What is the appropriate way to compute $z=frac{1}{sqrt{7+24i}}?$












0












$begingroup$



What is the appropriate way to compute $$z=frac{1}{sqrt{7+24i}}?$$




I'm asking because in 2 different books I found 2 different answers for the problem:



Book A: $(4-3i)/25$ and $(-4+3i)/25$;



Book B: $(-4+3i)/25$.



To add to the confusion Wolfram Alpha gives the single answer $(4-3i)/25$, which I believe is correct by my own development.



The difficulty, I think, is on how to deal with the signs in the positive root of a complex number. This is not, I believe, the same thing as solving $z^2=1/(7+24i)$. Like, on reals, the problem of solving $x^2=4$ or computing $x=sqrt{4}$.



I would appreciate some clarification, if possible.






algebra-precalculus complex-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is the appropriate way to compute $sqrt{4}$? Book A gives $2$ and $-2$, while Book B gives just $2$.
    $endgroup$
    – vadim123
    Dec 19 '18 at 16:34






  • 2




    $begingroup$
    @vadim123, I would pick only 2 because what is asked is the positive $sqrt{4}$, that is 2. 2 and -2 would be the solution for $x^2=4$. See below how both answers diverge....
    $endgroup$
    – bluemaster
    Dec 19 '18 at 16:36








  • 1




    $begingroup$
    I agree with @bluemaster, the main question is what is meant by $sqrt{x}$, the principal square root, or all possible solutions. Most commonly, it represents the principal square root.
    $endgroup$
    – kvantour
    Dec 19 '18 at 16:38










  • $begingroup$
    Wait? TWO books had the exact same example?????
    $endgroup$
    – fleablood
    Dec 19 '18 at 18:15










  • $begingroup$
    Could you explain how you say the books gave the answers they did?
    $endgroup$
    – fleablood
    Dec 19 '18 at 18:18
















0












$begingroup$



What is the appropriate way to compute $$z=frac{1}{sqrt{7+24i}}?$$




I'm asking because in 2 different books I found 2 different answers for the problem:



Book A: $(4-3i)/25$ and $(-4+3i)/25$;



Book B: $(-4+3i)/25$.



To add to the confusion Wolfram Alpha gives the single answer $(4-3i)/25$, which I believe is correct by my own development.



The difficulty, I think, is on how to deal with the signs in the positive root of a complex number. This is not, I believe, the same thing as solving $z^2=1/(7+24i)$. Like, on reals, the problem of solving $x^2=4$ or computing $x=sqrt{4}$.



I would appreciate some clarification, if possible.






algebra-precalculus complex-numbers







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is the appropriate way to compute $sqrt{4}$? Book A gives $2$ and $-2$, while Book B gives just $2$.
    $endgroup$
    – vadim123
    Dec 19 '18 at 16:34






  • 2




    $begingroup$
    @vadim123, I would pick only 2 because what is asked is the positive $sqrt{4}$, that is 2. 2 and -2 would be the solution for $x^2=4$. See below how both answers diverge....
    $endgroup$
    – bluemaster
    Dec 19 '18 at 16:36








  • 1




    $begingroup$
    I agree with @bluemaster, the main question is what is meant by $sqrt{x}$, the principal square root, or all possible solutions. Most commonly, it represents the principal square root.
    $endgroup$
    – kvantour
    Dec 19 '18 at 16:38










  • $begingroup$
    Wait? TWO books had the exact same example?????
    $endgroup$
    – fleablood
    Dec 19 '18 at 18:15










  • $begingroup$
    Could you explain how you say the books gave the answers they did?
    $endgroup$
    – fleablood
    Dec 19 '18 at 18:18














0












0








0


1



$begingroup$



What is the appropriate way to compute $$z=frac{1}{sqrt{7+24i}}?$$




I'm asking because in 2 different books I found 2 different answers for the problem:



Book A: $(4-3i)/25$ and $(-4+3i)/25$;



Book B: $(-4+3i)/25$.



