What is the factorization of $(da+dbi)^2$, $(da+dbi)^4$, and $(da+dbi)^8$,

Multi tool use
Multi tool use












1












$begingroup$


I need a clarification for factorization when the variable 'd' is included in $(a + ib)^n$, such that $(da + dib)^n$.



Specifically for three cases.



If $(a + ib)^2$ = $1*a^2 + 2iab − 1*b^2$,
then what would:
$(da + dib)^2$
= ??



For the first one I thought that $(da + dib)^2$ was either




  1. $d1*a^2 + d2iab − 1*b^2$


  2. $d1*a^2 + d^22iab − 1*b^2$



    however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$






If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$



then what would:
$(da + dib)^4$ = ??





If $(a + bi)^8$ = $a^8 + 8a^7bi − 28a^6b^2 − 56a^5b^3i + 70a^4b^4 + 56a^3b^5i − 28a^2b^6 − 8ab^7i + b^8$



then what would: $(da + dib)^8$ = ??













share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I need a clarification for factorization when the variable 'd' is included in $(a + ib)^n$, such that $(da + dib)^n$.



    Specifically for three cases.



    If $(a + ib)^2$ = $1*a^2 + 2iab − 1*b^2$,
    then what would:
    $(da + dib)^2$
    = ??



    For the first one I thought that $(da + dib)^2$ was either




    1. $d1*a^2 + d2iab − 1*b^2$


    2. $d1*a^2 + d^22iab − 1*b^2$



      however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$






    If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$



    then what would:
    $(da + dib)^4$ = ??





    If $(a + bi)^8$ = $a^8 + 8a^7bi − 28a^6b^2 − 56a^5b^3i + 70a^4b^4 + 56a^3b^5i − 28a^2b^6 − 8ab^7i + b^8$



    then what would: $(da + dib)^8$ = ??













    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I need a clarification for factorization when the variable 'd' is included in $(a + ib)^n$, such that $(da + dib)^n$.



      Specifically for three cases.



      If $(a + ib)^2$ = $1*a^2 + 2iab − 1*b^2$,
      then what would:
      $(da + dib)^2$
      = ??



      For the first one I thought that $(da + dib)^2$ was either




      1. $d1*a^2 + d2iab − 1*b^2$


      2. $d1*a^2 + d^22iab − 1*b^2$



        however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$






      If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$



      then what would:
      $(da + dib)^4$ = ??





      If $(a + bi)^8$ = $a^8 + 8a^7bi − 28a^6b^2 − 56a^5b^3i + 70a^4b^4 + 56a^3b^5i − 28a^2b^6 − 8ab^7i + b^8$



      then what would: $(da + dib)^8$ = ??













      share|cite|improve this question











      $endgroup$




      I need a clarification for factorization when the variable 'd' is included in $(a + ib)^n$, such that $(da + dib)^n$.



      Specifically for three cases.



      If $(a + ib)^2$ = $1*a^2 + 2iab − 1*b^2$,
      then what would:
      $(da + dib)^2$
      = ??



      For the first one I thought that $(da + dib)^2$ was either




      1. $d1*a^2 + d2iab − 1*b^2$


      2. $d1*a^2 + d^22iab − 1*b^2$



        however, according to an online calculator $(da + dib)^2$ = $d^2a^2+2d^2abi−d^2b^2$






      If $(a + bi)^4$ = $a^4 + 4a^3bi − 6a^2b^2 − 4ab^3i + b^4$



      then what would:
      $(da + dib)^4$ = ??





      If $(a + bi)^8$ = $a^8 + 8a^7bi − 28a^6b^2 − 56a^5b^3i + 70a^4b^4 + 56a^3b^5i − 28a^2b^6 − 8ab^7i + b^8$



      then what would: $(da + dib)^8$ = ??










      complex-numbers factoring






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 14 at 0:01







      Du'uzu Mes

















      asked Jan 13 at 23:53









      Du'uzu MesDu'uzu Mes

      135




      135






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          It looks like you simply need to factor out the variable $d$ from each expression. Note that
          $$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$






          share|cite|improve this answer









          $endgroup$














            Your Answer








            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072694%2fwhat-is-the-factorization-of-dadbi2-dadbi4-and-dadbi8%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            It looks like you simply need to factor out the variable $d$ from each expression. Note that
            $$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              It looks like you simply need to factor out the variable $d$ from each expression. Note that
              $$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                It looks like you simply need to factor out the variable $d$ from each expression. Note that
                $$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$






                share|cite|improve this answer









                $endgroup$



                It looks like you simply need to factor out the variable $d$ from each expression. Note that
                $$(da+dbi)^n = (d(a+bi))^n = d^n(a+bi)^n.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 13 at 23:56









                D.B.D.B.

                1,61029




                1,61029






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072694%2fwhat-is-the-factorization-of-dadbi2-dadbi4-and-dadbi8%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    vtgQ,sYoJjpvDBcqH8G2UebW6jzA lvaHZQChKG,YJz NAwRWFtJXD3uIY7UA37 dg6HP8R U G,xvGb nk TU
                    B3b T0EIGqdavyLw2LcSfNad,06b j,1RD r,IorC6Me4Xfr vt6bNfCFr,EHRr

                    Popular posts from this blog

                    Bressuire

                    Cabo Verde

                    Gyllenstierna