### Video Transcript

Given π§ is equal to six π minus
six multiplied by four π plus three π all over one plus two π squared, express
the complex number π§ in the form π₯ plus π¦π, and then determine its trigonometric
form.

We have been given a rather
complicated-looking complex number. And the first thing weβre asked to
do is represent it in the form π₯ plus π¦π. Now whilst π is not a variable,
remember itβs the square root of negative one. We can distribute the brackets on
the top and the bottom of our fraction as we would normally. Letβs look at using the FOIL
method.

F stands for first. Weβre going to multiply the first
term in the first bracket by the first term in the second bracket. Six π multiplied by four is
24π. O stands for outer. Weβre going to multiply the outer
two terms. Six π multiplied by three π is
18π squared. And remember, since π is equal to
the square root of negative one, π squared is just negative one. And we can say that 18π squared is
the same as negative 18.

Then, I stands for inner. Weβre going to multiply the inner
terms in each bracket. Negative six multiplied by four is
negative 24. And finally, L stands for last. We multiply the last term in each
bracket. Negative six multiplied by three π
is negative 18π. Collecting like terms, we can see
that the numerator of our fraction becomes six π minus 42.

Next, weβre going to repeat this
process for the denominator, one plus two π squared. And since squaring just means to
multiply it by itself, this becomes one plus two π multiplied by one plus two
π. Expanding this using the FOIL
method, we get one plus two π plus two π plus four π squared, which becomes one
plus two π plus two π minus four.

So, the denominator of our fraction
simplifies to negative three plus four π. And we can see that we can rewrite
our complex number as six π plus 42 over negative three plus four π. This still isnβt in the form π₯
plus π¦π though. So, how do we divide these two
numbers?

Weβre going to need to multiply
both the numerator and the denominator of this fraction by the conjugate of the
denominator. And remember, for a complex number
of the form π plus ππ, its conjugate, often denoted π§ bar, is π minus ππ. Essentially, we change the sign
between the two terms. So, the conjugate of negative three
plus four π is negative three minus four π. And weβre going to need to expand
the brackets on the top and the bottom of this fraction again.

And doing so, we get negative 18π
minus 24 squared plus 126 plus 168π all over nine plus 16. And, once again, we use the fact
that π squared is equal to negative one. And negative 24π squared is
positive 24. So, our complex number simplifies
to 150 plus 150π all over 25. And since 150 and 25 share a common
factor of 25, we divide through. And we can see that our complex
number is now in the form π₯ plus π¦π. Itβs six plus six π.

Our very last step is to write this
number in trigonometric form. For a complex number in algebraic
or rectangular form, π plus ππ, we write it in trigonometric form as π cos π
plus π sine π, where π is the modulus of the complex number and π is its
argument. And we can use these two-conversion
formulae to help us find the modulus and the argument. The modulus is the square root of
π squared plus π squared. And the argument is arctan, or
inverse tan, of π over π.

For our complex number, π is the
constant; Itβs six. And π is the coefficient of π;
Itβs also six. So, we can see that the modulus of
our complex number is the square root of six squared plus six squared, which is the
square root of 72, or six root two. And the argument is the arctan of
six over six, which is π by four.

And itβs always worth
double-checking this value on the Argand diagram. Remember, the horizontal axis
represents the real component and the vertical axis represents the imaginary
component of our complex number. Six plus six π lies in the first
quadrant as shown.

And we always measure of the
argument from the horizontal axis in a counterclockwise direction. So, we can see that π by four is
certainly a sensible value for our argument. All thatβs left is for us to
substitute the value of the modulus and argument into the general trigonometric
form. And we see that itβs six root two
multiplied by cos of π by four plus π sin of π by four.