To add to the confusion Wolfram Alpha gives the single answer $(4-3i)/25$, which I believe is correct by my own development.



The difficulty, I think, is on how to deal with the signs in the positive root of a complex number. This is not, I believe, the same thing as solving $z^2=1/(7+24i)$. Like, on reals, the problem of solving $x^2=4$ or computing $x=sqrt{4}$.



I would appreciate some clarification, if possible.






algebra-precalculus complex-numbers







share|cite|improve this question











$endgroup$





What is the appropriate way to compute $$z=frac{1}{sqrt{7+24i}}?$$




I'm asking because in 2 different books I found 2 different answers for the problem:



Book A: $(4-3i)/25$ and $(-4+3i)/25$;



Book B: $(-4+3i)/25$.



To add to the confusion Wolfram Alpha gives the single answer $(4-3i)/25$, which I believe is correct by my own development.



The difficulty, I think, is on how to deal with the signs in the positive root of a complex number. This is not, I believe, the same thing as solving $z^2=1/(7+24i)$. Like, on reals, the problem of solving $x^2=4$ or computing $x=sqrt{4}$.



I would appreciate some clarification, if possible.






algebra-precalculus complex-numbers




algebra-precalculus complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 16:45







bluemaster

















asked Dec 19 '18 at 16:27









bluemasterbluemaster

1,638519




1,638519








  • 2




    $begingroup$
    What is the appropriate way to compute $sqrt{4}$? Book A gives $2$ and $-2$, while Book B gives just $2$.
    $endgroup$
    – vadim123
    Dec 19 '18 at 16:34






  • 2




    $begingroup$
    @vadim123, I would pick only 2 because what is asked is the positive $sqrt{4}$, that is 2. 2 and -2 would be the solution for $x^2=4$. See below how both answers diverge....
    $endgroup$
    – bluemaster
    Dec 19 '18 at 16:36








  • 1




    $begingroup$
    I agree with @bluemaster, the main question is what is meant by $sqrt{x}$, the principal square root, or all possible solutions. Most commonly, it represents the principal square root.
    $endgroup$
    – kvantour
    Dec 19 '18 at 16:38










  • $begingroup$
    Wait? TWO books had the exact same example?????
    $endgroup$
    – fleablood
    Dec 19 '18 at 18:15










  • $begingroup$
    Could you explain how you say the books gave the answers they did?
    $endgroup$
    – fleablood
    Dec 19 '18 at 18:18














  • 2




    $begingroup$
    What is the appropriate way to compute $sqrt{4}$? Book A gives $2$ and $-2$, while Book B gives just $2$.
    $endgroup$
    – vadim123
    Dec 19 '18 at 16:34






  • 2




    $begingroup$
    @vadim123, I would pick only 2 because what is asked is the positive $sqrt{4}$, that is 2. 2 and -2 would be the solution for $x^2=4$. See below how both answers diverge....
    $endgroup$
    – bluemaster
    Dec 19 '18 at 16:36








  • 1




    $begingroup$
    I agree with @bluemaster, the main question is what is meant by $sqrt{x}$, the principal square root, or all possible solutions. Most commonly, it represents the principal square root.
    $endgroup$
    – kvantour
    Dec 19 '18 at 16:38










  • $begingroup$
    Wait? TWO books had the exact same example?????
    $endgroup$
    – fleablood
    Dec 19 '18 at 18:15










  • $begingroup$
    Could you explain how you say the books gave the answers they did?
    $endgroup$
    – fleablood
    Dec 19 '18 at 18:18








2




2




$begingroup$
What is the appropriate way to compute $sqrt{4}$? Book A gives $2$ and $-2$, while Book B gives just $2$.
$endgroup$
– vadim123
Dec 19 '18 at 16:34




$begingroup$
What is the appropriate way to compute $sqrt{4}$? Book A gives $2$ and $-2$, while Book B gives just $2$.
$endgroup$
– vadim123
Dec 19 '18 at 16:34




