### Video Transcript

Previously, we learned about the
polar form of a complex number. This is also known as the
trigonometric or modulus argument form. We saw how to recognise when the
complex number is written in this form and to how to convert between this form and
the form 𝑎 plus 𝑏𝑖, known as the algebraic rectangular or Cartesian form of a
complex number. But we haven’t seen why we should
ever write a complex number in polar form. What makes it better than algebraic
form? The answer or part of the answer is
that polar form makes multiplication easy.

Let’s recap the polar form of a
complex number. This is 𝑧 equals 𝑟 times cos 𝜃
plus 𝑖 sin 𝜃, where 𝑟 must be greater than or equal to zero. With 𝑧 written in this form, we
can just read off the modulus and argument. Its modulus is the value of 𝑟. And its argument is the value of
𝜃. This explains why polar form is
also known as modulus-argument form. This is best understood using an
Argand diagram. The modulus 𝑟 is the distance at
the point 𝑧 from the origin zero. And the argument 𝜃 is the measure
of the angle that the vector from zero to 𝑧 makes with the positive real axis, when
measured counterclockwise. I’d like to show you how the polar
form makes multiplication easy. So we’re going to need two complex
numbers.

Let’s call this one 𝑧 one then
instead of just 𝑧, with modulus 𝑟 one and argument 𝜃 one, and introduce the
second complex number, 𝑧 two, whose modulus is 𝑟 two and whose argument is 𝜃 two,
making it 𝑟 two times cos 𝜃 two plus 𝑖 sin 𝜃 two in polar form. Then, it turns out there’s a very
simple expression for the product 𝑧 one 𝑧 two in polar form. This is 𝑟 one 𝑟 two times cos 𝜃
one plus 𝜃 two plus 𝑖 sin 𝜃 one plus 𝜃 two. This is in polar form, with a
modulus of 𝑟 one times 𝑟 two. That’s the product of the moduluses
of 𝑧 one and 𝑧 two. And its argument is 𝜃 one plus 𝜃
two, which is the sum of the arguments of 𝑧 one and 𝑧 two. So we multiply the moduluses but
add the arguments. This gives a geometric
interpretation to the product of complex numbers on an Argand diagram.

We can think of the effect on 𝑧
one of multiplying by 𝑧 two. We end up with a complex number 𝑧
one 𝑧 two, whose modulus is 𝑟 two times as big as 𝑧 ones and whose argument is 𝜃
two more than 𝑧 ones. Multiplication by 𝑧 two therefore
transforms the complex plane by dilation scale factor 𝑟 two, followed by rotation
of 𝜃 two counterclockwise. It’s beyond the scope of this video
to discuss this any further. But you might like to think for
yourself what happens when 𝑧 two is a positive real number or a negative real
number or 𝑖 or an imaginary number.

We’ve made this claim about the
multiplication of complex numbers in polar form. And now we should prove it. We do this by first using the
expressions for 𝑧 one and 𝑧 two in polar form. We can reorder the factors so that
𝑟 two appears next to 𝑟 one. And now we just have to worry about
the terms in parentheses. And this is just the multiplication
of two complex numbers in the algebraic form. We get cos 𝜃 one cos 𝜃 two plus
𝑖 times cos 𝜃 one sin 𝜃 two, notice how we’ve reordered the factors so that 𝑖 is
at the front, plus 𝑖 sin 𝜃 one cos 𝜃 two plus 𝑖 squared sin 𝜃 one sin 𝜃
two. And 𝑖 squared is negative one. So this should be minus sin 𝜃 one
sin 𝜃 two at the end. Using this fact, we can group the
real terms and imaginary terms.

And now, for the main part of the
proof, we recognise the real and imaginary parts from the trigonometric angle sum
identities. The real part is exactly cos 𝜃 one
plus 𝜃 two. And the imaginary part after
swapping the terms is just sin of 𝜃 one plus 𝜃 two. And this is what we had to
prove. These trigonometric identities seem
to come from out of the blue to give us an unexpectedly nice answer. We get 𝑧 one times 𝑧 two in polar
form. And we can read off the modulus and
argument. The modulus is 𝑟 one times 𝑟
two. And the argument is 𝜃 one plus 𝜃
two.

