Question Video: Roots of a Complex Number Mathematics

Video Transcript

1) Solve 𝑧 to the power of five equals 16 root two plus 16 𝑖 root two. 2) By plotting the solutions on an Argand diagram, or otherwise, describe the geometric properties of the solutions.

For part one, we need to solve an equation that involves finding the roots for a complex number written in algebraic form. Remember though, De Moivre’s theorem for roots uses the polar and exponential form of a complex number instead of the algebraic form. So we’ll need to begin by calculating the modulus and argument of a complex number that’s denoted 𝑧 to the power of five.

The real part of this complex number is 16 root two. And similarly, its imaginary part is also 16 root two. So the modulus is fairly straightforward to calculate. It’s the square root of the sum of the squares of these two parts. That’s the square root of 16 root two squared plus 16 root two squared, which is simply 32. So the modulus of 𝑧 to the power of five is 32.

In exponential form, this is the value of 𝑟. Its argument is also fairly straightforward. The complex number has both positive real and imaginary parts. So it must lie in the first quadrant on the Argand diagram. This means we can use the formula arctan of 𝑏 divided by 𝑎, where 𝑏 is the imaginary part and 𝑎 is the real part, to find the argument of 𝑧 to the power of five. That’s arctan of 16 root two over 16 root two.

Well, in fact, 16 root two divided by 16 root two is one. So we need to find the arctan of one. And this is a value we should know by heart. We know that tan of 𝜋 by four is one. So the arctan of one must be 𝜋 by four. And the argument of 𝑧 to the power of five is 𝜋 by four. And we can say, in exponential form, we can write this equation as 𝑧 to the power of five equals 32𝑒 to the 𝜋 by four 𝑖.

To solve this equation, we’ll need to find the fifth root of both sides. Now the fifth root of 𝑧 to the power of five is simply 𝑧. And we can say that the fifth root of 32𝑒 to the 𝜋 by four 𝑖 is 32𝑒 to the 𝜋 by four 𝑖 to the power of one over five. Comparing this to the formula for De Moivre’s theorem, we see that 𝑟, the modulus, is 32. 𝜃, the argument, is 𝜋 by four. And 𝑛 must be equal to five, which means 𝑘 is going to take the values of zero, one, two, three, and four.

Applying this theorem with 𝑛 equals five, we get 𝑧 equals 32 to the power of one-fifth times 𝑒 to the 𝜋 by four plus two 𝜋𝑘 over five 𝑖. 32 to the power of one-fifth is two. And substituting 𝑘 is equal to zero into the equation, we see that the first solution must be two 𝑒 to the 𝜋 by 20𝑖.

Our second solution is two 𝑒 to the nine 𝜋 by 20𝑖. When 𝑘 is equal to two, we get two 𝑒 to the 17𝜋 by 20𝑖. Substituting 𝑘 is equal to three into our equation and then subtracting two 𝜋 from the argument so that it’s within the range for the principal argument, we see that the fourth solution is two 𝑒 to the negative three 𝜋 by four 𝑖. And similarly, the final solution is two 𝑒 to the negative seven 𝜋 by 20𝑖.

And there we have the five solutions to the equation 𝑧 to the power of five equals 16 root two plus 16𝑖 root two. And we’ve expressed them in exponential form.

For part two, we’re going to need to plot these solutions on an Argand diagram. Now one way to do this would be to convert these numbers back into their algebraic form. Once we know their real and imaginary parts, we can plot them fairly easily on the Argand diagram. Alternatively, we could spot that their modulus is two and then use their arguments to plot the roots. Either way, we see that they form the vertices of a regular pentagon, inscribed in a circle of radius two, whose centre is the origin.