Question

Complex numbers are often used when dealing with alternating current (AC) circuits. In the equation $V = IZ$, $V$ is voltage, $I$ is current, and $Z$ is a value known as impedance. If $V = 1-i$ and $Z=1+3i$, find $I$. Express your answer as a complex number in the form $a+bi$, where $a$ and $b$ are real numbers.

Answer 1

The current (
`I`) expressed in the form
`a+bi` is
`I=-0.2-0.4i`, where
`a=-0.2` and
`b=-0.4`.

Step-by-step explanation:

- The first step is to write the expression of the current (
`I`) in terms of voltage (
`V`) and impedance (
`Z`). As
`V=I\cdot Z`, you must pass
`Z` to divide to the other side. The expression would be:

`I=(V)/(Z)`

- The second step is to replace the values of
`V` and
`Z`:

`I=(V)/(Z)=(1-i)/(1+3i)`

The next step is very important: As long as you have a division of complex numbers and you want to convert it to the form
`a+bi`, you must multiply the numerator and denominator by the complex conjugate of the denominator. Keep in mind that the complex conjugate of a complex number is the same complex number but the sign of the imaginary part is opposite. Let's see two examples.

• Consider the complex number
  `1+3i`. Its complex conjugate is
  `1-3i`
• Consider the complex number
  `1-3i`. Its complex conjugate is
  `1+3i`

For this case, the complex conjugate is used because it allows to remove the imaginary part in the denominator. Let's multiply the complex conjugate of the denominator both in the numerator and in the denominator:

`I=(1-i)/(1+3i)\cdot (1-3i)/(1-3i)=(1-3i-i+3i^2)/(1-3i+3i-3^2i^2)`

Remember that
`i^2=-1`:

`I=(1-3i-i+3\cdot (-1))/(1-3i+3i-3^2\cdot (-1))`

It is time to simplify the expression:

`I=(1-3i-i-3)/(1-3i+3i+3^2)=(-2-4i)/(10)`

`I=-(2)/(10)-(4i)/(10)`

`I=-0.2-0.4i`

Thus, the current (
`I`) expressed in the form
`a+bi` is
`I=-0.2-0.4i`, where
`a=-0.2` and
`b=-0.4`.