Linear Algebra Solutions 56

Linear Algebra Solutions 56 - Geometrically, the solutions...

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Geometrically, the solutions lie equi–spaced on the unit circle at arguments π 8 , π 8 + π 2 = 5 π 8 , π 8 + π = 9 π 8 , π 8 + 3 π 2 = 13 π 8 . Also z 2 = - z 0 and z 3 = - z 1 . (iii) To solve the equation z 3 = - 8 i , we rewrite the equation as µ z - 2 i 3 = 1 . Then µ z - 2 i = 1 , - 1 + 3 i 2 , or - 1 - 3 i 2 . Hence z = - 2 i, 3 + i or - 3 + i . Geometrically, the solutions lie equi–spaced on the circle | z | = 2, at arguments π 6 , π 6 + 2 π 3 = 5 π 6 , π 6 + 2 2 π 3 = 3 π 2 . (iv) To solve z 4 = 2 - 2 i , we write 2 - 2 i in modulus–argument form: 2 - 2 i = 2 3 / 2 µ cos - π 4 + i sin - π 4 . Hence the solutions are z k = 2 3 / 8 cos µ - π 4 + 2 4 + i sin µ - π 4 + 2 4 , k = 0
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