Now, if we look back at our claim,
we can see that 𝑟 one was just the modulus of 𝑧 one and 𝑟 two was the modulus of
𝑧 two. So the modulus of 𝑧 one times 𝑧
two is the modulus of 𝑧 one times the modulus of 𝑧 two. The modulus of the product is the
product of the moduluses. And similarly, 𝜃 one was the
argument of 𝑧 one and 𝜃 two of 𝑧 two. So the argument of 𝑧 one times 𝑧
two is the argument of 𝑧 one plus the argument of 𝑧 two. And knowing the modulus and
argument of this product allows us to write it in polar form. So these two facts taken together
are equivalent to the claim that we proved. And now that we have proved this
claim, let’s apply it.

Given that 𝑧 one equals two times
cos 𝜋 by six plus 𝑖 sin 𝜋 by six and 𝑧 two equals one over root three times cos
𝜋 by three plus 𝑖 sin 𝜋 by three, find 𝑧 one times 𝑧 two.

If 𝑧 one equals 𝑟 one times cos
𝜃 one plus 𝑖 sin 𝜃 one and 𝑧 two equals 𝑟 two times cos 𝜃 two plus 𝑖 sin 𝜃
two, then 𝑧 one times 𝑧 two is 𝑟 one times 𝑟 two times cos 𝜃 one plus 𝜃 two
plus 𝑖 sin 𝜃 one plus 𝜃 two. Comparing what we have with our
formula then, we see that 𝑟 one is two. And 𝑟 two is one over root
three. And the modulus of our product, 𝑟
one times 𝑟 two, is therefore two times one over root three. How about the argument? Well, we can see that 𝜃 one is 𝜋
by six. And 𝜃 two is 𝜋 by three. Our argument is their sum, 𝜋 by
six plus 𝜋 by three. And now, we just need to
simplify. Two times one over root three is
two over root three. And we can rationalise this
denominator if we want to by multiplying both numerator and denominator by root
three, to get two root three over three. How about the argument? We can write 𝜋 by three as two 𝜋
by six. And hence, the argument is three 𝜋
by six or 𝜋 by two.

Making the substitution, we see
that the product we’re looking for is two root three over three times cos 𝜋 by two
plus 𝑖 sin 𝜋 by two. We’ll leave our answer in polar
form even though we might know the values of cos 𝜋 by two and sin 𝜋 by two. And hopefully, having gone through
this example, you will agree that multiplying two numbers in polar form is very
easy. You just have to multiply their
moduluses and add their arguments. That’s less work than multiplying
two numbers in the algebraic form.

Let’s see another example.

If 𝑧 one equals seven times cos 𝜃
one plus 𝑖 sin 𝜃 one, 𝑧 two equals 16 times cos 𝜃 two plus 𝑖 sin 𝜃 two, and 𝜃
one plus 𝜃 two equals 𝜋, then what is 𝑧 one times 𝑧 two?

Well, we know that the modulus of
𝑧 one times 𝑧 two will be the modulus of 𝑧 one, that’s seven, times the modulus
of 𝑧 two. That’s 16. And seven times 16 is 112. And we also know that the argument
of 𝑧 one times 𝑧 two is the argument of 𝑧 one plus the argument of 𝑧 two. The argument of 𝑧 one is 𝜃
one. And the argument of 𝑧 two is 𝜃
two. So this becomes 𝜃 one plus 𝜃 two,
which we’re told in the question is 𝜋. Now we know the modulus and
argument of this product. It’s easy to write it down in polar
form. If the modulus of 𝑧 is 𝑟 and the
argument of 𝑧 is 𝜃, then we can write 𝑧 in polar form as 𝑟 times cos 𝜃 plus 𝑖
sin 𝜃. In our case, 𝑟 is 112. And 𝜃 is 𝜋. Substituting then, we find that 𝑧
one times 𝑧 two is 112 times cos 𝜋 plus 𝑖 sin 𝜋. This is our answer in polar
form.

We could also write this answer as
negative 112. That will be acceptable too. And we could’ve found this value
either by using the fact that cos 𝜋 is negative one and sin 𝜋 is zero or by using
the fact that if a complex number has an argument of 𝜋, that means it’s a negative
real number. And of course, the only negative
real number with modulus 112 is negative 112.

We’ve seen that using the polar
form of the complex number makes multiplication very easy. And the same is true for
division. If 𝑧 one equals 𝑟 one times cos
𝜃 one plus 𝑖 sin 𝜃 one and 𝑧 two equals 𝑟 two times cos 𝜃 two plus 𝑖 sin 𝜃
two, then 𝑧 one over 𝑧 two, the quotient of 𝑧 one and 𝑧 two, is 𝑟 one over 𝑟
two times cos 𝜃 one minus 𝜃 two plus 𝑖 sin 𝜃 one minus 𝜃 two. Equivalently, the modulus of a
quotient of complex numbers is the quotient of their moduluses. And the argument of a quotient of
complex numbers is the difference of their arguments. And it’s this equivalent statement
that we’ll prove.

