hw1.sol

hw1.sol - 1(a e(b 3 i 2 n 2 29 e(c e(d 2(a n interger 5 i 2...

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e i 2 π n 3 π 2 ----- +   n: interger 29 e i 2 π n 5 2 -- atan + e i 2 π n π 2 + e i π n π 4 + = 1 2 cos i 1 2 sin e i 2 π n 1 2 = 32 (29 cos i 2 sin + 2 π cos i π sin + 2 = e ai b + e a e ib e a b cos sin + == i ---------- ia i + + ------------------------------- 1 a 2 1 + -------------- 1 ia + 2 6 e i 2 π n π 4 +  1 3 2 6 e i 2 π n 3 --------- π 12 + = n=0,1,2 The roots are π 12 3 π 4 17 π 12 -------- 2 6 The radius of circle is 0 Z 1 + 5 e i 2 π n = e i 2 π n 1 = Because Z 1 + e i 2 π n 5 = Ze i 2 π n 5 1 = 1.(a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) n=0,1,2,3,4
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2 π n 5 ---------   cos 1 i 2 π n 5 sin + Z = n=0,1,2,3,4 -1 -2 0 X Y i i e ii ln e ie i 2 π n π 2 -- +  ln e 2 π n π 2 + == = Z (29 cos 1 Z 2 2! ----- Z 4 4! Z 6 6! ++ = i cos 1 1 2! ---- 1 4! 1 2 n ! ------------ ++++ + = e x 1 x x 2 2! x 3 3! x n n ! + +++++ = Because e 1 11 1 2! 1 3! 1 n ! + = e 1 1 2! 1 3! 1 n n ! ------------- + + = i cos e 1 e + 2 ----------------
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This note was uploaded on 12/29/2011 for the course ENGR 5a taught by Professor Brewer during the Fall '09 term at UCSB.

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hw1.sol - 1(a e(b 3 i 2 n 2 29 e(c e(d 2(a n interger 5 i 2...

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