Answer
The standard form of the expression $\frac{3-4i}{4+3i}$ is $-i$.
Work Step by Step
Consider the expression,$\frac{3-4i}{4+3i}$
Multiply by complex conjugate of the denominator in the numerator and the denominator.
$\frac{3-4i}{4+3i}=\frac{\left( 3-4i \right)}{\left( 4+3i \right)}\cdot \frac{\left( 4-3i \right)}{\left( 4-3i \right)}$
Use the FOIL method.
\[\begin{align}
& \frac{3-4i}{4+3i}=\frac{\left( 3-4i \right)}{\left( 4+3i \right)}\cdot \frac{\left( 4-3i \right)}{\left( 4-3i \right)} \\
& =\frac{12-9i-16i+12{{i}^{2}}}{16-12i+12i-9{{i}^{2}}} \\
& =\frac{12-25i+12{{i}^{2}}}{16-9{{i}^{2}}}
\end{align}\]
Replace the value ${{i}^{2}}=-1$.
\[\begin{align}
& \frac{3-4i}{4+3i}=\frac{12-25i+12\left( -1 \right)}{16-9\left( -1 \right)} \\
& =\frac{12-25i-12}{16+9} \\
& =\frac{12-12-25i}{25} \\
& =\frac{-25i}{25}
\end{align}\]
Express the complex number in the standard form.
\[\frac{3-4i}{4+3i}=-i\]
Therefore, the standard form of the expression $\frac{3-4i}{4+3i}$ is $-i$.