To prove this, we’ll use what we
know about the modulus and argument of a product of complex numbers. Given two complex numbers, 𝑤 one
and 𝑤 two, we know that the modulus of their product is the product of their
moduluses. And the argument of their product
is the sum of their arguments. This we’ve proved already. The key idea of this proof is to
let 𝑤 one equal 𝑧 one over 𝑧 two and 𝑤 two equal 𝑧 two. Making this substitution, we find
that the modulus of 𝑧 one over 𝑧 two times 𝑧 two is the modulus of 𝑧 one over 𝑧
two times the modulus of 𝑧 two. And on the left-hand side, the 𝑧
twos cancel. And so we get just the modulus of
𝑧 one on the left-hand side. And now, we can rearrange this to
make the modulus of 𝑧 one over 𝑧 two the subject. And when we do this, we find that
it is the modulus of 𝑧 one over the modulus of 𝑧 two, as required.

Now, we just need to do the same
for the argument. Substituting 𝑧 one over 𝑧 two for
𝑤 one and 𝑧 two for 𝑤 two, we get that the argument of 𝑧 one over 𝑧 two times
𝑧 two is the argument of 𝑧 one over 𝑧 two plus the argument of 𝑧 two. And again, the 𝑧 twos on the
left-hand side cancel. So we get just the argument of 𝑧
one on the left-hand side. Rearranging, we find that the
argument of 𝑧 one over 𝑧 two is the argument of 𝑧 one minus the argument of 𝑧
two as required. We’ve managed to prove this
equivalent statement, finding the modulus and argument of our quotient. And with this modulus and argument,
we can write down the polar form of the quotient 𝑧 one over 𝑧 two. This can also be proved like fully
multiplication case just by writing the quotient of the two complex numbers in polar
form. This is more straightforward but
involves a lot of algebraic manipulation. Anyway, now that we’ve seen how we
can divide complex numbers in polar form, let’s apply it to some problems.

Given that 𝑧 one equals 20 times
cos 𝜋 by two plus 𝑖 sin 𝜋 by two and 𝑧 two equals four times cos 𝜋 by six plus
𝑖 sin 𝜋 by six, find 𝑧 one over 𝑧 two in polar form.

Well, we know in general that the
modulus of a quotient of complex numbers is the quotient of the moduluses. And we can just read off the
moduluses. They are 20 and four. So the of modulus of our quotient
is 20 divided by four, which is five. And we can use the fact that the
argument of a quotient is the difference of the arguments, which in our case are 𝜋
by two and 𝜋 by six. And we can write 𝜋 by two over the
denominator six. It’s three 𝜋 by six. So the difference is two 𝜋 by six,
which is 𝜋 by three. That’s the argument of our
quotient. Now that we have the modulus and
argument, we can write our quotient in polar form. 𝑧 one over 𝑧 two is five times
cos 𝜋 by three plus 𝑖 sin 𝜋 by three. Now, isn’t this much easier than
dividing complex numbers in algebraic form, where you have to multiply both
numerator and denominator by the complex conjugate of the denominator? I think so.

Given that 𝑧 equals cos seven 𝜋
over six plus 𝑖 sin seven 𝜋 over six, find one over 𝑧.

Now, we could divide in the way
that we divide a complex number in algebraic form, using the complex conjugate of
that number to make the denominator real. But 𝑧 is also in polar form. Its modulus one is not explicitly
written. But we can write it in. And we can see that the argument is
seven 𝜋 by six. Now, if you write one in polar
form, we’ll have a quotient of two complex numbers in polar form, which we know how
to evaluate. The modulus of one is one. And the argument of one and indeed
any positive real number is zero. So one in polar form is one times
cos zero plus 𝑖 sin zero.

Now, we have to find the reciprocal
of 𝑧. That’s one over 𝑧. And we can write both one and 𝑧 in
polar form. The modulus of their quotient is
then one divided by one, which is one. And their argument is the
difference of the arguments. That’s zero minus seven 𝜋 by six,
which is negative seven 𝜋 by six. We don’t need to write the modulus
of one explicitly. We can just write that one over 𝑧
is cos of negative seven 𝜋 over six plus 𝑖 sin of negative seven 𝜋 over six. This is our final answer.

