Question Video: Finding the Trigonometric Form of Roots of Complex Numbers Mathematics

Video Transcript

Find the cube roots of 64, giving your answers in trigonometric form.

This question is in effect asking us to solve the equation 𝑧 cubed is equal to 64. The method for solving this is very similar to the method we would use to solve the equation 𝑧 cubed is equal to one. We say that the equation 𝑧 to the 𝑛th power equals one has roots, which are called the 𝑛th roots of unity. They are in trigonometric form 𝑧 is equal to cos two 𝜋𝑘 over 𝑛 plus 𝑖 sin two 𝜋𝑘 over 𝑛, where 𝑘 is equal to zero, one, two, and so on up to 𝑛 minus one.

We begin by rewriting our equation as 𝑧 cubed is equal to 64 multiplied by one. Next, we’ll find the cube root of both sides of our equation. We recall that the 𝑛th root of the product of two real numbers 𝑎 and 𝑏 is equal to the product of the 𝑛th root of these numbers. So we write the cube root of 64 multiplied by one as the cube root of 64 multiplied by the cube root of one. Well, the cube root of 64 is four and the cube root of one, in other words, the cube roots of unity, are given by cos of two 𝑘𝜋 over three plus 𝑖 sin of two 𝑘𝜋 over three, where 𝑘 has values zero, one, and two.

To find our three roots, we will now let 𝑘 equal zero, one, and two. When 𝑘 is equal to zero, the root, which we will call 𝑧 sub one, is equal to four multiplied by cos of zero plus 𝑖 sin of zero. The cos of zero is equal to one and sin of zero is equal to zero. This means that 𝑧 sub one is simply equal to four. When 𝑘 is equal to one, our second root 𝑧 sub two is equal to four multiplied by cos two 𝜋 over three plus 𝑖 sin two 𝜋 over three. At this stage, we could evaluate cos of two 𝜋 over three and sin of two 𝜋 over three. However, as the question tells us to give our answers in trigonometric form, we will leave it as it is. Finally, when 𝑘 is equal to two, our third root 𝑧 sub three is equal to four multiplied by cos four 𝜋 over three plus 𝑖 sin four 𝜋 over three.

At this stage, we might think that we have the three roots. However, we recall that the principal value of the argument 𝜃 must be greater than negative 𝜋 and less than or equal to 𝜋. This is not true of four 𝜋 over three. Luckily, by adding or subtracting multiples of two 𝜋 from our argument, we can achieve one that is inside the range for the principal value. Four 𝜋 over three minus two 𝜋 is equal to negative two 𝜋 over three, which is indeed greater than negative 𝜋. We can therefore rewrite the third root 𝑧 sub three as four multiplied by cos of negative two 𝜋 over three plus 𝑖 sin negative two 𝜋 over three.

We now have the three cube roots of 64 as required. They are four, four multiplied by cos two 𝜋 over three plus 𝑖 sin two 𝜋 over three, and four multiplied by cos negative two 𝜋 over three plus 𝑖 sin negative two 𝜋 over three.