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If $z=(1-i)^3(x+i)$ is a purely imaginary number for $x=x_1$ and if $\mathrm{z}$ is a purely real number for $\mathrm{x}=\mathrm{x}_2$, then $\mathrm{x}_1 \mathrm{x}_2=$
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Verified Answer
The correct answer is:
$-1$
$\text {} \begin{aligned}
z & =(1-i)^3(x+i) \\
\Rightarrow \quad z & =(1+i-3 i(1-i))(x+i) \\
& =(-2 i-2)(x+i) \\
\Rightarrow \quad & z=(2-2 x)-i(2+2 x)
\end{aligned}$
If $z$ is purely imaginary: $2-2 x=0$
$\Rightarrow x=1 \quad \therefore x_1=1$
If $z$ is purely real:
$\begin{aligned}
& 2+2 x=0 \Rightarrow x=-1 \\
& \therefore \quad x_2=-1
\end{aligned}$
Now, $x_1 x_2=1 \times(-1)=-1$
z & =(1-i)^3(x+i) \\
\Rightarrow \quad z & =(1+i-3 i(1-i))(x+i) \\
& =(-2 i-2)(x+i) \\
\Rightarrow \quad & z=(2-2 x)-i(2+2 x)
\end{aligned}$
If $z$ is purely imaginary: $2-2 x=0$
$\Rightarrow x=1 \quad \therefore x_1=1$
If $z$ is purely real:
$\begin{aligned}
& 2+2 x=0 \Rightarrow x=-1 \\
& \therefore \quad x_2=-1
\end{aligned}$
Now, $x_1 x_2=1 \times(-1)=-1$
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