Solve for: (2-i)/(3+i)

Expression: $\frac{ 2-i }{ 3+i }$

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

$\frac{ \left( 2-i \right) \times \left( 3-i \right) }{ \left( 3+i \right) \times \left( 3-i \right) }$

Simplify the expression

$\frac{ 6-2i-3i+{i}^{2} }{ \left( 3+i \right) \times \left( 3-i \right) }$

Use $\left( a-b \right)\left( a+b \right)={a}^{2}-{b}^{2}$ to simplify the product

$\frac{ 6-2i-3i+{i}^{2} }{ 9-{i}^{2} }$

By definition ${i}^{2}=-1$

$\frac{ 6-2i-3i-1 }{ 9-{i}^{2} }$

By definition ${i}^{2}=-1$

$\frac{ 6-2i-3i-1 }{ 9-\left( -1 \right) }$

Subtract the numbers

$\frac{ 5-2i-3i }{ 9-\left( -1 \right) }$

Collect like terms

$\frac{ 5-5i }{ 9-\left( -1 \right) }$

When there is a $-$ in front of an expression in parentheses, change the sign of each term of the expression and remove the parentheses

$\frac{ 5-5i }{ 9+1 }$

Factor out $5$ from the expression

$\frac{ 5\left( 1-i \right) }{ 9+1 }$

Add the numbers

$\frac{ 5\left( 1-i \right) }{ 10 }$

Cancel out the common factor $5$

$\frac{ 1-i }{ 2 }$

Separate the real and the imaginary parts

$\begin{align*}&\frac{ 1 }{ 2 }-\frac{ 1 }{ 2 }i \\&0.5-0.5i\end{align*}$