If x+iy=(-1+isqrt{3})^{2010},then x is | vedclass

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(B) Given $x+iy=(-1+i\sqrt{3})^{2010}$.We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ is the cube root of unity,so $-1+i\sqrt{3} = 2\omega$.Substituti

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$2^{2010}$

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(B) Given $x+iy=(-1+i\sqrt{3})^{2010}$.We know that $\omega = \frac{-1+i\sqrt{3}}{2}$ is the cube root of unity,so $-1+i\sqrt{3} = 2\omega$.Substituting this into the expression:$x+iy = (2\omega)^{2010} = 2^{2010} \cdot \omega^{2010}$.Since $\omega^3 = 1$,we have $\omega^{2010} = (\omega^3)^{670} = 1^{670} = 1$.Therefore,$x+iy = 2^{2010} \cdot 1 = 2^{2010} + i(0)$.Comparing the real parts,we get $x = 2^{2010}$.

If $x+iy=(-1+i\sqrt{3})^{2010}$,then $x$ is

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