Complex

Manipulating the equation a little bit we get ${ (13-\frac { 1 }{ z } })^{ 10 }=-1$

So if ${ a }_{ 1 },{ b }_{ 1 },.......$ are the solutions of this equation

then $\frac { 1 }{ { a }_{ 1 } } ,\frac { 1 }{ { b }_{ 1 } } ,......$ are the solutions of the equation ${ (13-z) }^{ 10 }=-1$

hence we have $z=13-{ (-1) }^{ 1/10 }$

The question is asking $\sum { \frac { 1 }{ { a }_{ k }{ b }_{ k } } }$ and since they are conjugate of each other it can be written as $\sum { \frac { 1 }{ { \left| { a }_{ k } \right| }^{ 2 } } }$

Solving for z we get $\frac { 1 }{ { a }_{ k } } =13-\cos(\frac { (2k+1)\pi }{ 10 } )-i\sin(\frac { (2k+1)\pi }{ 10 } )$

$\Rightarrow \frac { 1 }{ { \left| { a }_{ k } \right| }^{ 2 } } =170-26\cos(\frac { (2k+1)\pi }{ 10 } )$

So the answer of our question is $\sum _{ k=1 }^{ 5 }{ 170-26\cos(\frac { (2k+1)\pi }{ 10 } ) } =850$ and we are done.