Let z be a complex number with a non-zero ima | vedclass

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(B) Given that $z 
eq \overline{z}$.Let $\alpha = \frac{2+3z+4z^2}{2-3z+4z^2} = \frac{(2-3z+4z^2)+6z}{2-3z+4z^2} = 1 + \frac{6z}{2-3z+4z^2}$.Since $\

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$0.50$

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(B) Given that $z 
eq \overline{z}$.Let $\alpha = \frac{2+3z+4z^2}{2-3z+4z^2} = \frac{(2-3z+4z^2)+6z}{2-3z+4z^2} = 1 + \frac{6z}{2-3z+4z^2}$.Since $\alpha$ is a real number,$\alpha = \overline{\alpha}$.This implies $\frac{z}{2-3z+4z^2} = \frac{\overline{z}}{2-3\overline{z}+4\overline{z}^2}$.Cross-multiplying,we get $z(2-3\overline{z}+4\overline{z}^2) = \overline{z}(2-3z+4z^2)$.$2z - 3z\overline{z} + 4z\overline{z}^2 = 2\overline{z} - 3z\overline{z} + 4z^2\overline{z}$.$2(z-\overline{z}) = 4z\overline{z}(z-\overline{z})$.Since $z 
eq \overline{z}$,we can divide by $(z-\overline{z})$:$2 = 4z\overline{z} \implies z\overline{z} = \frac{2}{4} = 0.5$.Thus,$|z|^2 = 0.50$.

Let $z$ be a complex number with a non-zero imaginary part. If $\frac{2+3z+4z^2}{2-3z+4z^2}$ is a real number,then the value of $|z|^2$ is:

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