## Precalculus (6th Edition) Blitzer

The standard form of the expression $\frac{3+4i}{4-2i}$ is $\frac{1}{5}+\frac{11}{10}i$.
Consider the expression, $\frac{3+4i}{4-2i}$ Since the imaginary part is in the denominator then multiply the numerator and denominator by the complex conjugate of the denominator -- that is, for the complex number $\left( 4-2i \right)$, its complex conjugate is $\left( 4+2i \right)$. Multiply the expression by $\frac{\left( 4+2i \right)}{\left( 4+2i \right)}$. \begin{align} & \frac{3+4i}{4-2i}=\frac{3+4i}{4-2i}\cdot \frac{4+2i}{4+2i} \\ & =\frac{\left( 3+4i \right)\left( 4+2i \right)}{\left( 4-2i \right)\left( 4+2i \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{3+4i}{4-2i}=\frac{\left( 3+4i \right)\left( 4+2i \right)}{{{4}^{2}}+{{2}^{2}}} \\ & =\frac{\left( 3+4i \right)\left( 4+2i \right)}{16+4} \\ & =\frac{\left( 3+4i \right)\left( 4+2i \right)}{20} \end{align} Apply FOIL method in the numerator. \begin{align} & \frac{\left( 3+4i \right)\left( 4+2i \right)}{20}=\frac{3\left( 4 \right)+3\left( 2i \right)+4i\left( 4 \right)+4i\left( 2i \right)}{20} \\ & =\frac{12+6i+16i+8{{i}^{2}}}{20} \\ & =\frac{12+22i+8{{i}^{2}}}{20} \end{align} As ${{i}^{2}}=-1$ Therefore, \begin{align} & \frac{12+22i+8{{i}^{2}}}{20}=\frac{12+22i+8\left( -1 \right)}{20} \\ & =\frac{12+22i-8}{20} \\ & =\frac{4+22i}{20} \end{align} Simplify the expression $\frac{4+22i}{20}$ \begin{align} & \frac{4+22i}{20}=\frac{2\left( 2+11i \right)}{2\left( 10 \right)} \\ & =\frac{2}{10}+\frac{11}{10}i \\ & =\frac{1}{5}+\frac{11}{10}i \end{align} Thus, $\frac{3+4i}{4-2i}=\frac{1}{5}+\frac{11}{10}i$ Hence, the standard form of the expression $\frac{3+4i}{4-2i}$ is $\frac{1}{5}+\frac{11}{10}i$.