Same Modulus Complex Roots Problem.

Consider the term. $\frac{{{w_3} \cdot \overline {{w_3}{w_2}} }}{{{{\left| {{w_1}} \right|}^3}{{\left( {\overline {{w_3}} } \right)}^{ - 1}}}}$ , As we know that $\left| {{w_1}} \right| = \left| {{w_2}} \right| = \left| {{w_3}} \right|$

• Change the term. $\frac{{{w_3} \cdot \overline {{w_3}{w_2}} }}{{{{\left| {{w_1}} \right|}^3}{{\left( {\overline {{w_3}} } \right)}^{ - 1}}}}$ $\to \frac{{{w_3}\overline {{w_3}} \cdot \overline {{w_3}{w_2}} }}{{{{\left| {{w_3}} \right|}^3}}}$ , As we know that for any complex number $k$ , $k\overline k = {\left| k \right|^2}$

• We can change the term through these steps. $\frac{{{w_3}\overline {{w_3}} \cdot \overline {{w_3}{w_2}} }}{{{{\left| {{w_3}} \right|}^3}}} \to \frac{{{{\left| {{w_3}} \right|}^2} \cdot \overline {{w_3}{w_2}} }}{{{{\left| {{w_3}} \right|}^3}}} \to \frac{{\overline {{w_3}{w_2}} }}{{\left| {{w_3}} \right|}}$

• Do those steps to the other two and we will get this as the result. ${\left| {\frac{{{w_3} \cdot \overline {{w_3}{w_2}} }}{{{{\left| {{w_1}} \right|}^3}{{\left( {\overline {{w_3}} } \right)}^{ - 1}}}} + \frac{{{w_1} \cdot \overline {{w_1}{w_3}} }}{{{{\left| {{w_2}} \right|}^3}{{\left( {\overline {{w_1}} } \right)}^{ - 1}}}} + \frac{{{w_2} \cdot \overline {{w_2}{w_1}} }}{{{{\left| {{w_3}} \right|}^3}{{\left( {\overline {{w_2}} } \right)}^{ - 1}}}}} \right|^2} = {\left| {\frac{{\overline {{w_3}{w_2}} }}{{\left| {{w_3}} \right|}} + \frac{{\overline {{w_1}{w_3}} }}{{\left| {{w_1}} \right|}} + \frac{{\overline {{w_2}{w_1}} }}{{\left| {{w_2}} \right|}}} \right|^2}$

• Again, consider the fact that $\left| {{w_1}} \right| = \left| {{w_2}} \right| = \left| {{w_3}} \right|$ ${\left| {\frac{{\overline {{w_3}{w_2}} }}{{\left| {{w_3}} \right|}} + \frac{{\overline {{w_1}{w_3}} }}{{\left| {{w_1}} \right|}} + \frac{{\overline {{w_2}{w_1}} }}{{\left| {{w_2}} \right|}}} \right|^2} \to {\left| {\frac{{\overline {{w_3}{w_2}} }}{{\left| {{w_1}} \right|}} + \frac{{\overline {{w_1}{w_3}} }}{{\left| {{w_1}} \right|}} + \frac{{\overline {{w_2}{w_1}} }}{{\left| {{w_1}} \right|}}} \right|^2} \to {\left| {\frac{{\overline {{w_3}{w_2}} + \overline {{w_1}{w_3}} + \overline {{w_2}{w_1}} }}{{\left| {{w_1}} \right|}}} \right|^2} \to \frac{{{{\left| {\overline {{w_3}{w_2}} + \overline {{w_1}{w_3}} + \overline {{w_2}{w_1}} } \right|}^2}}}{{{{\left| {{w_1}} \right|}^2}}}$

From the polynomial. ${z^2} + Az + \left( {25 + 30i} \right) = \frac{{125i}}{z}$ , As we know that $z \ne 0$ .

• Multiply by $z$ both sides. ${z^2} + Az + \left( {25 + 30i} \right) = \frac{{125i}}{z} \to {z^3} + A{z^2} + \left( {25 + 30i} \right)z - 125i = 0$

• From the Vieta's formular. We know that. $\begin{array}{c} {w_1}{w_2} + {w_2}{w_3} + {w_3}{w_1} = 25 + 30i\\ {w_1}{w_2}{w_3} = 125i \end{array}$

• Take the absolute both sides. $\begin{array}{c} \left| {{w_1}{w_2} + {w_2}{w_3} + {w_3}{w_1}} \right| = \left| {25 + 30i} \right| \to {\left| {{w_1}{w_2} + {w_2}{w_3} + {w_3}{w_1}} \right|^2} = 1525\\ {\rm{and}}\\ \left| {{w_1}{w_2}{w_3}} \right| = \left| {125i} \right| \to {\left| {{w_1}} \right|^3} = 125 \to {\left| {{w_1}} \right|^2} = 25 \end{array}$

• Plug in the value. $\frac{{{{\left| {\overline {{w_3}{w_2}} + \overline {{w_1}{w_3}} + \overline {{w_2}{w_1}} } \right|}^2}}}{{{{\left| {{w_1}} \right|}^2}}} = \frac{{{{\left| {{w_1}{w_2} + {w_2}{w_3} + {w_3}{w_1}} \right|}^2}}}{{{{\left| {{w_1}} \right|}^2}}} = \frac{{1525}}{{25}} = 61$

So, ${\left| {\frac{{{w_3} \cdot \overline {{w_3}{w_2}} }}{{{{\left| {{w_1}} \right|}^3}{{\left( {\overline {{w_3}} } \right)}^{ - 1}}}} + \frac{{{w_1} \cdot \overline {{w_1}{w_3}} }}{{{{\left| {{w_2}} \right|}^3}{{\left( {\overline {{w_1}} } \right)}^{ - 1}}}} + \frac{{{w_2} \cdot \overline {{w_2}{w_1}} }}{{{{\left| {{w_3}} \right|}^3}{{\left( {\overline {{w_2}} } \right)}^{ - 1}}}}} \right|^2} = 61$