If |z_1| = 2,|z_2| = 3,|z_3| = 4 and |2z_1 + | vedclass

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(A) Given $|z_1| = 2$,$|z_2| = 3$,$|z_3| = 4$ and $|2z_1 + 3z_2 + 4z_3| = 9$.We know that for any complex number $z$,$z\bar{z} = |z|^2$,so $\bar{z} =

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$216$

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(A) Given $|z_1| = 2$,$|z_2| = 3$,$|z_3| = 4$ and $|2z_1 + 3z_2 + 4z_3| = 9$.We know that for any complex number $z$,$z\bar{z} = |z|^2$,so $\bar{z} = \frac{|z|^2}{z}$.Consider the expression $|8z_2z_3 + 27z_3z_1 + 64z_1z_2|$.Factor out $z_1z_2z_3$ from the expression:$|z_1z_2z_3| \left| \frac{8}{z_1} + \frac{27}{z_2} + \frac{64}{z_3} ight|$.Since $|z_1|=2, |z_2|=3, |z_3|=4$,we have $|z_1|^2=4, |z_2|^2=9, |z_3|^2=16$.Substitute $8 = 2|z_1|^2$,$27 = 3|z_2|^2$,and $64 = 4|z_3|^2$:$|z_1z_2z_3| \left| \frac{2|z_1|^2}{z_1} + \frac{3|z_2|^2}{z_2} + \frac{4|z_3|^2}{z_3} ight| = |z_1z_2z_3| |2\bar{z}_1 + 3\bar{z}_2 + 4\bar{z}_3|$.Since $|z| = |\bar{z}|$,we have $|2\bar{z}_1 + 3\bar{z}_2 + 4\bar{z}_3| = |\overline{2z_1 + 3z_2 + 4z_3}| = |2z_1 + 3z_2 + 4z_3| = 9$.Thus,the value is $|z_1| |z_2| |z_3| 	imes 9 = 2 	imes 3 	imes 4 	imes 9 = 216$.

If $|z_1| = 2$,$|z_2| = 3$,$|z_3| = 4$ and $|2z_1 + 3z_2 + 4z_3| = 9$,then the value of $|8z_2z_3 + 27z_3z_1 + 64z_1z_2|$ is equal to:

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