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HW4 - 2 2 n πi = ⇒ log i 1 2 =(1 4 n πi And 1 2 log i =...

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MATH 185 SOLUTIONS OF HW IV (1) (24. 7) (a) f ( z ) is real-valued in D . = f ( z ) = f ( z ). By the main result in Ex 3, Sec 24, f is constant in D since f and f = f are both analytic in D . (b) If c = 0 then | f ( z ) | = 0, i.e. f ( z ) 0 in D . Hence f is constant. Otherwise f ( z ) = c 2 /f ( z ) is well-defined and analytic. By the main result in Ex 3, Sec 24, f is constant in D (2) (33.13) By using the formula (11),(12) on page 102, we can easily show that sin(¯ z ) = sin z and cos(¯ z ) = cos z . If sin(¯ z ) is analytic at some point p then there is an open neighborhood D of p such that sin(¯ z ) is analytic on D , i.e. sin z is analytic on D . By the main result in Ex 3, Sec 24, sin z is constant on D . But sin z is not constant in any open region, so this is contradiction, i.e. sin(¯ z ) is not analytic at any point. Similarly, cos(¯ z ) is not analytic. (3) (30.5) (a) i 1 / 2 = e 1 / 2 log i = e 1
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Unformatted text preview: 2+2 n ) πi . = ⇒ log( i 1 / 2 ) = (1 / 4 + n ) πi . And 1 / 2 log i = 1 / 2(1 / 2 + 2 n ) πi = (1 / 4 + n ) πi . Hence log( i 1 / 2 ) = 1 / 2 log i . (b) log( i 2 ) = log(-1) = (-1 + 2 n ) πi . But 2 log i = 2(1 / 2 + 2 n ) πi = (1 + 4 n ) πi . Hence log( i 2 ) 6 = 2 log i . (4) (32.2) (a) i i = e iLogi = e i ( πi/ 2) = e-π/ 2 . (b) e 2 (-1-√ 3 i ) = e · e (-2 πi/ 3) . = ⇒ [ e 2 (-1-√ 3 i )] 3 πi = e 3 πiLog [ e 2 (-1-√ 3 i )] = e 3 πi (1-2 πi/ 3) = e 3 πi · e 2 π 2 =-e 2 π 2 since e 3 πi =-1. (c) (1-i ) 4 i = e 4 iLog (1-i ) = e 4 i (ln √ 2-πi/ 4) = e π +2 ln √ 2 = e π (cos(2 ln 2) + i sin(2 ln 2)) since 4 ln √ 2 = 2 ln 2. Date : today. 1...
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