Maximum Modulus of Complex Function












0












$begingroup$


here's a question I'm working on that I'm a bit stuck on.

Let:
$f(z) = frac{z^2}{z + 2}$

Find the maximum value of $|f(z)|$ as $z$ varies over the unit disc.



Since $f(z)$ is analytic $forall z$ in the region,and $f(z)$ is non-constant, the maximum of this function will be on the boundary of the unit disc (Is this the correct application of the theorem?)

If we let $z = e^{it}$ then
$|f(e^{it})| = frac{|e^{it}|^2}{|e^{it} + 2|}$
$|f(e^{it})| = frac{1}{|e^{it} + 2|}$



However, I'm stuck here.

The answer given is $1$, but I'm not quite sure how to figure that out. There's quite a bit of confusion on my end as well, as I'm not fully certain if I'm on the right track.



Any help would be appreciated.

Thanks.






complex-analysis maximum-principle







share|cite|improve this question









$endgroup$












  • $begingroup$
    Uhm... I wonder how we could maximise $frac{1}{lvert e^{it}+2rvert}$ over $tin[0,2pi]$. If only it were a function from $Bbb R$ to $Bbb R$ maybe I could do it...
    $endgroup$
    – Saucy O'Path
    Dec 13 '18 at 0:50






  • 1




    $begingroup$
    If you want to maximize $frac {1}{|e^{it} + 2|}$ you will want to minimize $|e^{it} + 2|$
    $endgroup$
    – Doug M
    Dec 13 '18 at 0:52










  • $begingroup$
    Would it be: $frac{1}{|e^{it} + 2|} leq frac{1}{|e^{it}|} = 1$ ?
    $endgroup$
    – user2965071
    Dec 13 '18 at 0:53










  • $begingroup$
    @user2965071 Since $1$ is the answer, then that inequality is true. I don't see how you prove that $lvert e^{it}+2rvertge lvert e^{it}rvert$, though.
    $endgroup$
    – Saucy O'Path
    Dec 13 '18 at 1:00












  • $begingroup$
    $|2+e^{it}|=|2+cos t +i sin t| $ which means that for $t=pi$ you get $|2+e^{it}|=min=1$
    $endgroup$
    – G Cab
    Dec 13 '18 at 1:51
















0












$begingroup$


here's a question I'm working on that I'm a bit stuck on.

Let:
$f(z) = frac{z^2}{z + 2}$

Find the maximum value of $|f(z)|$ as $z$ varies over the unit disc.



Since $f(z)$ is analytic $forall z$ in the region,and $f(z)$ is non-constant, the maximum of this function will be on the boundary of the unit disc (Is this the correct application of the theorem?)

If we let $z = e^{it}$ then
$|f(e^{it})| = frac{|e^{it}|^2}{|e^{it} + 2|}$
$|f(e^{it})| = frac{1}{|e^{it} + 2|}$



However, I'm stuck here.

The answer given is $1$, but I'm not quite sure how to figure that out. There's quite a bit of confusion on my end as well, as I'm not fully certain if I'm on the right track.



Any help would be appreciated.

Thanks.






complex-analysis maximum-principle







share|cite|improve this question









$endgroup$












  • $begingroup$
    Uhm... I wonder how we could maximise $frac{1}{lvert e^{it}+2rvert}$ over $tin[0,2pi]$. If only it were a function from $Bbb R$ to $Bbb R$ maybe I could do it...
    $endgroup$
    – Saucy O'Path
    Dec 13 '18 at 0:50






  • 1




    $begingroup$
    If you want to maximize $frac {1}{|e^{it} + 2|}$ you will want to minimize $|e^{it} + 2|$
    $endgroup$
    – Doug M
    Dec 13 '18 at 0:52










  • $begingroup$
    Would it be: $frac{1}{|e^{it} + 2|} leq frac{1}{|e^{it}|} = 1$ ?
    $endgroup$
    – user2965071
    Dec 13 '18 at 0:53










  • $begingroup$
    @user2965071 Since $1$ is the answer, then that inequality is true. I don't see how you prove that $lvert e^{it}+2rvertge lvert e^{it}rvert$, though.
    $endgroup$
    – Saucy O'Path
    Dec 13 '18 at 1:00












  • $begingroup$
    $|2+e^{it}|=|2+cos t +i sin t| $ which means that for $t=pi$ you get $|2+e^{it}|=min=1$
    $endgroup$
    – G Cab
    Dec 13 '18 at 1:51














0












0








0





$begingroup$


here's a question I'm working on that I'm a bit stuck on.

Let:
$f(z) = frac{z^2}{z + 2}$

Find the maximum value of $|f(z)|$ as $z$ varies over the unit disc.



Since $f(z)$ is analytic $forall z$ in the region,and $f(z)$ is non-constant, the maximum of this function will be on the boundary of the unit disc (Is this the correct application of the theorem?)

