College Algebra (6th Edition)

$E=(47+13i)$ volts
$I=(4-5i)$ amperes $;$ $R=(3+7i)$ ohms Ohm's law is: $$E=IR$$ where $I$ is the current in a circuit, $R$ is the resistance of the circuit and $E$ is the voltage of the circuit. $I$ and $R$ are given and $E$ is to be found. Substitute the known values into the formula: $E=(4-5i)(3+7i)=...$ Evaluate the product: $...=12+28i-15i-35i^{2}=...$ Substitute $i^{2}$ by $-1$ and simplify: $...=12+28i-15i-35(-1)=...$ $...=12+28i-15i+35=...$ $...=47+13i$ The voltage in the circuit is $47+13i$ volts