### Video Transcript

Given π§ one equals five π to the
negative π by two π and π§ two equals six π to the π by three π, express π§ one
π§ two in the form π plus ππ.

Here, weβve been given two complex
numbers in exponential form and weβre being asked to find their product in algebraic
form. Itβs much simpler to multiply
complex numbers whilst they are in exponential form. So weβll do that bit first before
converting their product to algebraic form. To multiply two complex numbers in
exponential form, we multiply their moduli and add their arguments.

The modulus of our first complex
number is five and its argument is negative π by two. The modulus of our second complex
number is six and its argument is π by three. This means the modulus of the
product of these two complex numbers will be five times six which is 30. And the argument of π§ one π§ two
will be negative π by two plus π by three.

We can add these two fractions by
creating a common denominator. Thatβs six. And we get negative three π by six
plus two π by six which is negative π by six. And therefore, we see that π§ one
π§ two is 30π to the negative π by six π. And thatβs in exponential form. So how do we convert this into
algebraic form? The easiest way is to represent it
in polar form first. Itβs 30 cos negative π by six plus
π sin of negative π by six. Weβll distribute these
parentheses.

And we see that this is equivalent
to 30 cos of negative π by six plus 30 sin of negative π by six π. Now, these are standard
results. Cos of negative π by six is root
three over two and sin of negative π by six is negative one-half. And we can therefore see that π§
one π§ two β the product of these two complex numbers β simplifies to 15 root three
minus 15π. This is now in algebraic form as
required. If we compare it to the general
form in our question, we see that π is 15 root three and π is negative 15.