If we let $z = e^{it}$ then
$|f(e^{it})| = frac{|e^{it}|^2}{|e^{it} + 2|}$
$|f(e^{it})| = frac{1}{|e^{it} + 2|}$



However, I'm stuck here.

The answer given is $1$, but I'm not quite sure how to figure that out. There's quite a bit of confusion on my end as well, as I'm not fully certain if I'm on the right track.



Any help would be appreciated.

Thanks.






complex-analysis maximum-principle







share|cite|improve this question









$endgroup$




here's a question I'm working on that I'm a bit stuck on.

Let:
$f(z) = frac{z^2}{z + 2}$

Find the maximum value of $|f(z)|$ as $z$ varies over the unit disc.



Since $f(z)$ is analytic $forall z$ in the region,and $f(z)$ is non-constant, the maximum of this function will be on the boundary of the unit disc (Is this the correct application of the theorem?)

If we let $z = e^{it}$ then
$|f(e^{it})| = frac{|e^{it}|^2}{|e^{it} + 2|}$
$|f(e^{it})| = frac{1}{|e^{it} + 2|}$



However, I'm stuck here.

The answer given is $1$, but I'm not quite sure how to figure that out. There's quite a bit of confusion on my end as well, as I'm not fully certain if I'm on the right track.



Any help would be appreciated.

Thanks.






complex-analysis maximum-principle




complex-analysis maximum-principle






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 0:47









user2965071user2965071

1356




1356












  • $begingroup$
    Uhm... I wonder how we could maximise $frac{1}{lvert e^{it}+2rvert}$ over $tin[0,2pi]$. If only it were a function from $Bbb R$ to $Bbb R$ maybe I could do it...
    $endgroup$
    – Saucy O'Path
    Dec 13 '18 at 0:50






  • 1




    $begingroup$
    If you want to maximize $frac {1}{|e^{it} + 2|}$ you will want to minimize $|e^{it} + 2|$
    $endgroup$
    – Doug M
    Dec 13 '18 at 0:52










  • $begingroup$
    Would it be: $frac{1}{|e^{it} + 2|} leq frac{1}{|e^{it}|} = 1$ ?
    $endgroup$
    – user2965071
    Dec 13 '18 at 0:53










  • $begingroup$
    @user2965071 Since $1$ is the answer, then that inequality is true. I don't see how you prove that $lvert e^{it}+2rvertge lvert e^{it}rvert$, though.
    $endgroup$
    – Saucy O'Path
    Dec 13 '18 at 1:00












  • $begingroup$
    $|2+e^{it}|=|2+cos t +i sin t| $ which means that for $t=pi$ you get $|2+e^{it}|=min=1$
    $endgroup$
    – G Cab
    Dec 13 '18 at 1:51


















  • $begingroup$
    Uhm... I wonder how we could maximise $frac{1}{lvert e^{it}+2rvert}$ over $tin[0,2pi]$. If only it were a function from $Bbb R$ to $Bbb R$ maybe I could do it...
    $endgroup$
    – Saucy O'Path
    Dec 13 '18 at 0:50






  • 1




    $begingroup$
    If you want to maximize $frac {1}{|e^{it} + 2|}$ you will want to minimize $|e^{it} + 2|$
    $endgroup$
    – Doug M
    Dec 13 '18 at 0:52










  • $begingroup$
    Would it be: $frac{1}{|e^{it} + 2|} leq frac{1}{|e^{it}|} = 1$ ?
    $endgroup$
    – user2965071
    Dec 13 '18 at 0:53










  • $begingroup$
    @user2965071 Since $1$ is the answer, then that inequality is true. I don't see how you prove that $lvert e^{it}+2rvertge lvert e^{it}rvert$, though.
    $endgroup$
    – Saucy O'Path
    Dec 13 '18 at 1:00












  • $begingroup$
    $|2+e^{it}|=|2+cos t +i sin t| $ which means that for $t=pi$ you get $|2+e^{it}|=min=1$
    $endgroup$
    – G Cab
    Dec 13 '18 at 1:51
















$begingroup$
Uhm... I wonder how we could maximise $frac{1}{lvert e^{it}+2rvert}$ over $tin[0,2pi]$. If only it were a function from $Bbb R$ to $Bbb R$ maybe I could do it...
$endgroup$
– Saucy O'Path
Dec 13 '18 at 0:50




$begingroup$
Uhm... I wonder how we could maximise $frac{1}{lvert e^{it}+2rvert}$ over $tin[0,2pi]$. If only it were a function from $Bbb R$ to $Bbb R$ maybe I could do it...
$endgroup$
– Saucy O'Path
Dec 13 '18 at 0:50




1




1




$begingroup$
If you want to maximize $frac {1}{|e^{it} + 2|}$ you will want to minimize $|e^{it} + 2|$
$endgroup$
– Doug M
Dec 13 '18 at 0:52




