If z_1 and z_2 are the roots of the equation | vedclass

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(A) Given that $z_1$ and $z_2$ are the roots of $x^2+2x+2=0$. From the sum of roots,$z_1+z_2 = -2$,which implies $z_2 = -2-z_1$. Multiplying by $2$,we

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$64$

Vedclass · Answer

(A) Given that $z_1$ and $z_2$ are the roots of $x^2+2x+2=0$. From the sum of roots,$z_1+z_2 = -2$,which implies $z_2 = -2-z_1$. Multiplying by $2$,we get $2z_2 = -2z_1-4$,or $2z_2+2 = -2z_1-2$. Let the given expression be $E = \frac{-2^{11}(z_1+1+3i)^{11}}{2^5(z_2+1-3i)^{11}}$. We can rewrite this as $E = -2^6 \left( \frac{z_1+1+3i}{z_2+1-3i} \right)^{11}$. Multiply the numerator and denominator inside the bracket by $2$: $E = -2^6 \left( \frac{2z_1+2+6i}{2z_2+2-6i} \right)^{11}$. Substitute $2z_2+2 = -2z_1-2$ into the denominator: $E = -2^6 \left( \frac{2z_1+2+6i}{-2z_1-2-6i} \right)^{11} = -2^6 \left( \frac{2z_1+2+6i}{-(2z_1+2+6i)} \right)^{11}$. $E = -2^6 (-1)^{11} = -2^6 (-1) = 2^6 = 64$.

If $z_1$ and $z_2$ are the roots of the equation $x^2+2x+2=0$,then $\frac{-2^{11}(z_1+1+3i)^{11}}{2^5(z_2+1-3i)^{11}}$ is equal to

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