Results 1 to 5 of 5
Like Tree3Thanks
  • 1 Post By Prove It
  • 1 Post By romsek
  • 1 Post By Plato

Thread: Maximum Modulus of Complex Function

  1. #1
    wps
    wps is offline
    Newbie
    Joined
    Feb 2017
    From
    Earth
    Posts
    9
    Thanks
    2

    Maximum Modulus of Complex Function

    How do we find the maximum value of |z^2+2z-3| in |z|=1?
    Can we just simply use the triangle inequality?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,841
    Thanks
    1931

    Re: Maximum Modulus of Complex Function

    You can use the triangle inequality to get an upper bound for this, but I am unsure whether that is the maximum value though...
    Thanks from wps
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    5,774
    Thanks
    2417

    Re: Maximum Modulus of Complex Function

    $|z|=1 \Rightarrow z=\cos(\theta) + i \sin(\theta)$

    Find a formula for $|z^2 + 2z - 3|^2$ and plug the above $z$ into that.

    That gives you a formula in $\theta$ for the square of the modulus.

    You can maximize this and then double check that the answers correspond to a maximum for the modulus itself.

    Spoiler:
    Letting software do all the grubby work

    $|z^2 + 2z - 3|^2 = x^4+4 x^3+2 x^2 \left(y^2-1\right)+4 x \left(y^2-3\right)+y^4+10 y^2+9$

    with $z = \cos(\theta + i \sin(\theta)$ this becomes (after a ton of computer aided simpilfication)

    $|z^2 + 2z - 3|^2 = 8 \sin ^2\left(\frac{\theta }{2}\right) (3 \cos (\theta )+5)$

    $\dfrac{d}{d\theta} \left( 8 \sin ^2\left(\frac{\theta }{2}\right) (3 \cos (\theta )+5) \right)=8 \sin \left(\frac{\theta }{2}\right) \left(5 \cos \left(\frac{\theta }{2}\right)+3 \cos \left(\frac{3 \theta }{2}\right)\right)$

    setting this equal to 0 and solving for $\theta$ we get

    $\theta = 0, ~\pi, ~\pi - \tan^{-1}(2\sqrt{2}), -\pi - \tan^{-1}(2\sqrt{2})$

    I leave it to you to determine which of these are maxima and which are local minima using the 2nd derivative.
    Thanks from wps
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,364
    Thanks
    2682
    Awards
    1

    Re: Maximum Modulus of Complex Function

    Quote Originally Posted by wps View Post
    How do we find the maximum value of |z^2+2z-3| in |z|=1?
    Can we just simply use the triangle inequality?
    $|z^2|=1~,~|2z|=2~\&,~|-3|=3$ so $|z^2+2z-3|\le 6$ is true.

    Quote Originally Posted by Prove It View Post
    You can use the triangle inequality to get an upper bound for this, but I am unsure whether that is the maximum value though...
    Good point. If $z=\bf{i}$ then
    $|\bf{i}^2+2\bf{i}-3|=|-4+2\bf{i}|=\sqrt{16+4}=2\sqrt5<4.5$

    On the other hand, $z=-1$ then $\left|(-1)^2+2(-1)-3\right|=4$

    If $z=a+b\bf{i}$ such that |a^2+b^2=1| then $\left|z^2+2z-3\right|^2=\left(a^2-b^2+2a-3\right)^2+\left(2ab+2a\right)^2$.
    Can we maximize this?
    Last edited by Plato; Mar 27th 2017 at 05:37 PM.
    Thanks from wps
    Follow Math Help Forum on Facebook and Google+

  5. #5
    wps
    wps is offline
    Newbie
    Joined
    Feb 2017
    From
    Earth
    Posts
    9
    Thanks
    2

    Re: Maximum Modulus of Complex Function

    Quote Originally Posted by Prove It View Post
    You can use the triangle inequality to get an upper bound for this, but I am unsure whether that is the maximum value though...
    Quote Originally Posted by Plato View Post
    $|z^2|=1~,~|2z|=2~\&,~|-3|=3$ so $|z^2+2z-3|\le 6$ is true.


    Good point. If $z=\bf{i}$ then
    $|\bf{i}^2+2\bf{i}-3|=|-4+2\bf{i}|=\sqrt{16+4}=2\sqrt5<4.5$

    On the other hand, $z=-1$ then $\left|(-1)^2+2(-1)-3\right|=4$

    If $z=a+b\bf{i}$ such that |a^2+b^2=1| then $\left|z^2+2z-3\right|^2=\left(a^2-b^2+2a-3\right)^2+\left(2ab+2a\right)^2$ $\rightarrow$ $\mathbf{\left(2ab+2b\right)^2$}.
    Can we maximize this?
    Yes,got it.
    $\left|z^2+2z-3\right|^2=\left( a^2-(1-a^2)+2a-3\right)^2+(1-a^2)\left(2a+2\right)^2$
    So we need to maximize a one variable function.
    Thanks everyone
    Last edited by wps; Mar 27th 2017 at 06:33 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. maximum modulus (complex analysis)
    Posted in the Calculus Forum
    Replies: 6
    Last Post: May 2nd 2012, 04:37 PM
  2. Replies: 2
    Last Post: Dec 5th 2011, 10:47 PM
  3. Maximum Modulus Theorem
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Nov 15th 2011, 12:50 PM
  4. Maximum modulus and mean values
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: Nov 24th 2009, 09:49 AM
  5. by using Maximum AND minimum modulus principle
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Aug 5th 2009, 12:22 AM

Search tags for this page


/mathhelpforum @mathhelpforum