2




2




$begingroup$
@vadim123, I would pick only 2 because what is asked is the positive $sqrt{4}$, that is 2. 2 and -2 would be the solution for $x^2=4$. See below how both answers diverge....
$endgroup$
– bluemaster
Dec 19 '18 at 16:36






$begingroup$
@vadim123, I would pick only 2 because what is asked is the positive $sqrt{4}$, that is 2. 2 and -2 would be the solution for $x^2=4$. See below how both answers diverge....
$endgroup$
– bluemaster
Dec 19 '18 at 16:36






1




1




$begingroup$
I agree with @bluemaster, the main question is what is meant by $sqrt{x}$, the principal square root, or all possible solutions. Most commonly, it represents the principal square root.
$endgroup$
– kvantour
Dec 19 '18 at 16:38




$begingroup$
I agree with @bluemaster, the main question is what is meant by $sqrt{x}$, the principal square root, or all possible solutions. Most commonly, it represents the principal square root.
$endgroup$
– kvantour
Dec 19 '18 at 16:38












$begingroup$
Wait? TWO books had the exact same example?????
$endgroup$
– fleablood
Dec 19 '18 at 18:15




$begingroup$
Wait? TWO books had the exact same example?????
$endgroup$
– fleablood
Dec 19 '18 at 18:15












$begingroup$
Could you explain how you say the books gave the answers they did?
$endgroup$
– fleablood
Dec 19 '18 at 18:18




$begingroup$
Could you explain how you say the books gave the answers they did?
$endgroup$
– fleablood
Dec 19 '18 at 18:18










3 Answers
3






active

oldest

votes


















3












$begingroup$

Book A just gives the square root and its negative (both valid).



For a complex nonzero $w$ there are always two distinct solutions $z$ to $z^2=w.$ Though there may be some way to single out one of them as "the" square root, it's not that simple since complex numbers aren't ordered.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Note that $$(4+3i)^2=16-9+24i$$ and $$frac{1}{4+3i}=frac{4-3i}{25}$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      Correct answer is $ (4-3i)/25 $ as given by Wolfram Alpha. The real part of the principal square root is always nonnegative. ( Source: https://en.wikipedia.org/wiki/Square_root#Algebraic_formula )






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Notice that the problem did not ask the solution of a second degree equation $z^2=...$.
        $endgroup$
        – bluemaster
        Dec 19 '18 at 17:24






      • 1




        $begingroup$
        I get your point @bluemaster. Let me edit the answer.
        $endgroup$
        – Kshitij Mishra
        Dec 19 '18 at 17:54













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Book A just gives the square root and its negative (both valid).



      For a complex nonzero $w$ there are always two distinct solutions $z$ to $z^2=w.$ Though there may be some way to single out one of them as "the" square root, it's not that simple since complex numbers aren't ordered.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Book A just gives the square root and its negative (both valid).



        For a complex nonzero $w$ there are always two distinct solutions $z$ to $z^2=w.$ Though there may be some way to single out one of them as "the" square root, it's not that simple since complex numbers aren't ordered.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Book A just gives the square root and its negative (both valid).



          For a complex nonzero $w$ there are always two distinct solutions $z$ to $z^2=w.$ Though there may be some way to single out one of them as "the" square root, it's not that simple since complex numbers aren't ordered.






          share|cite|improve this answer











          $endgroup$



          Book A just gives the square root and its negative (both valid).



