roots of complex numbers

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 8^1/2 so the mod of the cube root = 2^1/2 and the arg of the original is 3/4 pi so the arg of the second is pi/4

so 2^{1/2 (}cos (pi/4) + isin(pi/4)) gives 2^1/2e^{pi/4 i} as the second root

I tried adding 2pi/3 to theta to get 2^1/2e^{11pi/12 i} but the answer says 2^3/2e^{3/4 pi i} could someone explain

thanks

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 8^1/2 so the mod of the cube root = 2^1/2 and the arg of the original is 3/4 pi so the arg of the second is pi/4

so 2^{1/2 (}cos (pi/4) + isin(pi/4)) gives 2^1/2e^{pi/4 i} as the second root

I tried adding 2pi/3 to theta to get 2^1/2e^{11pi/12 i} but the answer says 2^3/2e^{3/4 pi i} could someone explain

thanks

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