Prove if $|z-1| le frac12$ then $|fracz - 1| le |z - 1|sqrt2$ for z complex

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As the title, problem is to prove if $|z-1| le frac12$ then $|fracz - 1| le |z - 1|sqrt2$ for complex $z$.



I noticed using the condition we can get $|fracz - 1| le fracsqrt22$.



Then can square both sides to get $2(1 - fracoperatornameRez$) $le frac12$.



From here I try further manipulations but am stuck and unable to proceed.



How would you solve this problem?




inequality complex-numbers




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  • Let $z=re^itheta$, and then you need to prove $|e^itheta-1|leq|re^itheta- 1|sqrt2$. Here $1/2leq rleq 3/2$.
    – Riemann
    Aug 6 at 9:57















up vote
0
down vote

favorite












As the title, problem is to prove if $|z-1| le frac12$ then $|fracz - 1| le |z - 1|sqrt2$ for complex $z$.



I noticed using the condition we can get $|fracz - 1| le fracsqrt22$.



Then can square both sides to get $2(1 - fracoperatornameRez$) $le frac12$.



From here I try further manipulations but am stuck and unable to proceed.



How would you solve this problem?




inequality complex-numbers




share|cite|improve this question



















  • Let $z=re^itheta$, and then you need to prove $|e^itheta-1|leq|re^itheta- 1|sqrt2$. Here $1/2leq rleq 3/2$.
    – Riemann
    Aug 6 at 9:57













up vote
0
down vote

favorite









up vote
0
down vote

favorite











As the title, problem is to prove if $|z-1| le frac12$ then $|fracz - 1| le |z - 1|sqrt2$ for complex $z$.



I noticed using the condition we can get $|fracz - 1| le fracsqrt22$.



Then can square both sides to get $2(1 - fracoperatornameRez$) $le frac12$.



From here I try further manipulations but am stuck and unable to proceed.



How would you solve this problem?




inequality complex-numbers




share|cite|improve this question











As the title, problem is to prove if $|z-1| le frac12$ then $|fracz - 1| le |z - 1|sqrt2$ for complex $z$.



I noticed using the condition we can get $|fracz - 1| le fracsqrt22$.



Then can square both sides to get $2(1 - fracoperatornameRez$) $le frac12$.



From here I try further manipulations but am stuck and unable to proceed.



How would you solve this problem?





inequality complex-numbers





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 6 at 9:40









trynalearn

540213




540213











  • Let $z=re^itheta$, and then you need to prove $|e^itheta-1|leq|re^itheta- 1|sqrt2$. Here $1/2leq rleq 3/2$.
    – Riemann
    Aug 6 at 9:57

















  • Let $z=re^itheta$, and then you need to prove $|e^itheta-1|leq|re^itheta- 1|sqrt2$. Here $1/2leq rleq 3/2$.
    – Riemann
    Aug 6 at 9:57
















Let $z=re^itheta$, and then you need to prove $|e^itheta-1|leq|re^itheta- 1|sqrt2$. Here $1/2leq rleq 3/2$.
– Riemann
Aug 6 at 9:57





Let $z=re^itheta$, and then you need to prove $|e^itheta-1|leq|re^itheta- 1|sqrt2$. Here $1/2leq rleq 3/2$.
– Riemann
Aug 6 at 9:57











2 Answers
2






active

oldest

votes

















up vote
2
down vote













Let $z=re^itheta$, and then you need to prove $|e^itheta-1|^2leq2|re^itheta- 1|^2$. Here $1/2leq rleq 3/2,-pi/6leq thetaleqpi/6$.
$$LHS=2-2costheta;$$
$$RHS=2(r^2-2rcostheta+1);$$
$$frac12(RSH-LHS)=r^2-2rcostheta+costhetageq 0.$$






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    up vote
    -1
    down vote













    Let $arg z=theta.$



    Thus, it's enough to show that:
    $$|costheta+isintheta-1|leqfrac1sqrt2$$ or
    $$2-2costhetaleqfrac12$$ or
    $$costhetageqfrac34,$$ which is
    $$0^circleq thetaleqarccosfrac34$$ or
    $$360^circ-arccosfrac34leq theta<360^circ,$$
    which is true because by the given
    $$0^circleq thetaleq30^circ$$ or
    $$330^circleq theta<360^circ.$$






    share|cite|improve this answer





















    • Why someone down voted. Explain please.
      – Michael Rozenberg
      Aug 6 at 17:05










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    2 Answers
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    2 Answers
    2






    active

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    active

    oldest

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    up vote
    2
    down vote













