Question

example 6 multiply complex numbers
electricity the voltage v of an ac circuit can be found using
the formula ( v = cl ), where c is current and / is impedance. if
( c = 3 + 2j ) amps and ( l = 7 - 5j ) ohms, determine the voltage.
( \begin{aligned} v &= cl & & \text { voltage formula } \\ &= (3 + 2j)(7 - 5j) & & c = 3 + 2j \text { and } l = 7 - 5j \\ &= 3 ( ) + 3 ( ) + 2j ( ) + 2j ( ) & & \text { foil method } \\ &= - 15j + - 10j ^ { 2 } & & \text { multiply. } \\ &= 21 - - 10 ( ) & & j ^ { 2 } = - 1 \\ &= __ - j & & \text { add. } end{aligned} )
the voltage is ( 31 - j ) volts.

Explanation:

Step1: Aplicar el método FOIL

$$(3 + 2j)(7-5j)=3\times7+3\times(-5j)+2j\times7+2j\times(-5j)$$

Step2: Multiplicar

$$=21-15j + 14j-10j^{2}$$

Step3: Sustituir \(j^{2}=-1\)

$$=21 - j-10\times(-1)$$

Step4: Sumar

$$=21 - j + 10=31 - j$$

Answer:

Los espacios se completan de la siguiente manera:

• En el método FOIL: \(7\), \(-5j\), \(7\), \(-5j\)
• En la multiplicación: \(21\), \(14j\)
• En la sustitución de \(j^{2}=-1\): \(j\), \(-1\)
• En la adición: \(31\)

El voltaje es \(31 - j\) volts.