$\frac{ \left( 2-i \right) \times \left( 3-i \right) }{ \left( 3+i \right) \times \left( 3-i \right) }$
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$\begin{align*}&\frac{ 1 }{ 2 }-\frac{ 1 }{ 2 }i \\&0.5-0.5i\end{align*}$