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If $z_1$ and $z_2$ are the roots of the equation $x^2+2 x+2=0$, then $\frac{-2^{11}\left(z_1+1+3 i\right)^{11}}{2^5\left(z_2+1-3 i\right)^{11}}$ is equal to
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The correct answer is:
64
$z_1$ and $z_2$ are the roots of $x^2+2 x+2=0$
$$
\therefore z_1+z_2=-2 \Rightarrow z_2=-2-z_1
$$
$$
\begin{aligned}
& \text { Now, } 2^6\left[\frac{-2^{11}\left(z_1+1+3 i\right)^{11}}{2^{11}\left(z_2+1-3 i\right)^{11}}\right] \\
& =2^6\left[\frac{-2 z_1-2-6 i}{2 z_2+2-6 i}\right]^{11}=2^6\left[\frac{-2 z_1-2-6 i}{-2 z_1-2-6 i}\right]^{11} \\
& =2^6=64 \quad \text { [from Eq. (i)] }
\end{aligned}
$$
$$
\therefore z_1+z_2=-2 \Rightarrow z_2=-2-z_1
$$
$$
\begin{aligned}
& \text { Now, } 2^6\left[\frac{-2^{11}\left(z_1+1+3 i\right)^{11}}{2^{11}\left(z_2+1-3 i\right)^{11}}\right] \\
& =2^6\left[\frac{-2 z_1-2-6 i}{2 z_2+2-6 i}\right]^{11}=2^6\left[\frac{-2 z_1-2-6 i}{-2 z_1-2-6 i}\right]^{11} \\
& =2^6=64 \quad \text { [from Eq. (i)] }
\end{aligned}
$$
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