 15 '18 at 8:00
















              1












              1








              1





              $begingroup$

              Basic operations: $;alpha:=x+iyneq0iff x^2+y^2neq0;$ ( since $;x,yinBbb R;$) , so:



              $$alphaalpha^{-1}=1impliesalpha^{-1}=frac1alpha=frac1{x+iy}cdotfrac{x-iy}{x-iy}=frac x{x^2+y^2}-frac y{x^2+y^2}i$$



              and there you have a cartesian expression for $;alpha^{-1};$ whenever $;0neqalphainBbb C;$






              share|cite|improve this answer









              $endgroup$



              Basic operations: $;alpha:=x+iyneq0iff x^2+y^2neq0;$ ( since $;x,yinBbb R;$) , so:



              $$alphaalpha^{-1}=1impliesalpha^{-1}=frac1alpha=frac1{x+iy}cdotfrac{x-iy}{x-iy}=frac x{x^2+y^2}-frac y{x^2+y^2}i$$



              and there you have a cartesian expression for $;alpha^{-1};$ whenever $;0neqalphainBbb C;$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Apr 14 '18 at 9:32









              DonAntonioDonAntonio

              179k1494233




              179k1494233








              • 1




                $begingroup$
                Thank you for your help! What you did in the beginning by saying that $alpha^-1 = beta$ made it a lot easier for me to understand. I just want to be sure that I understand the proof: $alphaalpha^-1 = 1 Rightarrow alpha^1 = frac{1}{alpha} = frac{1}{x+iy}cdot1 Rightarrow frac{1}{x+iy}cdotfrac{x-iy}{x-iy}$ Since $1 = frac{x-iy}{x-iy}$
                $endgroup$
                – A39-A20
                Apr 15 '18 at 8:00
















              • 1




                $begingroup$
                Thank you for your help! What you did in the beginning by saying that $alpha^-1 = beta$ made it a lot easier for me to understand. I just want to be sure that I understand the proof: $alphaalpha^-1 = 1 Rightarrow alpha^1 = frac{1}{alpha} = frac{1}{x+iy}cdot1 Rightarrow frac{1}{x+iy}cdotfrac{x-iy}{x-iy}$ Since $1 = frac{x-iy}{x-iy}$
                $endgroup$
                – A39-A20
                Apr 15 '18 at 8:00










              1




              1




              $begingroup$
              Thank you for your help! What you did in the beginning by saying that $alpha^-1 = beta$ made it a lot easier for me to understand. I just want to be sure that I understand the proof: $alphaalpha^-1 = 1 Rightarrow alpha^1 = frac{1}{alpha} = frac{1}{x+iy}cdot1 Rightarrow frac{1}{x+iy}cdotfrac{x-iy}{x-iy}$ Since $1 = frac{x-iy}{x-iy}$
              $endgroup$
              – A39-A20
              Apr 15 '18 at 8:00






              $begingroup$
              Thank you for your help! What you did in the beginning by saying that $alpha^-1 = beta$ made it a lot easier for me to understand. I just want to be sure that I understand the proof: $alphaalpha^-1 = 1 Rightarrow alpha^1 = frac{1}{alpha} = frac{1}{x+iy}cdot1 Rightarrow frac{1}{x+iy}cdotfrac{x-iy}{x-iy}$ Since $1 = frac{x-iy}{x-iy}$
              $endgroup$
              – A39-A20
              Apr 15 '18 at 8:00













              1












              $begingroup$

              You start from $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$. Then impose the conditions $ac-bd = 1$ and $ad+bc=0$. From the second you get $d = -lambda b$ and $c = lambda a$ for some $lambda$. Plugging these in the first equation you end up with $lambda = frac{1}{a^2+b^2}$ and so $c = frac{a}{a^2+b^2}, d = -frac{b}{a^2+b^2}$. So your inverse is
              $$beta = frac{a}{a^2+b^2}-ifrac{b}{a^2+b^2}.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                You start from $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$. Then impose the conditions $ac-bd = 1$ and $ad+bc=0$. From the second you get $d = -lambda b$ and $c = lambda a$ for some $lambda$. Plugging these in the first equation you end up with $lambda = frac{1}{a^2+b^2}$ and so $c = frac{a}{a^2+b^2}, d = -frac{b}{a^2+b^2}$. So your inverse is
                $$beta = frac{a}{a^2+b^2}-ifrac{b}{a^2+b^2}.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You start from $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$. Then impose the conditions $ac-bd = 1$ and $ad+bc=0$. From the second you get $d = -lambda b$ and $c = lambda a$ for some $lambda$. Plugging these in the first equation you end up with $lambda = frac{1}{a^2+b^2}$ and so $c = frac{a}{a^2+b^2}, d = -frac{b}{a^2+b^2}$. So your inverse is
                  $$beta = frac{a}{a^2+b^2}-ifrac{b}{a^2+b^2}.$$






                  share|cite|improve this answer









                  $endgroup$



                  You start from $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$. Then impose the conditions $ac-bd = 1$ and $ad+bc=0$. From the second you get $d = -lambda b$ and $c = lambda a$ for some $lambda$. Plugging these in the first equation you end up with $lambda = frac{1}{a^2+b^2}$ and so $c = frac{a}{a^2+b^2}, d = -frac{b}{a^2+b^2}$. So your inverse is
                  $$beta = frac{a}{a^2+b^2}-ifrac{b}{a^2+b^2}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 14 '18 at 9:42









