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What additional information can be concluded about the function?

(a)A function analytic in the closed disk:$|z|\leq 4$ with $\inf|f(z)|=5$ on the circle $|z|=4$ and with $f(1)=i$

(b)A function analytic in the closed disk:$|z|\leq 1$ with $\sup|f(z)|=2$ on the circle $|z|=1$ and with $f(0)=-2i$

My Attempt.

(a) $f(z)=i, \forall z:|z|\leq 4$, By Minimum modulus principle.

(b) $f(z)=-2i, \forall z:|z|\leq 1$, By maximum modulus principle. Am I correct? Is there any flaw in applying Maximum Modulus and Minimum modulus theorem?

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  • $\begingroup$ (b) is correct, (a) is incorrect. To fix (a), think about the following. What are the hypotheses for the Minimum Modulus Principle and what is the conclusion? If the conclusion doesn't hold, what must be true about the hypotheses? $\endgroup$
    – zugzug
    Nov 3, 2020 at 1:00
  • $\begingroup$ Suppose f(z) is analytic in a domain D, and that $f(z) \neq 0$ in D. Then |f(z)| cannot attain a minimum in D unless f(z) is constant. If f(z) is also continuous on D, $\overline D$ compact, then |f(z)| attains a minimum on the boundary. $\endgroup$
    – Unknown x
    Nov 3, 2020 at 2:11
  • $\begingroup$ Here $f(z)\neq 0$ can't say. $\endgroup$
    – Unknown x
    Nov 3, 2020 at 2:11
  • $\begingroup$ @zugzug But what additional information can be drawn? $\endgroup$
    – Unknown x
    Nov 3, 2020 at 2:27
  • $\begingroup$ If $f(z)=i$, so $|f(z)|=1$ for all $|z|\leq 4$, how can the infinimum of the modulus be $5$? $\endgroup$
    – zugzug
    Nov 3, 2020 at 2:27

1 Answer 1

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I'm going to put our conversation into an answer that I hope will help for (a).

If you have a statement of the form $P$ implies $Q$. This is logically equivalent to not $Q$ implies not $P$.

Minimum modulus principle:: If $f$ is analytic on and inside $D$ and $f(z)\neq 0$ for all $z$, then $f$ attains its minimum modulus on the boundary.

P: $f$ is analytic on and inside $D$ and $f(z)\neq 0$ for all $z$.

Q: $f$ attains its minimum modulus on the boundary.

In your problem, $f$ does not attain its minimum modulus on the boundary. We know this since $\inf |f(z)|=5$ on the boundary, but for a point inside, $|f(1)|=|i|=1$.

Hence, you have not Q, which implies not P.

Not P: $f$ is not analytic on and inside $D$ or $f(z)=0$ for some $z$. What can you conclude?

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  • $\begingroup$ Henc, $f(z)=0$ for all $z$. right? $\endgroup$
    – Unknown x
    Nov 3, 2020 at 2:43
  • $\begingroup$ but it contradict the fact that $f(1)=i$. $\endgroup$
    – Unknown x
    Nov 3, 2020 at 2:44
  • $\begingroup$ The negation of $f(z)\neq 0$ for all $z$ is $f(z)=0$ for some $z$. $\endgroup$
    – zugzug
    Nov 3, 2020 at 2:48
  • $\begingroup$ okay. you mean conclusion is only, f has a zero in D. Right? $\endgroup$
    – Unknown x
    Nov 3, 2020 at 2:49
  • $\begingroup$ yes, at least one. $\endgroup$
    – zugzug
    Nov 3, 2020 at 2:55

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