Question Video: Using Properties of Cubic Roots of Unity to Simplify Expressions Mathematics

Evaluate (βˆ’5 βˆ’ 4/πœ” βˆ’ 4/πœ”Β²)(βˆ’2 + 3πœ” + 3πœ”Β²) where πœ” is a complex cube root of unity.


Video Transcript

Evaluate negative five minus four over πœ” minus four over πœ” squared multiplied by negative two plus three πœ” plus three πœ” squared, where πœ” is a complex cube root of unity.

The key to this question is in understanding the properties of the complex cube roots of unity. Firstly, when we refer to a cube root of unity, recall that this means a number πœ” such that πœ” cubed equals one. In particular, the complex part excludes πœ” from being one, which is known as the trivial root of unity. In the case of the cubic roots of unity, the complex roots are also instances of primitive roots, which satisfy some useful expressions.

The first property we want to use is that πœ” to the power of 𝑛 is equal to πœ” to the power of π‘Ž, for any integers 𝑛 and π‘Ž that are equivalent modulo three. So, for instance, seven is equivalent to one modulo three. And we know this because seven and one differ by a multiple of three. Hence, we can say that πœ” to the power of seven is equal to πœ” to the power of one.

Now, the second property we would like to use is that πœ” squared plus πœ” plus one is equal to zero. This quality in particular looks as though it will be very useful when we consider the terms inside the parentheses of the given expression since they appear to be of a similar form. The main challenge we now have is to manipulate the terms so that they are in a form where we can apply these properties.

Let us first consider the expression in the right parentheses. This almost looks to be three times our second property, since we have a three πœ” and a three πœ” squared term, but the last term is not three. However, we can get around this by adding five to the left of the expression and subtracting five from the right of it, which is a way of rewriting your expression without changing its value. We can then factor out three from the first three terms. And we can set the factor inside the parentheses to zero by using our second property. So, we actually end up with just negative five.

Next, we should simplify the expression in the parentheses on the left. In this case, we do not have an πœ” or an πœ” squared term explicitly. Instead, they appear in the denominators of the terms. To get this into the form we want, we can use the first property of primitive cube roots. To make this easier to see, we can rewrite the second and third terms using properties of exponents. That is to say, we have used negative exponents to express that the πœ”-terms involve reciprocals.

Now, let us consider the powers of πœ” here. First, we have negative one, which we note is equal to two minus three, which shows us that negative one is equivalent to two modulo three. Next, we have that negative two equals one minus three. So, negative two is equivalent to one modulo three. Thus, we can use the first property to rewrite πœ” to the power of negative one as πœ” squared and πœ” to the power of negative two as πœ” to the power of one.

We are now in a position to use the second property to simplify things. In a similar manner to above, we can add one to the left and subtract one from the right of the expression. We can then take out a factor of negative four from the first three terms. We can then see that the expression inside the parentheses is equal to zero due to the second property. So, we just have negative one.

Now that we have done all the heavy lifting, it remains to combine these expressions together. We have that the product of these two expressions in parentheses is the product of negative five and negative one, which gives us our final answer, five.

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