                  GibbsGibbs

                  5,4103827




                  5,4103827























                      1












                      $begingroup$

                      You have the following:



                      $ac-bd=1$, Eq. 1



                      $bc+ad=0$, Eq. 2



                      Recognize this as a linear system of two equations in two unknowns $c, d$ and solve using linear combinations. But before you do so check the matrix of known coefficients:



                      $left[begin{matrix}a&-b\b&aend{matrix}right]$



                      Observe that this guarantees a unique solution if and only if the determinant $a^2+b^2$ is nonzero, but for real numbers $a$ and $b$ this requirement is met unless $a=b=0$; the square of one nonzero number is (strictly) positive and the square of the other real number is never negative. Thereby, a unique solution is guaranteed for any nonzero complex number input $a+bi$.



                      Now for a linear combination solution. Setting the multipliers to make the terms containing $d$ cancel gives:



                      $a^2c-abd=a$, Eq. 1 × $a$



                      $b^2c+abd=0$, Eq. 2 × $b$



                      And then the sum contains only one unknown:



                      $(a^2+b^2)c=a, c=frac{a}{a^2+b^2}$



                      Substitute this for $c$ in either Eq. 1 or Eq. 2 and you can then solve for $d=-frac{b}{a^2+b^2}$.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        You have the following:



                        $ac-bd=1$, Eq. 1



                        $bc+ad=0$, Eq. 2



                        Recognize this as a linear system of two equations in two unknowns $c, d$ and solve using linear combinations. But before you do so check the matrix of known coefficients:



                        $left[begin{matrix}a&-b\b&aend{matrix}right]$



                        Observe that this guarantees a unique solution if and only if the determinant $a^2+b^2$ is nonzero, but for real numbers $a$ and $b$ this requirement is met unless $a=b=0$; the square of one nonzero number is (strictly) positive and the square of the other real number is never negative. Thereby, a unique solution is guaranteed for any nonzero complex number input $a+bi$.



                        Now for a linear combination solution. Setting the multipliers to make the terms containing $d$ cancel gives:



                        $a^2c-abd=a$, Eq. 1 × $a$



                        $b^2c+abd=0$, Eq. 2 × $b$



                        And then the sum contains only one unknown:



                        $(a^2+b^2)c=a, c=frac{a}{a^2+b^2}$



                        Substitute this for $c$ in either Eq. 1 or Eq. 2 and you can then solve for $d=-frac{b}{a^2+b^2}$.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          You have the following:



                          $ac-bd=1$, Eq. 1



                          $bc+ad=0$, Eq. 2



                          Recognize this as a linear system of two equations in two unknowns $c, d$ and solve using linear combinations. But before you do so check the matrix of known coefficients:



                          $left[begin{matrix}a&-b\b&aend{matrix}right]$



                          Observe that this guarantees a unique solution if and only if the determinant $a^2+b^2$ is nonzero, but for real numbers $a$ and $b$ this requirement is met unless $a=b=0$; the square of one nonzero number is (strictly) positive and the square of the other real number is never negative. Thereby, a unique solution is guaranteed for any nonzero complex number input $a+bi$.



                          Now for a linear combination solution. Setting the multipliers to make the terms containing $d$ cancel gives:



                          $a^2c-abd=a$, Eq. 1 × $a$



                          $b^2c+abd=0$, Eq. 2 × $b$



                          And then the sum contains only one unknown:



                          $(a^2+b^2)c=a, c=frac{a}{a^2+b^2}$



                          Substitute this for $c$ in either Eq. 1 or Eq. 2 and you can then solve for $d=-frac{b}{a^2+b^2}$.






                          share|cite|improve this answer











                          $endgroup$



                          You have the following:



                          $ac-bd=1$, Eq. 1



                          $bc+ad=0$, Eq. 2



                          Recognize this as a linear system of two equations in two unknowns $c, d$ and solve using linear combinations. But before you do so check the matrix of known coefficients:



                          $left[begin{matrix}a&-b\b&aend{matrix}right]$



                          Observe that this guarantees a unique solution if and only if the determinant $a^2+b^2$ is nonzero, but for real numbers $a$ and $b$ this requirement is met unless $a=b=0$; the square of one nonzero number is (strictly) positive and the square of the other real number is never negative. Thereby, a unique solution is guaranteed for any nonzero complex number input $a+bi$.



                          Now for a linear combination solution. Setting the multipliers to make the terms containing $d$ cancel gives:



                          $a^2c-abd=a$, Eq. 1 × $a$



                          $b^2c+abd=0$, Eq. 2 × $b$



                          And then the sum contains only one unknown:



                          $(a^2+b^2)c=a, c=frac{a}{a^2+b^2}$



                          Substitute this for $c$ in either Eq. 1 or Eq. 2 and you can then solve for $d=-frac{b}{a^2+b^2}$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Apr 14 '18 at 10:33

























                          answered Apr 14 '18 at 10:12









                          Oscar LanziOscar Lanzi

                          13.2k12136




                          13.2k12136






























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                              Dieringhausen

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                              Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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