{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

math185f09-hw8sol

# math185f09-hw8sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 8 SOLUTIONS 1. Let Ω be a region containing D (0 , 1). Let f : Ω → C be analytic. (a) Let M > 0 be a constant. Suppose | f ( z ) | ≥ M for all z ∈ ∂D (0 , 1) and | f (0) | < M . Show that f has at least one zero in D (0 , 1). Solution. Suppose not and f has no zeros in D (0 , 1). Then by the minimum modulus theorem (or Corollary 4.15 ), M ≤ min z ∈ ∂D (0 , 1) | f ( z ) | = min z ∈ D (0 , 1) | f ( z ) | ≤ | f (0) | < M, a contradiction. So f must have at least one zero in D (0 , 1). (b) Suppose | f ( z 2 ) | ≥ | f ( z ) | for all z ∈ D (0 , 1). Show that f is constant on D (0 , 1). Solution. Let 0 < r < 1. By the compactness of ∂D (0 ,r 2 ), there exists z ∈ ∂D (0 ,r 2 ) such that | f ( z ) | = max z ∈ ∂D (0 ,r 2 ) | f ( z ) | . By the maximum modulus theorem, | f ( z ) | = max z ∈ D (0 ,r 2 ) | f ( z ) | . But since | f ( z 2 ) | ≥ | f ( z ) | , | f ( z ) | = max z ∈ D (0 ,r 2 ) | f ( z ) | = max z ∈ D (0 ,r ) | f ( z 2 ) | ≥ max z ∈ D (0 ,r ) | f ( z ) | . (1.1) Note that r 2 < r and so z ∈ ∂D (0 ,r 2 ) ⊂ D (0 ,r ). So | f ( z ) | ≤ max z ∈ D (0 ,r ) | f ( z ) |...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

math185f09-hw8sol - MATH 185 COMPLEX ANALYSIS FALL 2009/10...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online