Show that $|z_1 + z_2|^2 < (1+C)|z_1|^2 + left(1 + frac{1}{C}right) |z_2|^2$












2















Let $z_1$ and $z_2$ be two complex numbers. Show that there exists $C > 0$ with
$$
|z_1 + z_2|^2 < (1+C)|z_1|^2 + left(1 + frac{1}{C}right) |z_2|^2.
$$




I tried to simplify the L.H.S and R.H.S, SNF I was finally left to compare between a real number and a complex number
I really couldn't think of anything else. Please help.






complex-analysis inequality complex-numbers







share|cite|improve this question




















  • 2




    $z_1 = z_2 = c = 1$ gives $4 < 4$, so probably the inequality should be $leq$ instead of $<$ or $c neq 1$.
    – Random Jack
    Jan 3 '16 at 13:44
















2















Let $z_1$ and $z_2$ be two complex numbers. Show that there exists $C > 0$ with
$$
|z_1 + z_2|^2 < (1+C)|z_1|^2 + left(1 + frac{1}{C}right) |z_2|^2.
$$




I tried to simplify the L.H.S and R.H.S, SNF I was finally left to compare between a real number and a complex number
I really couldn't think of anything else. Please help.






complex-analysis inequality complex-numbers







share|cite|improve this question




















  • 2




    $z_1 = z_2 = c = 1$ gives $4 < 4$, so probably the inequality should be $leq$ instead of $<$ or $c neq 1$.
    – Random Jack
    Jan 3 '16 at 13:44














2












2








2


2






Let $z_1$ and $z_2$ be two complex numbers. Show that there exists $C > 0$ with
$$
|z_1 + z_2|^2 < (1+C)|z_1|^2 + left(1 + frac{1}{C}right) |z_2|^2.
$$




I tried to simplify the L.H.S and R.H.S, SNF I was finally left to compare between a real number and a complex number
I really couldn't think of anything else. Please help.






complex-analysis inequality complex-numbers







share|cite|improve this question
















Let $z_1$ and $z_2$ be two complex numbers. Show that there exists $C > 0$ with
$$
|z_1 + z_2|^2 < (1+C)|z_1|^2 + left(1 + frac{1}{C}right) |z_2|^2.
$$




I tried to simplify the L.H.S and R.H.S, SNF I was finally left to compare between a real number and a complex number
I really couldn't think of anything else. Please help.






complex-analysis inequality complex-numbers




complex-analysis inequality complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 '16 at 13:54









Gyumin Roh

2,935820




2,935820










asked Jan 3 '16 at 13:34









TejusTejus

4051416




4051416








  • 2




    $z_1 = z_2 = c = 1$ gives $4 < 4$, so probably the inequality should be $leq$ instead of $<$ or $c neq 1$.
    – Random Jack
    Jan 3 '16 at 13:44














  • 2




    $z_1 = z_2 = c = 1$ gives $4 < 4$, so probably the inequality should be $leq$ instead of $<$ or $c neq 1$.
    – Random Jack
    Jan 3 '16 at 13:44








2




2




$z_1 = z_2 = c = 1$ gives $4 < 4$, so probably the inequality should be $leq$ instead of $<$ or $c neq 1$.
– Random Jack
Jan 3 '16 at 13:44




$z_1 = z_2 = c = 1$ gives $4 < 4$, so probably the inequality should be $leq$ instead of $<$ or $c neq 1$.
– Random Jack
Jan 3 '16 at 13:44










2 Answers
2






active

oldest

votes


















3














HINT: The given inequality can be rewritten as the obvious inequality



$$left|sqrt{c}z_1-frac{1}{sqrt{c}}z_2right|^2ge 0$$






share|cite|improve this answer



















  • 1




    I don't understand this .. Can you please explain
    – Tejus
    Jan 3 '16 at 13:42






  • 1




    @Tejus: Just multiply both inequalities out and check that they're equivalent.
    – Matt L.
    Jan 3 '16 at 13:45










