## Precalculus (6th Edition) Blitzer

The standard form of the expression$\frac{5}{2-i}$ is $2+i$.
Consider the expression, $\frac{5}{2-i}$ Since, the imaginary part is in the denominator, we multiply the numerator and denominator by the complex conjugate of the denominator -- that is, for the complex number $\left( 2-i \right)$, its complex conjugate is $\left( 2+i \right)$ Multiply the expression by $\frac{\left( 2+i \right)}{\left( 2+i \right)}$. \begin{align} & \frac{5}{2-i}=\frac{5}{2-i}\cdot \frac{2+i}{2+i} \\ & =\frac{5\left( 2+i \right)}{\left( 2-i \right)\left( 2+i \right)} \end{align} The product of the complex number $\left( a+bi \right)$ and its complex conjugate $\left( a-bi \right)$ results in a real number -- that is, $\left( a+bi \right)\left( a-bi \right)={{a}^{2}}+{{b}^{2}}$ \begin{align} & \frac{5\left( 2+i \right)}{\left( 2-i \right)\left( 2+i \right)}=\frac{5\left( 2+i \right)}{{{2}^{2}}+{{1}^{2}}} \\ & =\frac{5\left( 2+i \right)}{4+1} \end{align} Further simplify the expression. \begin{align} & \frac{5}{2-i}=\frac{5\left( 2+i \right)}{4+1} \\ & =\frac{5\left( 2+i \right)}{5} \\ & =2+i \end{align} Hence, the standard form of the expression $\frac{5}{2-i}$ is $2+i$.
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