### Video Transcript

Put π equals four times the square root of three all multiplied by cos of five π over six minus π sin of five π over six in exponential form.

The first thing we can do is to recognise that we have been given a complex number. Complex numbers can be expressed in a number of different forms. These are trigonometric form, exponential form, and algebraic form. For the purposes of this question, we wonβt be using algebraic form. So letβs put this aside. We should be able to see that the question expresses the complex number π in a form that is very close to the trigonometric form. One difference that we can immediately spot is that, instead of having a positive symbol between our cos π and our π sin π, we instead have a negative symbol.

In order to move forward, letβs work on changing this symbol. Due to the symmetries of the sine function, weβre able to use the following identity. The negative sin of π is equal to sin of negative π. This identity allows us to take the term from our question, negative π sin of five π over six, and equate it to positive π sin of negative five π over six. We then rewrite the complex number π given in the question as four times the square root of three all multiplied by cos of five π over six add π sin negative five π over six. And here we can see the term that has been substituted in.

Let us now look back at the trigonometric form that we are aiming for. We have matched the symbol to be positive in between our cos π term and our π sin π term. However, in doing so, we have introduced a new difference. Instead of having matching values of π between our cos and our sin term, we now have the positive and negative of a π-value.

In order to change this, we can use another identity, which arises because of the symmetry of the cosine function. The identity is that cos of π is equal to cos of negative π. This allows us to take the term from our question, cos of five π over six, and equate it to cos of negative five π over six. Substituting this into our equation, weβre able to say that our complex number, π, is equal to four times the square root of three all multiplied by cos of negative five π over six add π sin of negative five π over six. In doing this, we now have matched π-values for our two terms.

There is now one more thing that we need to take into account. Generally, when expressing a complex number in trigonometric or an exponential form, we want our value of π to be greater than or equal to zero and less than two π in radians. In order to achieve this, we can observe the periodicity of the sine and cosine functions. Both of these functions repeat with an interval of two π radians. And this allows us to say that sin of π is equal to sin of π plus two ππ, where π takes integer values.

To use this relationship, we must find the value of π that will give us a π-value within our desired range. As it turns out, we can use the simple case of π equals one. Taking cosine for an example, our identity then reduces to cos of π is equal to cos of π plus two π.

Let us see this substitution in action when performed on both the cos and the sin terms. The terms in our parentheses now become cos of negative five π over six add two π. And the same is true for our sine term. We can simplify this by recognising that negative five π over six add two π is the same as negative five π over six add 12π over six, which then simplifies to seven π over six. We can now confirm that this value does indeed lie within the range for π that we are looking for.

Letβs see what our complex number looks like after this simplification is performed. We now have π equals four times the square root of three all multiplied by cos of seven π over six add π sin of seven π over six. We can now see that we have reached the trigonometric form for a complex number that we are looking for.

In general, a complex number expressed in this way, π equals π times cos π plus π sin π, can also be expressed in exponential form. π equals π times π to the power of ππ. We can move between these forms by taking the values of π and π and putting them into the corresponding expression.

Looking at our complex number, we can see that our value for π is four times the square root of three. And our value for π is seven π over six. We are able to take our values for π and for π and use them to express our complex number in exponential form. We have therefore answered our question. And weβre able to say that our complex number, π, in exponential form is four times the square root of three times π to the power of seven π over six times π.