# 10 exercises these exercises are divided into three

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10 Exercises These exercises are divided into three groups. I suggest that you work through the questions in Part A in order, getting as far as you can in your alloted supervisions. Part B consists of questions for those who get through Part A quickly. They are not more difficult and will give you further prac- tice. Part C consists of a few questions which are a bit skew to the course but which are quite interesting. The notation (Q x , Paper X, 19AB) tells you that the question is based on question x on Paper X in 19AB, but I have often made slight changes. Part A Q 10.1. Prove Lemma 1.2. Q 10.2. Prove Theorem 1.4. Q 10.3. (This is Exercise 2.13.) We use the notation of Definition 2.10. Show that we can not choose θ 0 so that log z 1 z 2 = log z 1 + log z 2 for all z 1 , z 2 , z 1 z 2 Ω. 24

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Q 10.4. (i) Write out the standard properties of powers x α when x and α are real and x> 0. (For example ( xy ) α = x α y α .) Investigate the extent to which they remain true in the complex case. (ii) Show that z 1 / 3 has three possible branches on C \{ x : x real and x 0 } . Show that the same is true for z 2 / 3 . For each real α determine the number of branches (possibly infinite) of z α . Q 10.5. (This is Exercise 3.3.) We say that open subsets Ω and Γ of C are conformally equivalent if there exists a conformal map f : Ω Γ. Show that conformal equivalence is an equivalence relation. Q 10.6. Let Λ = { w | 2 π + θ 0 > w>θ 0 } , Ω = { z = re : r> 0 , 2 π + θ 0 >θ>θ 0 } . Let f ( z ) = exp z for z Λ. Show that f : Λ Ω is conformal and use Lemma 3.2 to establish the existence of a function log with properties given in Lemma 2.11. Q 10.7. (Q7(b), Paper I, 1993) For each of the following conformal maps f j and simply connected domains D j find the image of the domain under the map (as usual z = x + iy ). (i) f 1 ( z ) = 1 / (1 + z ) , D 1 = { x + iy : x 2 + y 2 < 1 , y> 0 } (ii) f 2 ( z ) = z 2 , D 2 = { x + iy : x> 0 , y< 0 } (iii) f 3 ( z ) = log z, D 3 = { x + iy : y> 0 } (You should make it clear which branch of log you choose for f 3 .) Hence, or otherwise, show that g ( z ) = 1 π log parenleftBigg 1 4 parenleftbigg 1 z 1 + z parenrightbigg 2 parenrightBigg is a conformal map of { x + iy : x 2 + y 2 < 1 , y> 0 } onto the infinite strip { x + iy : 0 <y< 1 } . Q 10.8. (Q7, Paper I, 2000) Let φ be a function of u ( x,y ) and v ( x,y ) which can also be regarded as a function of x and y . [I repeat the examiner’s wording without necessarily approving it.] Starting from the formula ∂φ ∂x = ∂φ ∂u ∂u ∂x + ∂φ ∂v ∂v ∂x , 25
obtain the formula 2 φ ∂x 2 = 2 φ ∂u 2 parenleftbigg ∂u ∂x parenrightbigg 2 + 2 φ ∂v 2 parenleftbigg ∂v ∂x parenrightbigg 2 + ∂φ ∂u 2 u ∂x 2 + ∂φ ∂v 2 v ∂x 2 + 2 2 φ ∂u∂v ∂u ∂x ∂v ∂x . By using the Cauchy-Riemann equations and associated results, deduce that, if w = u + iv is an analytic function of z = x + iy , then 2 φ ∂x 2 + 2 φ ∂y 2 = | w ( z ) | 2 parenleftbigg 2 φ ∂u 2 + 2 φ ∂v 2 parenrightbigg .

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• Fall '08
• Groah
• Math, Analytic function, Q7, Cauchy, Lemma

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