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Unformatted text preview: Mathematics 185 Intro to Complex Analysis Fall 2009 M. Christ Solutions to Selecta from Problem Set 2 1.3.3(a) Solve cos( z ) = 3 4 + 1 4 i . Solution. Substitute w = e iz . The strategy is first to solve for w , then for z in terms of w . Set = 3 4 + 1 4 i . By the definition cos( z ) = 1 2 e iz + 1 2 e iz , the equation to be solved is equivalent to w + w 1 = 2 . Since w 6 = 0 (it equals e iz ), this is equivalent to w 2 2 w + 1 = 0. By the quadratic formula, the solutions are w = 2 p 4 2 4 2 = p 2 1 Now 2 = (3+ i ) 2 / 4 2 = (8+6 i ) / 16, so 2 1 = (8+6 i 16) / 16 = ( 8+6 i ) / 16. Were supposed to take the square root. That looks awful, unless we notice that ( 2 1) = (8 6 i ) / 16, which is the complex conjugate of 2 , hence equals ( ) 2 ! (And there are no other square roots, of course.) Therefore w = i = ( 3 4 + 1 4 i ) ( i 3 4 + 1 4 ) The two possible values w 1 ,w 2 for w are thus w 1 = 1 + i , and w 2 = 1 2 1 2 i . Now our final answer will be all solutions of e iz = w 1 , together with all solutions of e iz = w 2 . Write w 1 ,w 2 in polar coordinates: w 1 = 2 1 / 2 e i/ 4 = e ln(2)+ i/ 4 , while w 2 = 2 1 / 2 e i/ 4 = e ln(2) i/ 4 . We are solving e iz = w j , so we need to divide these exponents by i . And then because of the periodicity of the exponential, we need to add arbitrary....
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This note was uploaded on 10/10/2009 for the course MATH 185 taught by Professor Lim during the Spring '07 term at University of California, Berkeley.
 Spring '07
 Lim
 Math

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