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Unformatted text preview: 1 Physics 214 Spring 2008 Solutions to Assignment ] 1 Problem A.1. ( a ). Given z = re i ( t + ) with real r , , and , we have Re [ z ] = r cos( t + ) and Im [ z ] = r sin( t + ). ( b ). Now z = re i (( + i 1 ) t + ) = re 1 t e i ( t + ) . Therefore, Re [ z ] = re 1 t cos( t + ) and Im [ z ] = re 1 t sin( t + ). ( c ). (i). dz/dt = iz , (ii). dx/dt = d/dtRe [ z ] = r sin( t + ), (ii). Re [ dz/dt ] = r sin( t + ). ( d ). With x = Re [ z ] and a z + b z + cz = ( a 2 + ib + c ) z = 0, we have ( a 2 + ib + c ) = 0 which gives = i b 2 a b 2 a p 1 + 4 ac/b 2 . Assuming a , b and c are real, we have Re [ ] = b 2 a p 1 + 4 ac/b 2 and Im [ ] = b 2 a for 4 ac/b 2 > 1. For 4 ac/b 2 < 1, we have Re [ ] = 0 and Im [ ] = b 2 a b 2 a p 1 4 ac/b 2 . For 4 ac/b 2 = 1, we have Re [ ] = 0 and Im [ ] = b 2 a ....
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This note was uploaded on 03/31/2009 for the course PHYS 214 taught by Professor Thorne during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 THORNE
 Physics

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