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s2_95-100aSOLUTION

# s2_95-100aSOLUTION - ACM95a/100a Last updated Problem Set...

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Unformatted text preview: ACM95a/100a Last updated October 14, 2009 Problem Set II Solutions Prepared by Faisal Amlani 1. (10 pts.) For a given real number φ , 0 ≤ φ < 2 π , find the image of the sector 0 ≤ arg( z ) < φ under the transformation w = z 4 . How large should φ be so that the w plane is covered exactly once? SOLUTION : In polar coordinates, the transformation is given as w = z 4 = ( r e iθ ) 4 = r 4 e i 4 θ . So we see that w : { r e iθ | r ≥ , ≤ θ < φ } → { r 4 e i 4 θ | r ≥ , ≤ θ < φ } = { r e iθ | r ≥ , ≤ θ < 4 φ } . So the image of the sector 0 ≤ arg( z ) < φ is w : { z | ≤ arg( z ) < φ } → { z | ≤ arg( z ) < 4 φ } If φ = π/ 2 , then { z | ≤ arg( z ) < 4 φ } → { z | ≤ arg( z ) < 2 π } and the map of the sector will cover exactly once. 2. (10 pts.) Let α negationslash = 0, β negationslash = 0 be two complex numbers. Show that α = tβ for some real number t (i.e. the vectors defined by α and β are parallel) if and only if ℑ ( α β ) = 0. SOLUTION : ⇐ : If α = tβ , then α β = t | β | 2 . This is a real number which implies ℑ ( α β ) = 0. ⇒ : Assume that ℑ ( α β ) = 0. Then α β = r for some r ∈ R . Multiplying by β gives α | β | 2 = rβ = ⇒ α = r | β | 2 β = ⇒ α = tβ for t = r | β | 2 . 3. (25 pts.) Show that cos- 1 ( z ) = − i log ( z + i (1 − z 2 ) 1 / 2 ) . Explain how to introduce cuts to give a branch with cos- 1 (0) = π/ 2. SOLUTION : Cos- 1 ( x ) is shown in Figure 1 for real variables in the range [ − 1 , 1]....
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s2_95-100aSOLUTION - ACM95a/100a Last updated Problem Set...

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