Solve for: (2i)/(5-i)

Expression: $\frac{ 2i }{ 5-i }$

Multiply the numerator and denominator by the complex conjugate of the denominator

$\frac{ 2i \times \left( 5+i \right) }{ \left( 5-i \right) \times \left( 5+i \right) }$

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

$-\frac{ 1 }{ 13 }+\frac{ 5 }{ 13 }i$