If z1,z2,z3 be three copmlex numbers whose moduli are 1,2,3 respectively and |z1+z2+z3|=1. then find the value of |9z1z2+z2z3+4z3z1|.
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Solution
|z1+z2+z3|=1 Squaring (z1+z2+z3)(¯z1+¯z2+¯z3)=1 |z1|2+|z2|2+|z3|2+z1(¯z1+¯z3)+z2(¯z3+¯z1)+z3(¯z1+¯z2)=1 or ∑z1(¯z2+¯z3)=1−(1+4+9)=−13(1) Now |9z1z2+z2z3+4z3z1|2=(9¯z1¯z2+¯z2¯z3+4¯z3¯z1) =81(1.4)+1(4.9)+16(9.1)+9|¯z2|2.z1¯z3+36|¯z1|2z2¯.z3+9|¯z2|2z3¯z1+4|¯z3|2z2¯z1+36|¯z1|2z3¯z2+4|z3|2z1¯z2 Now put |¯z1|2=1,|z2|2=4 and |z2|2=9 =324+36+144+36∑z1(¯z2+¯z3) =36+(9+1+4)+36(−13)=36 by (1) ∴|9z1z2+z2z3+4z3z1|=6