## Precalculus (6th Edition) Blitzer

The voltage of the circuit $E$ in volts is $\left( 21+i \right)$.
Consider the equation, $I=\left( 2-3i \right)$ and $R=\left( 3+5i \right)$ The voltage of the circuit in volts is, $E=IR$ Substitute $I=\left( 2-3i \right)$ and $R=\left( 3+5i \right)$ in the formula $E=IR$. $E=\left( 2-3i \right)\left( 3+5i \right)$ Use the FOIL method. \begin{align} & E=2\cdot 3+2\cdot 5i-3i\cdot 3-3i\cdot 5i \\ & =6+10i-9i-15{{i}^{2}} \\ & =6+i-15{{i}^{2}} \end{align} Replace the value ${{i}^{2}}=-1$. \begin{align} & E=6+i-15\left( -1 \right) \\ & =6+15+i \\ & =21+i \end{align} Therefore, the voltage of the circuit $E$ in volts is $\left( 21+i \right)$.