Let z1,z2,z3 be complex numbers of equal magnitude satisfying z1+z2+z3=−32−i5 and z1z2z3=3+i5. If z1=x1+iy1,z2=x2+iy2,z3=x3+iy3;x1,x2,x3,y1,y2,y3∈R then 16(x1y1+x2y2+x3y3)25=

|z1z2z3|=22 so|z1|=|z2|=|z3|=2 Now, z¯1+z¯2+z¯3=−32+i5 ⇒ 2(z1z2+z2z3+z3z1)z1z2z3=−32+i5 ⇒ z1z2+z2z3+z3z1=(−34+i52)(3+i5) Now, x1y1+x2y2+x3y3=Im(z12+z22+z32)2 =Im(z1+z2+z3)22−Im(z1z2+z2z3+z3z1) =152−(−154+152)=154

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Let $z_{1}, z_{2}, z_{3}$ be complex numbers of equal magnitude satisfying $z_{1} + z_{2} + z_{3} = - \frac{\sqrt{3}}{2} - i \sqrt{5}$ and $z_{1} z_{2} z_{3} = \sqrt{3} + i \sqrt{5} .$ If $z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}, z_{3} = x_{3} + i y_{3}; x_{1}, x_{2}, x_{3}, y_{1}, y_{2}, y_{3} \in R$ then $\frac{16 {(x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3})}^{2}}{5} =$

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Detailed Solution

$| z_{1} z_{2} z_{3} | = 2 \sqrt{2} s o | z_{1} | = | z_{2} | = | z_{3} | = \sqrt{2} N o w, {\bar{z}}_{1} + {\bar{z}}_{2} + {\bar{z}}_{3} = - \frac{\sqrt{3}}{2} + i \sqrt{5} \Rightarrow \frac{2 (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1})}{z_{1} z_{2} z_{3}} = - \frac{\sqrt{3}}{2} + i \sqrt{5} \Rightarrow z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} = (- \frac{\sqrt{3}}{4} + \frac{i \sqrt{5}}{2}) (\sqrt{3} + i \sqrt{5}) N o w, x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} = \frac{Im (z_{1}^{2} + z_{2}^{2} + z_{3}^{2})}{2} = \frac{I_{m} {(z_{1} + z_{2} + z_{3})}^{2}}{2} - Im (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}) = \frac{\sqrt{15}}{2} - (- \frac{\sqrt{15}}{4} + \frac{\sqrt{15}}{2}) = \frac{\sqrt{15}}{4}$