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ps1sol

# ps1sol - ACM 95a/100a Problem Set 1 Solutions Prepared by...

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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 + ¯ z 1 + ¯ z = z + ¯ z 1 + ¯ z 1 + z 1 + z = z + ¯ z + z 2 + | z | 2 | 1 + z | 2 = ( x + iy ) + ( x - iy ) + ( x 2 - y 2 + i 2 xy ) + ( x 2 + y 2 ) (1 + x ) 2 + y 2 = 2 x (1 + x ) + i 2 xy (1 + x ) 2 + y 2 (1) b) (1 + i ) 7 = | 1 + i | e i (Arg(1+ i )+2 πk ) · 7 , k Z = 2 e i ( π 4 +2 πk ) · 7 = 8 2 e i ( 7 π 4 +14 πk ) = 8 2 1 - i 2 = 8 - 8 i (2) Or, since z n is single-valued for every n Z , we may write (1 + i ) 7 = (1 + i ) 2 (1 + i ) 2 (1 + i ) 2 (1 + i ) = (2 i ) 3 (1 + i ) = - 8 i (1 + i ) = 8 - 8 i (3) 1

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Problem 2 (2 × 10 points) Sketch the regions of the complex plane corresponding to fl fl fl fl z + 1 z - 1 fl fl fl fl < α for α = 1 and α = 2. Solution 2 We rewrite our inequality: | z + 1 | < α | z - 1 | (4) Since both | z + 1 | and α | z - 1 | are nonnegative, -→ | z + 1 | 2 < α 2 | z - 1 | 2 (5) -→ ( z + 1) (¯ z + 1) < α 2 ( z - 1) (¯ z - 1) (6) -→ | z | 2 + z + ¯ z + 1 < α 2 | z | 2 - z - ¯ z + 1 · . (7) For α = 1: -→ 2 ( z + ¯ z ) < 0 (8) -→ 4 ( z ) < 0 (9) -→ ( z ) < 0 (10) For α = 2: -→ 0 < 3 | z | 2 - 5 ( z + ¯ z ) + 3 (11) -→ 0 < | z | 2 - 5 3 ( z + ¯ z ) + 1 (12) -→ 0 < z - 5 3 ¯ z - 5 3 - 25 9 + 1 (13) -→ 0 < fl fl fl fl z - 5 3 fl fl fl fl 2 - 16 9 (14) -→ 4 3 < fl fl fl fl z - 5 3 fl fl fl fl (15) Alternatively, we may substitute z = x + iy at some point, e.g., | z + 1 | 2 < α 2 | z - 1 | 2 -→ ( x + 1) 2 + y 2 < α 2 h ( x - 1) 2 + y 2 i . (16) For α = 1: 2
Figure 1: The region given by | z + 1 | < | z - 1 | -→ 4 x < 0 (17) -→ x < 0 (18) For α = 2: -→ 0 < 3 x 2 - 10 x + 3 y 2 + 3 (19) -→ 0 < 3 x 2 - 10 3 x + 25 9 - 3 · 25 9 + 3 y 2 + 3 (20) -→ 4 3 2 < x - 5 3 2 + y 2 (21) In summary, α = 1: ( z ) < 0, i.e., x < 0 (see Figure 1 for plot) (22) α = 2: 4 3 < fl fl z - 5 3 fl fl , i.e., ( 4 3 ) 2 < ( x - 5 3 ) 2 + y 2 (see Figure 2 for plot) (23) Note: we can understand these results geometrically. For example, the α = 1 inequality describes the set of z -points such that the distance between - 1 and z is less than the distance between 1 and z . The imaginary axis is equidistant from - 1 and 1; points to the left of the imaginary axis are closer to - 1 than to 1. Problem 3 (20 points) Consider two complex variables z and w .

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