If z=1-sqrt{3} i,then z^3-3 z^2+3 z= | vedclass

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(B) Given $z=1-\sqrt{3} i$. We know that $(z-1)^3 = z^3 - 3z^2 + 3z - 1$. Therefore,$z^3 - 3z^2 + 3z = (z-1)^3 + 1$. Substitute $z = 1 - \sqrt{3} i$ i

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$1+3 \sqrt{3} i$

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(B) Given $z=1-\sqrt{3} i$. We know that $(z-1)^3 = z^3 - 3z^2 + 3z - 1$. Therefore,$z^3 - 3z^2 + 3z = (z-1)^3 + 1$. Substitute $z = 1 - \sqrt{3} i$ into the expression: $(z-1) = (1 - \sqrt{3} i - 1) = -\sqrt{3} i$. Now,calculate $(z-1)^3$: $(-\sqrt{3} i)^3 = -(\sqrt{3})^3 	imes i^3 = -3\sqrt{3} 	imes (-i) = 3\sqrt{3} i$. Finally,$z^3 - 3z^2 + 3z = 3\sqrt{3} i + 1 = 1 + 3\sqrt{3} i$.

If $z=1-\sqrt{3} i$,then $z^3-3 z^2+3 z=$

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