  • Got it!!! thanksss
    – Tejus
    Jan 3 '16 at 13:45



















2














Hint:
$$lvert z_1+ z_2rvert^2 +lvert z_1 - z_2rvert^2=2lvert z_1rvert^2 +2lvert z_2rvert^2. $$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    HINT: The given inequality can be rewritten as the obvious inequality



    $$left|sqrt{c}z_1-frac{1}{sqrt{c}}z_2right|^2ge 0$$






    share|cite|improve this answer



















    • 1




      I don't understand this .. Can you please explain
      – Tejus
      Jan 3 '16 at 13:42






    • 1




      @Tejus: Just multiply both inequalities out and check that they're equivalent.
      – Matt L.
      Jan 3 '16 at 13:45










    • Got it!!! thanksss
      – Tejus
      Jan 3 '16 at 13:45
















    3














    HINT: The given inequality can be rewritten as the obvious inequality



    $$left|sqrt{c}z_1-frac{1}{sqrt{c}}z_2right|^2ge 0$$






    share|cite|improve this answer



















    • 1




      I don't understand this .. Can you please explain
      – Tejus
      Jan 3 '16 at 13:42






    • 1




      @Tejus: Just multiply both inequalities out and check that they're equivalent.
      – Matt L.
      Jan 3 '16 at 13:45










    • Got it!!! thanksss
      – Tejus
      Jan 3 '16 at 13:45














    3












    3








    3






    HINT: The given inequality can be rewritten as the obvious inequality



    $$left|sqrt{c}z_1-frac{1}{sqrt{c}}z_2right|^2ge 0$$






    share|cite|improve this answer














    HINT: The given inequality can be rewritten as the obvious inequality



    $$left|sqrt{c}z_1-frac{1}{sqrt{c}}z_2right|^2ge 0$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 3 '16 at 13:46

























    answered Jan 3 '16 at 13:41









    Matt L.Matt L.

    8,871822




    8,871822








    • 1




      I don't understand this .. Can you please explain
      – Tejus
      Jan 3 '16 at 13:42






    • 1




      @Tejus: Just multiply both inequalities out and check that they're equivalent.
      – Matt L.
      Jan 3 '16 at 13:45










    • Got it!!! thanksss
      – Tejus
      Jan 3 '16 at 13:45














    • 1




      I don't understand this .. Can you please explain
      – Tejus
      Jan 3 '16 at 13:42






    • 1




      @Tejus: Just multiply both inequalities out and check that they're equivalent.
      – Matt L.
      Jan 3 '16 at 13:45










    • Got it!!! thanksss
      – Tejus
      Jan 3 '16 at 13:45








    1




    1




    I don't understand this .. Can you please explain
    – Tejus
    Jan 3 '16 at 13:42




    I don't understand this .. Can you please explain
    – Tejus
    Jan 3 '16 at 13:42




    1




    1




    @Tejus: Just multiply both inequalities out and check that they're equivalent.
    – Matt L.
    Jan 3 '16 at 13:45




    @Tejus: Just multiply both inequalities out and check that they're equivalent.
    – Matt L.
    Jan 3 '16 at 13:45












    Got it!!! thanksss
    – Tejus
    Jan 3 '16 at 13:45




    Got it!!! thanksss
    – Tejus
    Jan 3 '16 at 13:45











    2














    Hint:
    $$lvert z_1+ z_2rvert^2 +lvert z_1 - z_2rvert^2=2lvert z_1rvert^2 +2lvert z_2rvert^2. $$






    share|cite|improve this answer


























      2














      Hint:
      $$lvert z_1+ z_2rvert^2 +lvert z_1 - z_2rvert^2=2lvert z_1rvert^2 +2lvert z_2rvert^2. $$






      share|cite|improve this answer
























        2












        2








        2






        Hint:
        $$lvert z_1+ z_2rvert^2 +lvert z_1 - z_2rvert^2=2lvert z_1rvert^2 +2lvert z_2rvert^2. $$






        share|cite|improve this answer












        Hint:
        $$lvert z_1+ z_2rvert^2 +lvert z_1 - z_2rvert^2=2lvert z_1rvert^2 +2lvert z_2rvert^2. $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 '16 at 14:04









        BernardBernard

        118k639112




        118k639112






























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