Question Video: Solving an Equation with Complex Roots

Find the fourth roots of 625, giving your answers in trigonometric form.

03:02

Video Transcript

Find the fourth roots of 625, giving your answers in trigonometric form.

What this question is really asking us is solve the equation π‘₯ to the fourth power equals 625. So, how do we do this? Well, the method we use for solving this is very similar to the method we use to solve the equation π‘₯ to the fourth power is equal to one. We say that the equation π‘₯ to the 𝑛th power equals one has roots which are called the 𝑛th root of unity. They are, in trigonometric form, π‘₯ equals cos of two πœ‹π‘˜ over 𝑛 plus 𝑖 sin of two πœ‹π‘˜ over 𝑛 for values of π‘˜ between zero and 𝑛 minus one.

We’re going to write our equation as π‘₯ to the fourth power equals 625 times one. And then, we’re going to find the fourth roots of both sides of our equation. We recall that the 𝑛th root of the product of two real numbers π‘Ž and 𝑏 is equal to the product of the 𝑛th roots of these numbers. So, we write the fourth root of 625 times one as the fourth root of 625 times the fourth root of one. Well, the fourth root of 625 is five. And the fourth root of one, in other words, the fourth roots of unity, are given by cos of two πœ‹π‘˜ over four plus 𝑖 sin of two πœ‹π‘˜ over four for values of π‘˜ between zero and three.

Let’s simplify two-quarters to one-half. And we see that the fourth root of unity are cos of πœ‹π‘˜ over two plus 𝑖 sin of πœ‹π‘˜ over two for those same values of π‘˜. To find each of our roots, we let π‘˜ be equal to zero, one, two, and three in turn. Now, when π‘˜ is equal to zero, our first root is π‘₯ equals five times cos of zero πœ‹ over two plus 𝑖 sin of zero πœ‹ over two. Of course, zero πœ‹ over two is simply zero. So, we have five times cos of zero plus 𝑖 sin of zero. Cos of zero is one, whilst sin of zero is zero. So, we have five times one plus zero 𝑖, which is simply five.

Our second root is found by letting π‘˜ be equal to one. And we get five times cos of πœ‹ by two plus 𝑖 sin of πœ‹ by two. Now, at this stage, we could actually evaluate cos of πœ‹ by two and sin of πœ‹ by two as we did when π‘˜ was equal to zero. But in fact, this question wants our answers in trigonometric form, so we’ll leave it as it is.

When π‘˜ is equal to two, we have five times cos of two πœ‹ by two plus 𝑖 sin of two πœ‹ by two, which is simply five times cos of πœ‹ plus 𝑖 sin of πœ‹. Our fourth and final solution is given by five times cos of three πœ‹ by two plus 𝑖 sin of three πœ‹ by two. Now, of course, this is outside of the range for the principal value of the argument, which is πœƒ is greater than negative πœ‹ or less than or equal to πœ‹. Luckily, by adding or subtracting multiples of two πœ‹ from our argument, we can achieve one that is inside the range for the principal value.

We’re going to subtract two πœ‹ from three πœ‹ by two, and we get negative πœ‹ by two. And so, our fourth root is π‘₯ equals five times cos of negative πœ‹ by two plus 𝑖 sin of negative πœ‹ by two. And so, we have the fourth roots of 625. In no particular order, they are five, five times cos of πœ‹ by two plus 𝑖 sin of πœ‹ by two, five times cos of πœ‹ plus 𝑖 sin of πœ‹, and five times cos of negative πœ‹ by two plus 𝑖 sin of negative πœ‹ by two.

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