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# roots of complex numbers watch

1. 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
2. (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
3. (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
4. (Original post by TeeEm)
It's in the first line of the original post.

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