Video Transcript
Evaluate five plus five 𝜔 plus seven 𝜔 squared all raised to the power of negative four , where 𝜔 is a complex cube root of unity.
In this question, we are told that the term 𝜔 represents one of the complex cube roots of unity. The cube roots of unity are the solutions to the cube root of one. In other words, they are the numbers that are equal to one when they are cubed. The obvious solution to the cube root of one is one. However, you may be familiar with the other two solutions, which are negative one plus root three 𝑖 all divided by two and negative one minus root three 𝑖 all divided by two, where 𝑖 is equal to the square root of negative one.
For this video, we won’t go into the calculation of these roots; however, we can see that we have one real number and two complex numbers. In order to answer this question, we will first demonstrate an interesting property of these two complex roots. This property is that the square of either of the complex cube roots of unity is equal to the other complex cube root of unity.
In order to prove this, let’s pick one of our complex cube roots of unity and define it as 𝜔. We begin by taking out a factor of one-half and then square each of the parentheses to find an equation for 𝜔 squared. We can now simplify, as a half squared is one-quarter, and by expanding out the terms of our binomial squared for the right-hand parentheses. Simplifying further, we have one-quarter multiplied by one minus two root three 𝑖 plus three 𝑖 squared.
Recalling that 𝑖 is the square root of negative one, 𝑖 squared must be equal to negative one. And three 𝑖 squared must therefore be equal to negative three. We now have 𝜔 squared is equal to one-quarter multiplied by negative two minus two root three 𝑖. Taking out a factor of two from our parentheses, we can rewrite 𝜔 squared as negative one minus root three 𝑖 all divided by two.
This does indeed match the other complex cube root of unity. At this stage, it’s worth noting that the relationship would still be true if we’d instead chosen our other complex cube root of unity as the original 𝜔. Whilst we won’t go through the same calculation steps in this video, if 𝜔 was equal to negative one minus root three 𝑖 all divided by two, then 𝜔 squared would be equal to negative one plus root three 𝑖 all divided by two.
We can therefore conclude that the three complex cube roots of unity can be defined as one, 𝜔, and 𝜔 squared.
Let’s now return to the expression given in the question. The first thing we can do is to take out a factor of five from the first two terms in our parentheses. Next, we will try to find a meaningful substitution that will help us move forward. We have now defined 𝜔 and 𝜔 squared as our two complex cube roots of unity. Since we know the value for both of these roots, we’re able to write out an equation for the sum of 𝜔 plus 𝜔 squared.
Splitting our two fractions into their individual terms allows us to simplify. And we find that 𝜔 plus 𝜔 squared is equal to negative one. We can now rearrange this equation to make it more useful for our substitution. Adding one and subtracting 𝜔 squared from both sides, we have 𝜔 plus one or one plus 𝜔 is equal to negative 𝜔 squared.
We can then use this relationship to substitute into our expression. We have five multiplied by negative 𝜔 squared plus seven 𝜔 squared all raised to the power of negative four. Our expression now only contains terms of 𝜔 squared. And this simplifies to two 𝜔 squared raised to the power of negative four.
Next, we can raise the contents of our parentheses to the power of negative four. This gives us two to the power of negative four multiplied by 𝜔 squared to the power of negative four. Using our power rule of exponents, we can multiply two and negative four, leaving us with two to the power of negative four multiplied by 𝜔 to the power of negative eight.
We can simplify two to the power of negative four as one over two to the power of four, which is equal to one over 16. Simplifying 𝜔 to the power of negative eight is more complicated. By definition, we know that cubing any of our cube roots of unity will give us one. Since 𝜔 is one of these roots, we know that 𝜔 cubed is equal to one. Since this is true, let’s see what happens when we raise 𝜔 cubed to any integer power 𝑛. Since one raised to the power of 𝑛 is still equal to one, we’re able to say that 𝜔 cubed raised to any power 𝑛 will also still be equal to one.
Once again, using the power rule of exponents, we have 𝜔 to the power of three 𝑛 is equal to one. Given this fact we know that multiplying any number by 𝜔 to the power of three 𝑛 is the same as multiplying by one, and the number will remain unchanged. If we consider the case when 𝑛 equals three, we know that 𝜔 to the power of three multiplied by three, or 𝜔 to the power of nine, is equal to one. And we can use this to simplify 𝜔 to the power of negative eight.
We will construct an equation where we multiply 𝜔 to the power of negative eight by 𝜔 to the power of nine. This is the same as multiplying 𝜔 to the power of negative eight by one and is therefore equal to 𝜔 to the power of negative eight.
Using another of our laws of exponents, we can add the powers on the left-hand side. And since negative eight plus nine is equal to one, we see that 𝜔 to the power of negative eight is equal to 𝜔. We can now substitute this into our original expression. Two to the power of negative four multiplied by 𝜔 to the power of negative eight is equal to one sixteenth 𝜔.
Using our knowledge of the complex cube roots of unity, we have found that five plus five 𝜔 plus seven 𝜔 squared all raised to the power of negative four is equal to one sixteenth 𝜔.
