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Math Help - maximum modulus (complex analysis)

  1. #1
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    maximum modulus (complex analysis)

    For complex number z, how to show that max|z|<=1 |azn+b|=|a|+|b| where a, b are real number?
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    Re: maximum modulus (complex analysis)

    Quote Originally Posted by shyuan View Post
    For complex number z, how to show that max|z|<=1 |azn+b|=|a|+|b| where a, b are real number?
    Use the triangle inequality: \displaystyle \begin{align*} |x + y| \leq |x| + |y| \end{align*}. So

    \displaystyle \begin{align*} \left|a\,z^n + b\right| &\leq \left|a\,z^n\right| + |b| \\ &= |a|\left|z^n\right| + |b| \\ &= |a||z|^n + |b| \\ &\leq |a| \cdot 1^n + |b| \textrm{ since } |z| \leq 1 \\ &= |a| + |b|  \end{align*}

    Therefore \displaystyle \begin{align*} \max_{|z| \leq 1} \left|a\,z^n + b\right| = |a| + |b| \end{align*}.
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    Re: maximum modulus (complex analysis)

    but this only prove that |a|+|b| is on eof the upper bound, not necessary to be the maximum value right?
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    Re: maximum modulus (complex analysis)

    Quote Originally Posted by shyuan View Post
    but this only prove that |a|+|b| is on eof the upper bound, not necessary to be the maximum value right?
    If it's the least upper bound (which it is) and a possible value (which it also is), then it is the maximum value.
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    Re: maximum modulus (complex analysis)

    thx for ur help bt cn i ask, hw to confirm that it is already the least upper bound?
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    Re: maximum modulus (complex analysis)

    Quote Originally Posted by shyuan View Post
    this only prove that |a|+|b| is on eof the upper bound, not necessary to be the maximum value right?
    Given a, b and n, try to find a z with |z| = 1 such that |ax^n + b| = |a| + |b|.
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    Re: maximum modulus (complex analysis)

    Thank you very much. I solved my problem with the helps from you guys. Thanks.
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