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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2007/08 PROBLEM SET 7 SOLUTIONS As in the lectures, for an a ∈ D (0 , 1), we will write ϕ a for the function defined by ϕ a ( z ) = z- a 1- az . 1. Let f : C → C be an entire function. Show that if | f ( z ) | ≤ 1 | Re z | for all z ∈ C , then f ≡ 0. What happens if we replace the condition by | f ( z ) | ≤ 1 | Im z | for all z ∈ C ? Solution. This is of course almost identical to Proposition 7.3 in the text and an almost identical proof works. Let R > 0. Consider the function g ( z ) = ( z 2 + R 2 ) f ( z ) = ( z- Ri )( z + Ri ) f ( z ) . When | z | = R and Im z ≥ 0, let θ be the angle between the line perpendicular to the imaginary axis and the line passing through z and Ri . Then 0 ≤ θ ≤ π/ 4, and z- Ri Re z = sec θ ≤ √ 2 . When | z | = R and Im z < 0, let θ be the angle between the line perpendicular to the imaginary axis and the line passing through z and- Ri . Then 0 ≤ θ ≤ π/ 4, and z + Ri Re z = sec θ < √ 2 . It then follows that when | z | = R , | g ( z ) | = | ( z- Ri )( z + Ri ) f ( z ) | ≤ ( z- Ri )( z + Ri ) Re z ≤ 2 √ 2 R. In fact, this holds for | z | ≤ R by the Maximum Modulus Theorem applied to the disc D (0 ,R ). For any z ∈ C , choose R > | z | and apply the above argument to get | f ( z ) | = g ( z ) ( z- Ri )( z + Ri ) ≤ 2 √ 2 R R 2- | z | 2 . Now let R → ∞ to get that f ( z ) = 0. Hence f ≡ 0 on C . The second part is exactly Proposition 7.3 in the text. But we may deduce it here by applying the argument to the entire function g ( z ) = f ( iz ). Date : November 14, 2007 (Version 1.0)....
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- Fall '07
- Math, GB, maximum modulus theorem, ωB