Problem in complex number multiplication [duplicate]












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This question already has an answer here:




  • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

    9 answers




I want to know $sqrt{-m}sqrt{-n}=$? I tried in the following ways:



Way 1:$$sqrt{-m}sqrt{-n}=sqrt{(-m)(-n)}=sqrt{mn}.$$
Way 2:$$sqrt{-m}sqrt{-n}=sqrt{m}isqrt{n}i=sqrt{mn}i^2=-sqrt{mn}$$



So I got two different values for the same $sqrt{-m}sqrt{-n}$. How can that be possible?






complex-numbers complex-multiplication







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marked as duplicate by Saad, Lord Shark the Unknown, Watson, Cesareo, Brahadeesh Dec 1 '18 at 10:18


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.




















    1












    $begingroup$



    This question already has an answer here:




    • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

      9 answers




    I want to know $sqrt{-m}sqrt{-n}=$? I tried in the following ways:



    Way 1:$$sqrt{-m}sqrt{-n}=sqrt{(-m)(-n)}=sqrt{mn}.$$
    Way 2:$$sqrt{-m}sqrt{-n}=sqrt{m}isqrt{n}i=sqrt{mn}i^2=-sqrt{mn}$$



    So I got two different values for the same $sqrt{-m}sqrt{-n}$. How can that be possible?






    complex-numbers complex-multiplication







    share|cite|improve this question











    $endgroup$



    marked as duplicate by Saad, Lord Shark the Unknown, Watson, Cesareo, Brahadeesh Dec 1 '18 at 10:18


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















      1












      1








      1





      $begingroup$



      This question already has an answer here:




      • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

        9 answers




      I want to know $sqrt{-m}sqrt{-n}=$? I tried in the following ways:



      Way 1:$$sqrt{-m}sqrt{-n}=sqrt{(-m)(-n)}=sqrt{mn}.$$
      Way 2:$$sqrt{-m}sqrt{-n}=sqrt{m}isqrt{n}i=sqrt{mn}i^2=-sqrt{mn}$$



      So I got two different values for the same $sqrt{-m}sqrt{-n}$. How can that be possible?






      complex-numbers complex-multiplication







      share|cite|improve this question











      $endgroup$





      This question already has an answer here:




      • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

        9 answers




      I want to know $sqrt{-m}sqrt{-n}=$? I tried in the following ways:



      Way 1:$$sqrt{-m}sqrt{-n}=sqrt{(-m)(-n)}=sqrt{mn}.$$
      Way 2:$$sqrt{-m}sqrt{-n}=sqrt{m}isqrt{n}i=sqrt{mn}i^2=-sqrt{mn}$$



      So I got two different values for the same $sqrt{-m}sqrt{-n}$. How can that be possible?





      This question already has an answer here:




      • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

        9 answers







      complex-numbers complex-multiplication




      complex-numbers complex-multiplication






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 8:18









      Saad

      19.7k92352




      19.7k92352










      asked Dec 1 '18 at 8:02









      Asif IqubalAsif Iqubal

      1397




      1397




      marked as duplicate by Saad, Lord Shark the Unknown, Watson, Cesareo, Brahadeesh Dec 1 '18 at 10:18


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Saad, Lord Shark the Unknown, Watson, Cesareo, Brahadeesh Dec 1 '18 at 10:18


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
























          1 Answer
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          3












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          The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
          Therefore, the 1st way is incorrect.






          share|cite|improve this answer









          $endgroup$




















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
            Therefore, the 1st way is incorrect.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
              Therefore, the 1st way is incorrect.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
                Therefore, the 1st way is incorrect.






                share|cite|improve this answer









                $endgroup$



                The rule $sqrt{a} cdot sqrt{b} = sqrt{a cdot b}$ doesn't necessarily apply when $a < 0$ or $b < 0$. For example, $$-1 = icdot i = sqrt{-1} cdot sqrt{-1} neq sqrt{(-1)cdot(-1)} = sqrt{1} = 1$$
                Therefore, the 1st way is incorrect.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 '18 at 8:40









                AlephZeroAlephZero

                22616




                22616















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