    Let $z=re^itheta$, and then you need to prove $|e^itheta-1|^2leq2|re^itheta- 1|^2$. Here $1/2leq rleq 3/2,-pi/6leq thetaleqpi/6$.
    $$LHS=2-2costheta;$$
    $$RHS=2(r^2-2rcostheta+1);$$
    $$frac12(RSH-LHS)=r^2-2rcostheta+costhetageq 0.$$






    share|cite|improve this answer

























      up vote
      2
      down vote













      Let $z=re^itheta$, and then you need to prove $|e^itheta-1|^2leq2|re^itheta- 1|^2$. Here $1/2leq rleq 3/2,-pi/6leq thetaleqpi/6$.
      $$LHS=2-2costheta;$$
      $$RHS=2(r^2-2rcostheta+1);$$
      $$frac12(RSH-LHS)=r^2-2rcostheta+costhetageq 0.$$






      share|cite|improve this answer























        up vote
        2
        down vote










        up vote
        2
        down vote









        Let $z=re^itheta$, and then you need to prove $|e^itheta-1|^2leq2|re^itheta- 1|^2$. Here $1/2leq rleq 3/2,-pi/6leq thetaleqpi/6$.
        $$LHS=2-2costheta;$$
        $$RHS=2(r^2-2rcostheta+1);$$
        $$frac12(RSH-LHS)=r^2-2rcostheta+costhetageq 0.$$






        share|cite|improve this answer













        Let $z=re^itheta$, and then you need to prove $|e^itheta-1|^2leq2|re^itheta- 1|^2$. Here $1/2leq rleq 3/2,-pi/6leq thetaleqpi/6$.
        $$LHS=2-2costheta;$$
        $$RHS=2(r^2-2rcostheta+1);$$
        $$frac12(RSH-LHS)=r^2-2rcostheta+costhetageq 0.$$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 6 at 10:07









        Riemann

        2,2941217




        2,2941217




















            up vote
            -1
            down vote













            Let $arg z=theta.$



            Thus, it's enough to show that:
            $$|costheta+isintheta-1|leqfrac1sqrt2$$ or
            $$2-2costhetaleqfrac12$$ or
            $$costhetageqfrac34,$$ which is
            $$0^circleq thetaleqarccosfrac34$$ or
            $$360^circ-arccosfrac34leq theta<360^circ,$$
            which is true because by the given
            $$0^circleq thetaleq30^circ$$ or
            $$330^circleq theta<360^circ.$$






            share|cite|improve this answer





















            • Why someone down voted. Explain please.
              – Michael Rozenberg
              Aug 6 at 17:05














            up vote
            -1
            down vote













            Let $arg z=theta.$



            Thus, it's enough to show that:
            $$|costheta+isintheta-1|leqfrac1sqrt2$$ or
            $$2-2costhetaleqfrac12$$ or
            $$costhetageqfrac34,$$ which is
            $$0^circleq thetaleqarccosfrac34$$ or
            $$360^circ-arccosfrac34leq theta<360^circ,$$
            which is true because by the given
            $$0^circleq thetaleq30^circ$$ or
            $$330^circleq theta<360^circ.$$






            share|cite|improve this answer





















            • Why someone down voted. Explain please.
              – Michael Rozenberg
              Aug 6 at 17:05












            up vote
            -1
            down vote










            up vote
            -1
            down vote









            Let $arg z=theta.$



            Thus, it's enough to show that:
            $$|costheta+isintheta-1|leqfrac1sqrt2$$ or
            $$2-2costhetaleqfrac12$$ or
            $$costhetageqfrac34,$$ which is
            $$0^circleq thetaleqarccosfrac34$$ or
            $$360^circ-arccosfrac34leq theta<360^circ,$$
            which is true because by the given
            $$0^circleq thetaleq30^circ$$ or
            $$330^circleq theta<360^circ.$$






            share|cite|improve this answer













            Let $arg z=theta.$



            Thus, it's enough to show that:
            $$|costheta+isintheta-1|leqfrac1sqrt2$$ or
            $$2-2costhetaleqfrac12$$ or
            $$costhetageqfrac34,$$ which is
            $$0^circleq thetaleqarccosfrac34$$ or
            $$360^circ-arccosfrac34leq theta<360^circ,$$
            which is true because by the given
            $$0^circleq thetaleq30^circ$$ or
            $$330^circleq theta<360^circ.$$







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Aug 6 at 10:06









            Michael Rozenberg

            88.2k1579180




            88.2k1579180











            • Why someone down voted. Explain please.
              – Michael Rozenberg
              Aug 6 at 17:05
















            • Why someone down voted. Explain please.
              – Michael Rozenberg
              Aug 6 at 17:05















            Why someone down voted. Explain please.
            – Michael Rozenberg
            Aug 6 at 17:05




            Why someone down voted. Explain please.
            – Michael Rozenberg
            Aug 6 at 17:05












             

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