          For a complex nonzero $w$ there are always two distinct solutions $z$ to $z^2=w.$ Though there may be some way to single out one of them as "the" square root, it's not that simple since complex numbers aren't ordered.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 16:46

























          answered Dec 19 '18 at 16:31









          coffeemathcoffeemath

          2,7381415




          2,7381415























              2












              $begingroup$

              Note that $$(4+3i)^2=16-9+24i$$ and $$frac{1}{4+3i}=frac{4-3i}{25}$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Note that $$(4+3i)^2=16-9+24i$$ and $$frac{1}{4+3i}=frac{4-3i}{25}$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Note that $$(4+3i)^2=16-9+24i$$ and $$frac{1}{4+3i}=frac{4-3i}{25}$$






                  share|cite|improve this answer









                  $endgroup$



                  Note that $$(4+3i)^2=16-9+24i$$ and $$frac{1}{4+3i}=frac{4-3i}{25}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 16:31









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  74.5k42865




                  74.5k42865























                      2












                      $begingroup$

                      Correct answer is $ (4-3i)/25 $ as given by Wolfram Alpha. The real part of the principal square root is always nonnegative. ( Source: https://en.wikipedia.org/wiki/Square_root#Algebraic_formula )






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Notice that the problem did not ask the solution of a second degree equation $z^2=...$.
                        $endgroup$
                        – bluemaster
                        Dec 19 '18 at 17:24






                      • 1




                        $begingroup$
                        I get your point @bluemaster. Let me edit the answer.
                        $endgroup$
                        – Kshitij Mishra
                        Dec 19 '18 at 17:54


















                      2












                      $begingroup$

                      Correct answer is $ (4-3i)/25 $ as given by Wolfram Alpha. The real part of the principal square root is always nonnegative. ( Source: https://en.wikipedia.org/wiki/Square_root#Algebraic_formula )






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Notice that the problem did not ask the solution of a second degree equation $z^2=...$.
                        $endgroup$
                        – bluemaster
                        Dec 19 '18 at 17:24






                      • 1




                        $begingroup$
                        I get your point @bluemaster. Let me edit the answer.
                        $endgroup$
                        – Kshitij Mishra
                        Dec 19 '18 at 17:54
















                      2












                      2








                      2





                      $begingroup$

                      Correct answer is $ (4-3i)/25 $ as given by Wolfram Alpha. The real part of the principal square root is always nonnegative. ( Source: https://en.wikipedia.org/wiki/Square_root#Algebraic_formula )






                      share|cite|improve this answer











                      $endgroup$



                      Correct answer is $ (4-3i)/25 $ as given by Wolfram Alpha. The real part of the principal square root is always nonnegative. ( Source: https://en.wikipedia.org/wiki/Square_root#Algebraic_formula )







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 19 '18 at 17:59

























                      answered Dec 19 '18 at 17:10









                      Kshitij MishraKshitij Mishra

                      413




                      413












                      • $begingroup$
                        Notice that the problem did not ask the solution of a second degree equation $z^2=...$.
                        $endgroup$
                        – bluemaster
                        Dec 19 '18 at 17:24






                      • 1




                        $begingroup$
                        I get your point @bluemaster. Let me edit the answer.
                        $endgroup$
                        – Kshitij Mishra
                        Dec 19 '18 at 17:54




















                      • $begingroup$
                        Notice that the problem did not ask the solution of a second degree equation $z^2=...$.
                        $endgroup$
                        – bluemaster
                        Dec 19 '18 at 17:24






                      • 1




                        $begingroup$
                        I get your point @bluemaster. Let me edit the answer.
                        $endgroup$
                        – Kshitij Mishra
                        Dec 19 '18 at 17:54


















                      $begingroup$
                      Notice that the problem did not ask the solution of a second degree equation $z^2=...$.
                      $endgroup$
                      – bluemaster
                      Dec 19 '18 at 17:24




                      $begingroup$
                      Notice that the problem did not ask the solution of a second degree equation $z^2=...$.
                      $endgroup$
                      – bluemaster
                      Dec 19 '18 at 17:24




                      1




                      1




                      $begingroup$
                      I get your point @bluemaster. Let me edit the answer.
                      $endgroup$
                      – Kshitij Mishra
                      Dec 19 '18 at 17:54






                      $begingroup$
                      I get your point @bluemaster. Let me edit the answer.
                      $endgroup$
                      – Kshitij Mishra
                      Dec 19 '18 at 17:54




















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