ps03_solns

ps03_solns - 18.03 PS3 Solutions—Spring 09 1a 2 1 i = 2(1...

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Unformatted text preview: 18.03 PS3 Solutions—Spring 09 1a. 2 1 + i = 2(1 − i ) (1 + i )(1 − i ) = 2 − 2 i 2 = 1 − i , and 1 − i = √ 2 e − iπ/ 4 . 1b. e 2 − πi = e 2 e − πi = ( e 2 )( − 1), so the real part is − e 2 and the imaginary part is 0. e 2+ πi/ 2 = e 2 e πi/ 2 = e 2 i , so the real part is 0 and the imaginary part is e 2 . 1c. z 3 − 1 = ( z − 1)( z 2 + z + 1) = 0 since z 3 = 1. As z − 1 negationslash = 0, we have z 2 + z + 1 = 0. 1d. Polar representations: e 2 πi/ 3 , e 4 πi/ 3 . Rectangular representations: − 1 2 ± √ 3 2 i . parenleftBigg − 1 2 + √ 3 2 i parenrightBiggparenleftBigg − 1 2 + √ 3 2 i parenrightBigg + parenleftBigg − 1 2 + √ 3 2 i parenrightBigg + 1 = parenleftBigg − 1 2 − √ 3 2 i parenrightBigg + parenleftBigg − 1 2 + √ 3 2 i parenrightBigg + 1 = 0 parenleftBigg − 1 2 − √ 3 2 i parenrightBiggparenleftBigg − 1 2 − √ 3 2 i parenrightBigg + parenleftBigg − 1 2 − √ 3 2 i parenrightBigg + 1 = parenleftBigg − 1 2 + √ 3 2 i parenrightBigg + parenleftBigg − 1 2 − √ 3 2 i parenrightBigg + 1 = 0 1e. cos4 t = Re( e 4 it ) and e 4 it = ( e it ) 4 = (cos t + i sin t ) 4 . This last expands to cos 4 t + 4 i cos 3 t sin t + 6 i 2 cos 2 t sin 2 t + 4 i 3 cos t sin 3 t + i 4 sin 4 t = ( cos 4 t − 6 cos 2 t sin 2 t + sin 4 t ) + i ( 4 cos 3 t sin t − 4 cos t sin 3 t ) which means that cos4 t = cos 4 t − 6 cos 2 t sin 2 t + sin 4 t ....
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ps03_solns - 18.03 PS3 Solutions—Spring 09 1a 2 1 i = 2(1...

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