Finding magnitude of a complex number











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$$z = biggr(dfrac{2+2i}{4-2i}biggr)$$



$$|z| = ? $$




My attempt:



$$dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = dfrac{4+12i}{20} = dfrac{4}{20}+dfrac{12}{20}i = dfrac{1}{5} + dfrac{3}{5}i$$



Now taking its magnitude and we have that



$$|z| = sqrt{biggr (dfrac 1 5 biggr ) ^2 +biggr (dfrac 3 5 biggr )^2} = sqrt {dfrac 2 5 }$$



Am I right?






complex-numbers







share|cite|improve this question


























    up vote
    8
    down vote

    favorite













    $$z = biggr(dfrac{2+2i}{4-2i}biggr)$$



    $$|z| = ? $$




    My attempt:



    $$dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = dfrac{4+12i}{20} = dfrac{4}{20}+dfrac{12}{20}i = dfrac{1}{5} + dfrac{3}{5}i$$



    Now taking its magnitude and we have that



    $$|z| = sqrt{biggr (dfrac 1 5 biggr ) ^2 +biggr (dfrac 3 5 biggr )^2} = sqrt {dfrac 2 5 }$$



    Am I right?






    complex-numbers







    share|cite|improve this question
























      up vote
      8
      down vote

      favorite









      up vote
      8
      down vote

      favorite












      $$z = biggr(dfrac{2+2i}{4-2i}biggr)$$



      $$|z| = ? $$




      My attempt:



      $$dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = dfrac{4+12i}{20} = dfrac{4}{20}+dfrac{12}{20}i = dfrac{1}{5} + dfrac{3}{5}i$$



      Now taking its magnitude and we have that



      $$|z| = sqrt{biggr (dfrac 1 5 biggr ) ^2 +biggr (dfrac 3 5 biggr )^2} = sqrt {dfrac 2 5 }$$



      Am I right?






      complex-numbers







      share|cite|improve this question














      $$z = biggr(dfrac{2+2i}{4-2i}biggr)$$



      $$|z| = ? $$




      My attempt:



      $$dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = dfrac{4+12i}{20} = dfrac{4}{20}+dfrac{12}{20}i = dfrac{1}{5} + dfrac{3}{5}i$$



      Now taking its magnitude and we have that



      $$|z| = sqrt{biggr (dfrac 1 5 biggr ) ^2 +biggr (dfrac 3 5 biggr )^2} = sqrt {dfrac 2 5 }$$



      Am I right?






      complex-numbers




      complex-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 8 hours ago









      Hamilton

      1627




      1627






















          2 Answers
          2






          active

          oldest

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          up vote
          7
          down vote













          Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$






          share|cite|improve this answer




























            up vote
            3
            down vote













            It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$



            $|2+2i|=sqrt{2^2+2^2}=2sqrt2$






            share|cite|improve this answer





















            • I first made the denominator real and then did my calculations to avoid any errors .
              – Hamilton
              8 hours ago












            • @Hamilton, At least for modulus, the formula I believe is more suitable
              – lab bhattacharjee
              8 hours ago










            • However, is it a bad way?
              – Hamilton
              8 hours ago










            • Rationalization may be costly in some cases
              – lab bhattacharjee
              8 hours ago










            • @Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
              – timtfj
              6 hours ago











            Your Answer





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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            7
            down vote













            Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$






            share|cite|improve this answer

























              up vote
              7
              down vote













              Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$






              share|cite|improve this answer























                up vote
                7
                down vote










                up vote
                7
                down vote









                Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$






                share|cite|improve this answer












                Yes, you are. You can do it also like this: $$Big|{2+2iover 4-2i}Big|=Big|{1+iover 2-i}Big|={|1+i|over |2-i|}= {sqrt{2}over sqrt{5}}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 8 hours ago









                greedoid

                35.6k114590




                35.6k114590






















                    up vote
                    3
                    down vote













                    It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$



                    $|2+2i|=sqrt{2^2+2^2}=2sqrt2$






                    share|cite|improve this answer





















                    • I first made the denominator real and then did my calculations to avoid any errors .
                      – Hamilton
                      8 hours ago












                    • @Hamilton, At least for modulus, the formula I believe is more suitable
                      – lab bhattacharjee
                      8 hours ago










                    • However, is it a bad way?
                      – Hamilton
                      8 hours ago










                    • Rationalization may be costly in some cases
                      – lab bhattacharjee
                      8 hours ago










                    • @Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
                      – timtfj
                      6 hours ago















                    up vote
                    3
                    down vote













                    It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$



                    $|2+2i|=sqrt{2^2+2^2}=2sqrt2$






                    share|cite|improve this answer





















                    • I first made the denominator real and then did my calculations to avoid any errors .
                      – Hamilton
                      8 hours ago












                    • @Hamilton, At least for modulus, the formula I believe is more suitable
                      – lab bhattacharjee
                      8 hours ago










                    • However, is it a bad way?
                      – Hamilton
                      8 hours ago










                    • Rationalization may be costly in some cases
                      – lab bhattacharjee
                      8 hours ago










                    • @Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
                      – timtfj
                      6 hours ago













                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$



                    $|2+2i|=sqrt{2^2+2^2}=2sqrt2$






                    share|cite|improve this answer












                    It is better to use $$left|dfrac abright|=dfrac{|a|}{|b|}$$



                    $|2+2i|=sqrt{2^2+2^2}=2sqrt2$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 8 hours ago









                    lab bhattacharjee

                    221k15154271




                    221k15154271












                    • I first made the denominator real and then did my calculations to avoid any errors .
                      – Hamilton
                      8 hours ago












                    • @Hamilton, At least for modulus, the formula I believe is more suitable
                      – lab bhattacharjee
                      8 hours ago










                    • However, is it a bad way?
                      – Hamilton
                      8 hours ago










                    • Rationalization may be costly in some cases
                      – lab bhattacharjee
                      8 hours ago










                    • @Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
                      – timtfj
                      6 hours ago


















                    • I first made the denominator real and then did my calculations to avoid any errors .
                      – Hamilton
                      8 hours ago












                    • @Hamilton, At least for modulus, the formula I believe is more suitable
                      – lab bhattacharjee
                      8 hours ago










                    • However, is it a bad way?
                      – Hamilton
                      8 hours ago










                    • Rationalization may be costly in some cases
                      – lab bhattacharjee
                      8 hours ago










                    • @Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
                      – timtfj
                      6 hours ago
















                    I first made the denominator real and then did my calculations to avoid any errors .
                    – Hamilton
                    8 hours ago






                    I first made the denominator real and then did my calculations to avoid any errors .
                    – Hamilton
                    8 hours ago














                    @Hamilton, At least for modulus, the formula I believe is more suitable
                    – lab bhattacharjee
                    8 hours ago




                    @Hamilton, At least for modulus, the formula I believe is more suitable
                    – lab bhattacharjee
                    8 hours ago












                    However, is it a bad way?
                    – Hamilton
                    8 hours ago




                    However, is it a bad way?
                    – Hamilton
                    8 hours ago












                    Rationalization may be costly in some cases
                    – lab bhattacharjee
                    8 hours ago




                    Rationalization may be costly in some cases
                    – lab bhattacharjee
                    8 hours ago












                    @Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
                    – timtfj
                    6 hours ago




                    @Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$.
                    – timtfj
                    6 hours ago


















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