$begingroup$
If you want to maximize $frac {1}{|e^{it} + 2|}$ you will want to minimize $|e^{it} + 2|$
$endgroup$
– Doug M
Dec 13 '18 at 0:52












$begingroup$
Would it be: $frac{1}{|e^{it} + 2|} leq frac{1}{|e^{it}|} = 1$ ?
$endgroup$
– user2965071
Dec 13 '18 at 0:53




$begingroup$
Would it be: $frac{1}{|e^{it} + 2|} leq frac{1}{|e^{it}|} = 1$ ?
$endgroup$
– user2965071
Dec 13 '18 at 0:53












$begingroup$
@user2965071 Since $1$ is the answer, then that inequality is true. I don't see how you prove that $lvert e^{it}+2rvertge lvert e^{it}rvert$, though.
$endgroup$
– Saucy O'Path
Dec 13 '18 at 1:00






$begingroup$
@user2965071 Since $1$ is the answer, then that inequality is true. I don't see how you prove that $lvert e^{it}+2rvertge lvert e^{it}rvert$, though.
$endgroup$
– Saucy O'Path
Dec 13 '18 at 1:00














$begingroup$
$|2+e^{it}|=|2+cos t +i sin t| $ which means that for $t=pi$ you get $|2+e^{it}|=min=1$
$endgroup$
– G Cab
Dec 13 '18 at 1:51




$begingroup$
$|2+e^{it}|=|2+cos t +i sin t| $ which means that for $t=pi$ you get $|2+e^{it}|=min=1$
$endgroup$
– G Cab
Dec 13 '18 at 1:51










2 Answers
2






active

oldest

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0












$begingroup$

The minimum of $mid e^{it}+2mid$ pretty clearly occurs when $t=pi$. That is, the maximum modulus is $1$.



You did the problem correctly (up to there).






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint: do this by common sense: you want to maximize $frac{1}{|e^{it} + 2|}$, which means minimizing $|e^{it} + 2|$, so draw the picture, and find a complex number on the unit circle whose distance from $z=-2$ is minimum.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      I like this kind of geometrical argument.
      $endgroup$
      – Lubin
      Dec 13 '18 at 3:45











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    The minimum of $mid e^{it}+2mid$ pretty clearly occurs when $t=pi$. That is, the maximum modulus is $1$.



    You did the problem correctly (up to there).






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The minimum of $mid e^{it}+2mid$ pretty clearly occurs when $t=pi$. That is, the maximum modulus is $1$.



      You did the problem correctly (up to there).






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The minimum of $mid e^{it}+2mid$ pretty clearly occurs when $t=pi$. That is, the maximum modulus is $1$.



        You did the problem correctly (up to there).






        share|cite|improve this answer









        $endgroup$



        The minimum of $mid e^{it}+2mid$ pretty clearly occurs when $t=pi$. That is, the maximum modulus is $1$.



        You did the problem correctly (up to there).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 2:40









        Chris CusterChris Custer

        12.9k3826




        12.9k3826























            1












            $begingroup$

            Hint: do this by common sense: you want to maximize $frac{1}{|e^{it} + 2|}$, which means minimizing $|e^{it} + 2|$, so draw the picture, and find a complex number on the unit circle whose distance from $z=-2$ is minimum.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I like this kind of geometrical argument.
              $endgroup$
              – Lubin
              Dec 13 '18 at 3:45
















            1












            $begingroup$

            Hint: do this by common sense: you want to maximize $frac{1}{|e^{it} + 2|}$, which means minimizing $|e^{it} + 2|$, so draw the picture, and find a complex number on the unit circle whose distance from $z=-2$ is minimum.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I like this kind of geometrical argument.
              $endgroup$
              – Lubin
              Dec 13 '18 at 3:45














            1












            1








            1





            $begingroup$

            Hint: do this by common sense: you want to maximize $frac{1}{|e^{it} + 2|}$, which means minimizing $|e^{it} + 2|$, so draw the picture, and find a complex number on the unit circle whose distance from $z=-2$ is minimum.






            share|cite|improve this answer









            $endgroup$



            Hint: do this by common sense: you want to maximize $frac{1}{|e^{it} + 2|}$, which means minimizing $|e^{it} + 2|$, so draw the picture, and find a complex number on the unit circle whose distance from $z=-2$ is minimum.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 13 '18 at 1:34









            MatematletaMatematleta

            10.7k2918




            10.7k2918








            • 1




              $begingroup$
              I like this kind of geometrical argument.
              $endgroup$
              – Lubin
              Dec 13 '18 at 3:45














            • 1




              $begingroup$
              I like this kind of geometrical argument.
              $endgroup$
              – Lubin
              Dec 13 '18 at 3:45








            1




            1




            $begingroup$
            I like this kind of geometrical argument.
            $endgroup$
            – Lubin
            Dec 13 '18 at 3:45




            $begingroup$
            I like this kind of geometrical argument.
            $endgroup$
            – Lubin
            Dec 13 '18 at 3:45


















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