We’ve seen that using the polar
form of complex numbers makes multiplication and division much easier. They don’t require as much
work. However, we haven’t seen how to add
or subtract complex numbers in polar form. This is because addition and
subtraction of complex numbers in polar form is in general much harder than of
complex numbers in algebraic form. To add two complex numbers in
algebraic form, you just add the real parts and the imaginary parts and similarly
for subtraction. It couldn’t get much easier.

If you do have to add or subtract
complex numbers in polar form, it’s generally best to convert them to algebraic
form, before adding or subtracting and then converting back to polar form if
required. Before the video ends, we’re going
to see another operation which polar form makes easier. This is exponentiation, which
involves finding powers of a complex number. This is a bit of a sneak peek of a
topic which will be covered in more depth later.

Consider the complex number 𝑧
equals one plus 𝑖 root three.

Our first task is to find the
modulus of 𝑧. That’s straightforward. We know that the modulus of 𝑎 plus
𝑏𝑖 is the square root of 𝑎 squared plus 𝑏 squared. Substituting one for 𝑎 and root
three for 𝑏, we get the square root of one plus three, which is the square root of
four, which is two. Now, for the argument of 𝑧, we see
that 𝑧 is in the first quadrant. And so its argument 𝜃 can be found
using arctan. Substituting again, we find that
the argument is arctan root three over one, which a calculator will tell us is 𝜋 by
three. Hence, that means using the
previous results, use the properties of multiplication of complex numbers in polar
form to find the modulus and the argument of 𝑧 cubed.

Let’s not rush straight to 𝑧 cubed
but first consider 𝑧 squared. 𝑧 squared is just 𝑧 times 𝑧. And we know that the modulus of the
product of complex numbers is the product of their moduluses. So we find that the modulus of the
square of 𝑧 is the modulus of 𝑧 squared. How about the argument of 𝑧 times
𝑧? This is the argument of 𝑧 plus the
argument of 𝑧 or two times the argument of 𝑧. Now, we’re ready to consider 𝑧
cubed. We use the fact that 𝑧 cubed is 𝑧
times 𝑧 squared. And so the modulus of 𝑧 cubed is
the modulus of 𝑧 times the modulus of 𝑧 squared, which we know to be the modulus
of 𝑧 squared. And so we see that the modulus of
the cube of a complex number is the cube of the modulus of that complex number. To find the argument of 𝑧 cubed,
we use the same trick, writing 𝑧 cubed as 𝑧 times 𝑧 squared, using what we know
about the argument of a product and the argument of 𝑧 squared, finding that the
argument of the cube of a complex number is three times the argument of that complex
number.

Now, we just need to substitute the
known values of the modulus of 𝑧 and the argument of 𝑧. The modulus of 𝑧 is two. So the modulus of 𝑧 cubed is
eight. And the argument of 𝑧 is 𝜋 by
three. So the argument of 𝑧 cubed is
𝜋. Finally, we find the value of 𝑧
cubed. We know the modulus and argument of
𝑧 cubed. So we can just write 𝑧 cubed down
in polar form. It’s eight times cos 𝜋 plus 𝑖 sin
𝜋. Or using the fact that cos 𝜋 is
negative one and sin 𝜋 is zero, we could write it as negative eight.

More interesting than the final
answer is what we found on the way to it, those being the expressions for the
modulus and argument of 𝑧 cubed. This leads to a very compact
expression for the cube of the complex number in polar form. This is a special case of de
Moivre’s theorem, which will be explored in more detail at another time.

The key points that we’ve covered
in this video are as follows. Calculations involving
multiplication and division of complex numbers are often simpler when we work in
polar form. However, for addition and
subtraction, this is definitely not true. For complex numbers, 𝑧 one equals
𝑟 one times cos 𝜃 one plus 𝑖 sin 𝜃 one and 𝑧 two equals 𝑟 two times cos 𝜃 two
plus 𝑖 sin 𝜃 two, the following rules hold. Their product, 𝑧 one 𝑧 two, is 𝑟
one times 𝑟 two times cos 𝜃 one plus 𝜃 two plus 𝑖 sin 𝜃 one plus 𝜃 two. And their quotient, 𝑧 one over 𝑧
two, is 𝑟 one over 𝑟 two times cos 𝜃 one minus 𝜃 two plus 𝑖 sin 𝜃 one minus 𝜃
two. From these formulas, we can read
off the modulus and argument of a product and quotient of complex numbers. The modulus of the product of
complex numbers is the product of their moduluses. And the argument of the product of
complex numbers is the sum of their arguments. The modulus of a quotient of
complex numbers is the quotient of their moduluses. And the argument of the quotient is
the difference of the arguments.