$\frac{ 2i \times \left( 5+i \right) }{ \left( 5-i \right) \times \left( 5+i \right) }$
Distribute $2i$ through the parentheses$\frac{ 10i+2{i}^{2} }{ \left( 5-i \right) \times \left( 5+i \right) }$
Use $\left( a-b \right)\left( a+b \right)={a}^{2}-{b}^{2}$ to simplify the product$\frac{ 10i+2{i}^{2} }{ 25-{i}^{2} }$
By definition ${i}^{2}=-1$$\frac{ 10i+2 \times \left( -1 \right) }{ 25-{i}^{2} }$
By definition ${i}^{2}=-1$$\frac{ 10i+2 \times \left( -1 \right) }{ 25-\left( -1 \right) }$
Any expression multiplied by $-1$ equals its opposite$\frac{ 10i-2 }{ 25-\left( -1 \right) }$
When there is a $-$ in front of an expression in parentheses, change the sign of each term of the expression and remove the parentheses$\frac{ 10i-2 }{ 25+1 }$
Factor out $2$ from the expression$\frac{ 2\left( 5i-1 \right) }{ 25+1 }$
Add the numbers$\frac{ 2\left( 5i-1 \right) }{ 26 }$
Cancel out the common factor $2$$\frac{ 5i-1 }{ 13 }$
Separate the real and the imaginary parts$-\frac{ 1 }{ 13 }+\frac{ 5 }{ 13 }i$