### Video Transcript

Find the solution set of the equation π§ cubed equals two plus two root three π in exponential form.

First, we note that if π§ cubed is equal to two plus two root three π, then to find π§, we would need to cube root both sides, giving π§ is equal to the cubed root of two plus two root three π. We could also write this cube root as a fractional power, π§ is equal to two plus two root three π to the power of one-third. Because a cube root is equivalent to a power of one-third.

The solution set of an equation just means the values of the variable, in this case π§, which satisfy the equation. So, what weβre actually being asked to find are the cube roots of the complex number two plus two root three π. These cube roots will themselves be complex numbers. And weβre asked to give them in exponential form. Thatβs the form ππ to the power of ππ, where π is the magnitude of the complex number and π is its amplitude, also known as itβs argument.

Itβs actually relatively straightforward to find the cube roots of a complex number in exponential form if that number itself is in exponential form. So, our first step is going to be to take the complex number two plus two root three π and write it in exponential form. We can introduce the notation π¦ to represent this complex number.

First, we need to recall how to find the magnitude of a complex number. And if we have the general complex number π¦, which is equal to π plus ππ, where both π and π are real values, then the magnitude of this complex number π¦, which is also referred to as π, is the square root of π squared plus π squared. Thatβs the real and imaginary parts squared added together and then square rooted.

In our complex number π¦, the values of π and π are two and two root three. So, we have that the magnitude of π¦ is equal to the square root of two squared plus two root three squared. Two squared is four. And to work out what two root three squared is, we can just reorder the terms in the product. We have two times two times root three times root three. Two multiplied by two is four. And root three multiplied by root three is three. So, we have four multiplied by three, which is just equal to 12.

So, the magnitude of π¦ then is the square root of four plus 12. Four plus 12 is 16. And the square root of 16 is four. So, weβve found the magnitude of our complex number π¦. And now we need to find the value of π, its amplitude, or argument, which we can do by considering the position of this complex number on the complex plane.

The real part of our complex number is two and the coefficient of the imaginary part is two root three, both of which are positive numbers. So, if we plot this complex number π¦ on the complex plane, it will be in the first quadrant. The angle π that weβre looking for, the amplitude, or argument of this complex number, is the angle made between the positive real axis and the line joining our complex number to the origin. We can sketch in a right-angled triangle below this line and then use trigonometry in order to find the value of angle π.

We know that, in a right-angled triangle, tan of an angle π is equal to the length of the opposite divided by the length of the adjacent. In our right-angled triangle, the opposite is two root three and the adjacent is two. So, we have that tan of π is equal to two root three over two. The factors of two in the numerator and denominator cancel out, leaving us with tan of π is equal to root three. π is, therefore, equal to the inverse tan of root three.

And at this point, we recall that thereβs a special angle whose tangent is root three. Itβs the angle 60 degrees, or π by three radians. Itβs far more usual to work with angles in radians rather than degrees when weβre working with complex numbers in exponential form. So, we have that the value of π is π by three. Our complex number π¦ then in exponential form is π¦ equals four π to the ππ by three. And now we need to find the cube roots of this number.

We can do this using DeMoivreβs theorem or a consequence of DeMoivreβs Theorem, which is that the πth roots of the complex number ππ to the ππ are given by π to the power of one over π multiplied by π to the power of ππ plus two πππ all over π. Where π takes values from zero, one, all the way up to π minus one. In our case, the value of π is three. So, the π will take the values zero, one, and two. And we will have three cube roots of this complex number π¦, one for each of these values of π. In general, a complex number has π πth root.

So, substituting the value three for π first of all, we have that the cube root of π¦, which will in turn be equal to the values of π§ weβre looking for, are equal to four to the power of one-third multiplied by π to the power of ππ by three plus two πππ all over three. We then need to substitute π equals zero, one, and two in turn to give us three cube roots.

First, then, we substitute π equals zero. And we get four to the power of one-third multiplied by π to the power of ππ by three plus two ππ zero all over three. Now this second part in the power is just equal to zero. So, our power becomes ππ by three divided by three. ππ by three divided by three is equal to ππ by nine. So, our first root is four to the power of one-third π to the ππ by nine. Now notice that four is not a cube number, so we canβt simplify four to the power of one-third easily. Weβll leave it in this form involving a power.

Substituting our next value of π, π equals one, we have four to the power of one-third multiplied by π to the power of ππ by three plus two ππ one all over three. Letβs simplify that power. Ignoring the factor of π for now, we have one-third of π over three plus two π. Two π can be written as six π over three. And then, adding six π over three and π over three gives seven π by three. So, we have one-third of seven π by three, which is equal to seven π over nine. Adding the π back into the power, we have that our second cube root is four to the power of one-third multiplied by π to the π seven π by nine.

Substituting π equals two gives our final root four to the power of one-third π to the power of ππ by three plus two ππ two all over three. Letβs simplify that power, again ignoring the factor of π for now. We have one-third of π over three plus four π. Four π can be written as 12π over three. So, we have one-third of 13π over three which will be equal to 13π over nine. Now notice though that this value of 13π over nine is bigger than π. And whenever we give a complex number in exponential form, we always want to give its principal amplitude, or principal argument, which is the value between negative π and π.

To achieve this, we can add or subtract multiples of two π until we get an amplitude which is in the desired range. Our value of 13π over nine is just a little bit too big. So, letβs try subtracting just one lot of two π first of all. Two π can be written as 18π over nine. And subtracting this from 13π over nine gives negative five π over nine, which is between negative π and π. So, this is the principal amplitude of this complex number.

The solution set then of the equation π§ cubed equals two plus two root three π are the cube roots of two plus two root three π, which, in exponential form, are four to the power of one-third π to the ππ over nine, four to the power of one-third π to the ππ seven π over nine, and four to the power of one-third π to the negative π five π over nine.