### Video Transcript

Given that π§ one is equal to eight
times cos of 240 degrees plus π sin of 240 degrees. π§ two is equal to four times cos
of five π over four plus π sin of five π over four. And π§ three is equal to eight
times cos of 45 degrees plus π sin of 45 degrees. Find π§ one multiplied by π§ two to
the power of six divided by π§ three to the power of four, giving your answer in
exponential form.

Now for this question, weβve been
given three complex numbers. And we have to perform various
operations, such as multiplication, division, and raising to some powers. We could approach this using De
Moivreβs theorem. However, you may notice that the
question asked for an answer in exponential form. Because of this, the method that
weβre going to explore will first be to convert our complex numbers to exponential
form and only then to perform the operations.

Let us first recall the
trigonometric and exponential form for a complex number. Youβll notice that, in both of
these forms, the complex number π§ is expressed in terms of two main parameters: the
magnitude π and the angle π. Weβre able to move between
trigonometric and exponential form by taking the values of π and π from one
expression and inputting them into the other expression.

Before we do this, let us first
align the units of π. Looking at our complex numbers, we
can see that π is expressed in terms of degrees for π§ one and π§ three and in
radians for π§ two. We can convert all of our degree
values to radians by remembering the following relationship. 360 degrees is two π radians. This allows us to calculate that
240 degrees is four π over three radians. And 90 degrees is π over four
radians.

Now that we found these values, let
us rewrite our complex numbers with π expressed in terms of radians. π§ one is equal to eight times cos
of four π over three plus π sin of four π over three. π§ two remains unchanged, since we
already had π in terms of radians. And π§ three is equal to eight
times cos of π over four plus π sin of π over four.

Now that weβve done this, letβs
work on converting our three complex numbers from trigonometric form to exponential
form. Remember, this can be achieved by
taking the parameters of π and π and inputting them into this expression. Starting with π§ one, we can see
that our value for π is eight, and our value for π is four π over three. π§ one is therefore equal to eight
times π to the power of four π over three times π.

Going through the same process for
π§ two, we find that this is equal to four times π to the power of five π over
four times π. And finally, we find that π§ three
is equal to eight times π to the power of π over four times π.

Now that we have our three complex
numbers in exponential form, letβs look at the operations that weβre going to be
performing. Here we see that we have π§ one
multiplied by π§ two to the power of six divided by π§ three to the power of
four. We can make our future calculations
slightly easier by remembering that one over some number π₯ to the power of π is
equal to π₯ to the power of negative π. This allows us to rewrite our
question as π§ one multiplied by π§ two to the power of six multiplied by π§ three
to the power of negative four.

First, letβs tidy things up by
getting rid of the trigonometric form of our complex numbers. Weβll leave some gaps to make room
for our future calculations. Now letβs work on applying the
powers to our complex numbers.

Our first complex number, π§ one,
has no power to apply. However, weβre gonna choose to
express the multiple of eight in terms of powers of two. The reasons for this will become
apparent later. Eight is equal to two to the power
of three. We can also do the same for π§ two
and π§ three, finding that four is equal to two squared and eight once again is
equal to two to the power of three.

Letβs sub in these values. We now need to raise π§ two to the
power of six. This can be done by raising two
squared to the power of six, which is multiplied by π to the power of five π over
four times π, all raised to the power of six. We can simplify this using the
following relationship. Some number π₯ to the power of π
raised to the power of π is equal to π₯ to the power of π times π.

Using this relationship, we find
that π§ two to the power of six is equal to two to the power of two times six times
π to the power of five π over four times six times π. Simplifying this further, we find
that two times six is of course equal to 12. And five π over four times six is
equal to 15π over two. Letβs replace these simplified
values.

We can now perform the same process
on π§ three, which is raised to the power of negative four. As with the previous complex
number, we can work on simplifying our powers. Here we have two to the power of
three times negative four, which is of course equal to negative 12. And here in parentheses we have π
over four times negative four, which is equal to negative four π over four or
negative π. Letβs put in these simplified
values.

We have now raised π§ two and π§
three to the relevant powers. But before we proceed, weβre going
to perform one final simplification. You may notice that our complex
numbers contain different fractional powers of π. Ignoring the multiple of π leaves
our four π over three, 15π over two, and negative π.

The next step of our calculation
will be much easier if each of these are expressed in terms of a fraction with the
same denominator. Since three, two, and one are all
multiples of six, weβre gonna choose this as our denominator. Four π over three is the same as
eight π over six. 15π over two is the same as 45π
over six. And negative π is equal to
negative six π over six.

Letβs rewrite each of our complex
numbers, replacing the powers of π. Letβs now see why weβve done this
by moving on to the multiplication step. Since we have found π§ one, π§ two,
and π§ three to the relevant powers, we can input these into our equation. Here we have input our three
complex numbers.

But we can reduce the calculation
by remembering the following relationship. Some number π₯ raised to the power
of π multiplied by the same number π₯ raised to a different power π is equal to π₯
to the power of π plus π. This relationship allows us to form
a sum for the power of two and the power of π. This is the same reason that we
expressed our multiples all in terms of powers of two. And we change the fractional powers
of π all to have the same denominators.

Reducing the sums, we find the
following answer. And we can now convert our two to
the power of three back to an eight. Doing so, we find that π§ one
multiplied by π§ two to the power of six divided by π§ three to the power of four is
equal to eight π to the power of 47π over six times π.

Although we have seemingly found
the answer to our question, we do have one more step to perform. Here we can clearly see that our
answer is in exponential form, with an π-value of eight and a π-value of 47π over
six. Although we have a valid answer,
generally, when working with complex numbers in exponential or in trigonometric
form, we want our value of π to be greater than or equal to zero and less than two
π.

By looking at this range in terms
of sixths, we can see that our π-value is very clearly greater than 12π over
six. In order to fix this, we can use
the following relationship, which takes advantage of the periodicity that occurs in
the complex plane. In the complex plane, π to the
power of ππ is equal to π to the power of π times π plus two ππ, where π is
an integer value.

For our complex number, the value
of π is 47π over six. By choosing a negative value of π
β in this case, π equals negative three β we can reduce our π-value to lie within
the desired range. π plus two π times negative three
is equal to the following. Working through the calculation, we
find the answer to this is 11π over six, which does indeed lie within the desired
range.

To recap, using this relationship
for π, we have found that, in the complex plane, a π-value of 47π over six is
equivalent to a π-value of 11π over six. This allows us to replace the
π-value in our answer. Since π is now in the desired
range, we have now reached the end of our journey. And we have found that π§ one
multiplied by π§ two to the power of six divided by π§ three to the power of four is
equal to eight π to the power of 11π over six times π.