Ps1sol

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Unformatted text preview: ACM 95a/100a Problem Set 1 Solutions Prepared by: Zhiyi Li October 3, 2007 Total: 105 points Include grading section: 5 points Problem 1 (2×5 points) Suppose z = x + iy with x, y ∈ and write the following in the form a + ib where a, b ∈ : a) z+z ¯ , 1+z ¯ b) (1 + i)7 Solution 1 a) z+z1+z ¯ z+z ¯ = 1+z ¯ 1+z 1+z ¯ 2 z + z + z 2 + |z | ¯ = 2 |1 + z | (x + iy ) + (x − iy ) + x2 − y 2 + i2xy + x2 + y 2 = 2 (1 + x) + y 2 = 2x (1 + x) + i2xy (1 + x) + y 2 2 (1) b) (1 + i) = |1 + i| ei(Arg(1+i)+2πk) = √ 2ei( 4 +2πk) π 7 7 , k∈Z 7 √ 7π = 8 2ei( 4 +14πk) √ 1−i =8 2 √ 2 = 8 − 8i (2) Or, since z n is single-valued for every n ∈ Z, we may write (1 + i) = (1 + i) (1 + i) (1 + i) (1 + i) = (2i) (1 + i) = −8i (1 + i) = 8 − 8i 3 7 2 2 2 (3) 1 Problem 2 (2×10 points) Sketch the regions of the complex plane corresponding to z+1 <α z−1 for α = 1 and α = 2. Solution 2 We rewrite our inequality: |z + 1| < α |z − 1| Since both |z + 1| and α |z − 1| are nonnegative, −→ |z + 1| < α2 |z − 1| 2 2 2 2 2 (4) (5) (6) (7) −→ (z + 1) (¯ + 1) < α (z − 1) (¯ − 1) z z −→ |z | + z + z + 1 < α2 |z | − z − z + 1 . ¯ ¯ For α = 1: −→ 2 (z + z ) < 0 ¯ −→ 4 (z ) < 0 −→ (z ) < 0 For α = 2: −→ 0 < 3 |z | − 5 (z + z ) + 3 ¯ 5 2 −→ 0 < |z | − (z + z ) + 1 ¯ 3 5 5 25 −→ 0 < z − z− ¯ − +1 3 3 9 5 3 4 5 −→ < z − 3 3 −→ 0 < z − 2 2 (8) (9) (10) (11) (12) (13) (14) (15) − 16 9 Alternatively, we may substitute z = x + iy at some point, e.g., |z + 1| < α2 |z − 1| −→ (x + 1) + y 2 < α2 (x − 1) + y 2 . For α =...
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This note was uploaded on 02/22/2010 for the course ACM 95A taught by Professor Nilesa.pierce during the Fall '06 term at Caltech.

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