Linear Algebra Solutions 55

Linear Algebra Solutions 55 - (b 1i)2 = i so 2 6 1i 2 2 1i...

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(b) ( 1 - i 2 ) 2 = - i , so µ 1 - i 2 - 6 = ˆ µ 1 - i 2 2 ! - 3 = ( - i ) - 3 = - 1 i 3 = - 1 - i = 1 i = - i. 8. (i) To solve the equation z 2 = 1 + 3 i , we write 1 + 3 i in modulus– argument form: 1 + 3 i = 2(cos π 3 + i sin π 3 ) . Then the solutions are z k = 2 µ cos µ π 3 + 2 2 + i sin µ π 3 + 2 2 ¶¶ , k = 0 , 1 . Now k = 0 gives the solution z 0 = 2(cos π 6 + i sin π 6 ) = 2 ˆ 3 2 + i 2 ! = 3 2 + i 2 . Clearly z 1 = - z 0 . (ii) To solve the equation z 4 = i , we write i in modulus–argument form: i = cos π 2 + i sin π 2 . Then the solutions are
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