If z=(1-i)^3(x+i) is a purely imaginary numbe | vedclass

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(A) Given $z = (1-i)^3(x+i)$.First,expand $(1-i)^3$:$(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 = 1 - 3i - 3 + i = -2 - 2i$.Now,substitute this into

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(A) Given $z = (1-i)^3(x+i)$.First,expand $(1-i)^3$:$(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 = 1 - 3i - 3 + i = -2 - 2i$.Now,substitute this into the expression for $z$:$z = (-2 - 2i)(x + i) = -2x - 2i - 2ix - 2i^2 = -2x - 2i - 2ix + 2 = (2 - 2x) - i(2 + 2x)$.For $z$ to be purely imaginary,the real part must be zero:$2 - 2x_1 = 0 \Rightarrow x_1 = 1$.For $z$ to be purely real,the imaginary part must be zero:$-(2 + 2x_2) = 0 \Rightarrow x_2 = -1$.Therefore,$x_1 x_2 = 1 	imes (-1) = -1$.

If $z=(1-i)^3(x+i)$ is a purely imaginary number for $x=x_1$ and if $z$ is a purely real number for $x=x_2$,then $x_1 x_2=$

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