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Let $f(z)=2z^2-1$.Then what is the maximum value of $|f(z)|$ on the unit disc

$D=\{z\in C:|z|\le1\} $ equals

  1. $2$
  2. $3$
  3. $1$

  4. $3$ more than minimum value

    This question can have more than 1 answer? I dont know how to proceed

amit
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  • The maximum is attained on the boundary. You can proceed from there. There are intuitive arguments as well, though I don't think the ones I'm thinking of can be made rigorous in any more efficient way. – zibadawa timmy Aug 27 '14 at 13:08

1 Answers1

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There cannot be more than one maximum (or minimum) value. There can be more than one place in which such values occur, but not more than one maximum (or minimum) value.

The maximum occurs on the boundary (why? use a well known theorem). Try to maximize $z^2-1$ first on $D$ to see what is going on. This is asking, for $|z|=1$ how far can $z^2$ be from $1$. Think geometrically on the unit circle, using the fact that when multiplying complex numbers, you multiply their magnitudes and add their angles. What is the number on the unit circle farthest from $1$? Can we square some number $z$ with $|z|=1$ to obtain this value?

Once you have solved the $z^2-1$ case you will see the $2z^2-1$ case is similar.

Seth
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