Question Video: Geometric Applications of the 𝑛th Roots of a Complex Number Mathematics

1) Find the roots of 𝑧⁸ + 16 = 0. 2) The complex numbers representing the roots of 𝑧⁸ + 16 = 0 are each squared to form the vertices of a new shape. What is the area of the shape?

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Video Transcript

1) Find the roots of 𝑧 to the power of eight plus 16 equals zero. 2) The complex numbers representing the roots of 𝑧 to the power of eight plus 16 equals zero are each squared to form the vertices of a new shape. What is the area of the shape?

Let’s begin with part one. To solve this equation, we’ll subtract 16 from both sides to get 𝑧 to the power of eight equals negative 16. And then we’ll find the eighth roots of both sides. But to apply De Moivre’s theorem for roots, negative 16 is going to need to be expressed in exponential or polar form.

Let’s write it in exponential form. Its modulus is 16. And since it’s a purely real number, which would lie on the negative real axis of an Argand diagram, its argument is 𝜋 radians. And therefore, the solutions to our equation are the eighth roots of 16𝑒 to the 𝜋𝑖. Applying De Moivre’s theorem and we see that the roots are given by 16 to the power of one-eighth times 𝑒 to the 𝜋 plus two 𝜋𝑘 over eight 𝑖, where 𝑘 takes values from zero through to seven.

So our roots whose arguments are expressed in the range for the principal argument are root two 𝑒 to the 𝜋 by eight 𝑖, root two 𝑒 to the three 𝜋 by eight 𝑖, all the way through to root two 𝑒 to the negative 𝜋 by eight 𝑖. And since these are the eighth roots of a complex number, it follows that they will form the vertices of a regular octagon inscribed in a circle whose radius is root two and centre is the origin.

And what about part two? Well, to square a complex number in exponential form, we square its modulus and we double its argument. Notice how our eight roots have diminished to just four. These roots represent the vertices of a square this time, inscribed in a circle radius two units. Its area can be found by using the Pythagorean theorem. And that will generate a side length of two root two units. So its area is two root two squared. It’s eight square units.

Now did you predict what might happen when we squared the roots? It actually makes a lot of sense that the number of roots would halve. Essentially, it’s a little like simply finding the fourth roots of our original equation, which we know would form the vertices of a square.