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    can someone help me find the third cube root of 2(-1+i)

    One is given as 1+ i, the modulus of 2(-1 + i) is 81/2 so the mod of the cube root = 21/2 and the arg of the original is 3/4 pi so the arg of the second is pi/4

    so 21/2 (cos (pi/4) + isin(pi/4)) gives 21/2epi/4 i as the second root

    I tried adding 2pi/3 to theta to get 21/2e11pi/12 i but the answer says 23/2e3/4 pi i could someone explain

    thanks
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    (Original post by bl64)
    can someone help me find the third cube root of 2(-1+i)

    One is given as 1+ i, the modulus of 2(-1 + i) is 81/2 so the mod of the cube root = 21/2 and the arg of the original is 3/4 pi so the arg of the second is pi/4

    so 21/2 (cos (pi/4) + isin(pi/4)) gives 21/2epi/4 i as the second root

    I tried adding 2pi/3 to theta to get 21/2e11pi/12 i but the answer says 23/2e3/4 pi i could someone explain

    thanks
    The book is wrong. Your answer is right.
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    (Original post by bl64)
    can someone help me find the third cube root of 2(-1+i)

    One is given as 1+ i, the modulus of 2(-1 + i) is 81/2 so the mod of the cube root = 21/2 and the arg of the original is 3/4 pi so the arg of the second is pi/4

    so 21/2 (cos (pi/4) + isin(pi/4)) gives 21/2epi/4 i as the second root

    I tried adding 2pi/3 to theta to get 21/2e11pi/12 i but the answer says 23/2e3/4 pi i could someone explain

    thanks
    please post a photo of your question
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    (Original post by TeeEm)
    please post a photo of your question
    It's in the first line of the original post.
 
 
 
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