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If $$\left| { z }_{ 1 } \right| =2,\left| { z }_{ 2 } \right| =3,\left| { z }_{ 3 } \right| =4$$ and $$\left| 2{ z }_{ 1 }+3{ z }_{ 2 }+4{ z }_{ 3 } \right| =10$$, then absolute value $$8{ z }_{ 2 }{ z }_{ 3 }+27{ z }_{ 3 }{ z }_{ 1 }+64{ z }_{ 1 }{ z }_{ 2 }$$ must be equal to-


Solution

$$|z_1|=2,z_1\bar{z_1}=4$$

$$|z_2|=3,z_2\bar{z_2}=9$$

$$|z_3|=4,z_3\bar{z_3}=16$$

$$|8z_2z_3+27z_3z_1+64z_1z_2|$$

$$\implies|2z_1\bar{z_1}z_2z_3+3z_2\bar{z_2}z_3z_1+4z_3\bar{z_3}z_1z_2|$$

$$\implies|z_1z_2z_3||2\bar{z_1}+3\bar{z_2}+4\bar{z_3}|$$

$$\implies 2\times 3\times 4|2{z_1}+3{z_2}+4{z_3}|$$

$$\implies 24\times 10=240$$


Mathematics

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