### Video Transcript

Which of the following complex numbers is equivalent to four minus three 𝑖 over two plus four 𝑖?
Note: 𝑖 is equal to the square root of negative one.
Is it option A) negative one-fifth minus 11𝑖 over 10, B) negative one-fifth plus 11𝑖 over 10, C) two minus three 𝑖 over four, or D) two plus three 𝑖 over four?

In order to simplify the complex number, we need to multiply the numerator and denominator by the complex conjugate of the denominator.
This will result in the denominator being a real number.
It will be noncomplex.
The complex conjugate of 𝑥 plus 𝑦𝑖 is 𝑥 minus 𝑦𝑖.
The real part of the complex number stays the same, and the complex part changes from positive to negative or vice versa.

In our question, the complex conjugate of two plus four 𝑖 is two minus four 𝑖.
On the numerator, we need to multiply four minus three 𝑖 by two minus four 𝑖.
On the denominator, we need to multiply two plus four 𝑖 by two minus four 𝑖.

We can distribute the two parentheses on the top using the FOIL method.
Multiplying the first terms, four and two, gives us eight.
Multiplying the outside terms gives us negative 16𝑖.
Multiplying the insight terms gives us negative six 𝑖.
And finally, multiplying the last terms gives us 12𝑖 squared.
Negative three multiplied by negative four is 12. And 𝑖 multiplied by 𝑖 is equal to 𝑖 squared.

We were told in the question that 𝑖 is equal to the square root of negative one.
If we square both sides of this equation, we can work out the value of 𝑖 squared.
The square root of negative one squared is equal to negative one, as the square root and the squared terms cancel.
This is because they are inverse operations. Therefore, 𝑖 squared is equal to negative one.

Substituting this value into our equation gives us eight minus 16𝑖 minus six 𝑖 minus 12, as 12 multiplied by negative one is equal to negative 12.
We can then group or collect the real and the imaginary parts.
Eight minus 12 is equal to negative four.
Negative 16𝑖 minus six 𝑖 is equal to negative 22𝑖.
This means that the numerator of our complex number can be simplified to negative four minus 22𝑖.

Our will next step is to repeat this process with the denominator.
We need to distribute the parentheses two plus four 𝑖 and two minus four 𝑖.
As mentioned previously, this should result in a noncomplex number.
We might notice at this stage that this expression is the difference of two squares.
So the middle two terms will cancel.
However, if we hadn’t noticed this, we can once again use the FOIL method to expand or distribute the parentheses.

Multiplying the first terms gives us four, as two multiplied by two equals four.
Multiplying the outside terms gives us negative eight 𝑖.
Multiplying the inside terms gives just positive eight 𝑖.
Multiplying the last terms gives us negative 16𝑖 squared.
Once again, we can replace 𝑖 squared with negative one.
Negative 16 multiplied by negative one is equal to 16.
So the denominator simplifies to four minus eight 𝑖 plus eight 𝑖 plus 16.
Negative eight 𝑖 plus eight 𝑖 is equal to zero. So these two terms cancel.
This leaves us with four plus 16, which is equal to 20.
Two plus four 𝑖 multiplied by two minus four 𝑖 is equal to 20.

Our next step is to separate the real and complex parts of this expression.
The real part is negative four over 20, and the complex part is negative 22𝑖 over 20.
Both of these fractions can be simplified by dividing the numerator and denominator by the same number.

The highest common factor of four and 20 is four.
So we can divide the numerator and denominator of the real part by four.
Four divided by four is equal to one, and 20 divided by four is equal to five.
Therefore, the real part becomes negative one-fifth.
In the same way, we can divide the numerator and denominator of the complex part by two.
This gives us negative 11𝑖 over 10, as 22 divided by two is equal to 11 and 20 divided by two is equal to 10.

The complex number four minus three 𝑖 over two plus four 𝑖 is equivalent to negative one-fifth minus 11𝑖 over 10.
This means that the correct answer of our four options was option A.