### Video Transcript

In this video, weβre going to learn
how to work with purely imaginary numbers. Knowing how to work with these
numbers is an important foundation towards working effectively and confidently with
complex numbers. Weβll begin by learning how to
evaluate and simplify imaginary numbers including when finding the product of these
numbers. Then, weβll discover how to solve
equations which have imaginary solutions.

Rafael Bombelli was the
mathematician regarded as the inventor of complex numbers. Whilst other mathematicians were
solving equations by recognizing purely real solutions, Bombelli saw the usefulness
of working with the square root of negative numbers and he introduced the rules of
arithmetic for imaginary numbers that we still use today. Interestingly, Bombelli shied away
from giving a special name to the square root of negative numbers, instead choosing
to deal with them as he would any other radical. He called what we now know as π
βplus of minus.β And he used the term βminus of
minusβ to describe negative π.

But what is the definition of this
imaginary number that we now call π? In its most basic form, π is
defined as the solution to the equation π₯ squared equals negative one. This means that π squared is equal
to negative one. And we can in turn say that π is
equal to the square root of negative one. π is called an imaginary number,
essentially because itβs not part of the set of real numbers. What this means is any real
multiple of π β in other words, ππ, where π is a real number β is also a purely
imaginary number.

So why do we use them? Why donβt we just stick with a set
of real numbers that we already know so well? Well, as weβve already seen in the
definition for π, there are some equations which have no real solutions. Imaginary numbers allow us to
actually solve these equations. Letβs look at an example of
this. Weβll begin by considering an
equation with very little in the way of rearranging required.

Solve the equation π₯ squared
equals negative 16.

To solve an equation like this, we
do begin by solving as we would any equation with real solutions by performing a
series of inverse operations. In this case, weβre going to find
the square root of both sides of the equation. Before we do though, we choose to
rewrite negative 16 slightly. Weβre going to write it as 16π
squared. And weβll see why we do this in a
moment. But for now, it works because π
squared is equal to negative one. And this means that 16π squared is
16 times negative one which is negative 16.

And now that we have the equation
π₯ squared equals 16π squared, we can now find the square root of both sides of the
equation, remembering that we can take both the positive and negative root of 16π
squared. The square root of π₯ squared is
π₯. So π₯ is equal to the positive and
negative square root of 16π squared. And during this next step, itβs
going to become evident why we chose to write negative 16 as 16π squared. We can split the square root of
16π squared into the square root of 16 times the square root of π squared.

The square root of 16 is four and
the square root of π squared is simply π so in turn we can see that π₯ is equal to
plus or minus four π. The solutions to the equation π₯
squared equals negative 16 are four π and negative four π. And we should now be able to see
why we did write negative 16 as 16π squared. It made these final steps a little
easier to deal with.

And of course, we can check these
solutions by substituting them back into the original equation. Letβs try this for π₯ equals four
π first. π₯ squared is four π squared. And of course, thatβs four π times
four π. Multiplication is commutative. It can be performed in any
order. So we can rewrite this as four
times four times π times π. Four multiplied by four is 16 and
π times π is π squared. And since π squared is negative
one, π₯ squared is negative 16 as required.

We can repeat this process for π₯
equals negative four π. π₯ squared is negative four π
times negative four π, which can in turn be written as negative four times negative
four times π times π. Once again, since negative four
times negative four is positive 16, we get 16π squared which is negative 16 as
required.

Next, weβll have a look at an
equation which requires just a little more work to solve it.

Solve the equation two π₯ squared
equals negative 50.

To begin solving this equation,
weβre going to divide both sides by two. Negative 50 divided by two is
negative 25. So π₯ squared is equal to negative
25. We now rewrite negative 25 as 25π
squared. And remember we can do that because
π squared is negative one. And then, we find the square root
of both sides of this equation.

Of course, we can find both the
positive and negative square root of 25π squared. So π₯ is equal to plus or minus
root 25π squared. We can then write the square root
of 25π squared as the square root of 25 multiplied by the square root of π
squared, which is simply five π. So π₯ is equal to plus or minus
five π. The solutions to the equation two
π₯ squared equals negative 50 are five π and negative five π.

In these next examples, weβll look
at how to extend the rules of arithmetic and algebra for real numbers to help us
solve problems involving purely imaginary numbers.

Simplify two π squared multiplied
by negative two π cubed.

When we square a number, we
multiply it by itself. So two π squared is the same as
two π multiplied by two π. And since multiplication is
commutative, we can write this as two times two times π times π. And in fact, this is a little bit
like evaluating an algebraic expression. We multiply two by two to get four
and we multiply π by π to get π squared. But remember π is not a
variable. Itβs the solution to the equation
π₯ squared equals negative one such that π squared is negative one. So four π squared is four
multiplied by negative one which is negative four.

Next, weβre going to evaluate
negative two π cubed. But weβre not going to write it out
as negative two π times negative two π times negative two π. Instead, weβre going to use the
rules for exponents that weβre used to. And weβre going to write it as
negative two cubed times π cubed. Negative two cubed is negative
eight. But what about π cubed? Now it might look a little
scary. But itβs just the same as writing
π squared times π. And π squared is negative one. So our expression becomes negative
eight multiplied by negative one times π which is simply eight π.

Our final step is to replace two π
squared and negative two π cubed with negative four and eight π, respectively. And then, weβll evaluate that just
as we would any algebraic expression. It becomes negative four times
eight π which is negative 32π.

Weβve just seen that we can apply
some of the rules for manipulating algebraic expressions to help us evaluate those
involving imaginary numbers. And we just saw the result that π
cubed is equal to negative π.

At this stage, it might be useful
to consider what happens with other powers of π, π to the power of four or five
for example. We can evaluate π to the power of
four by thinking of it as π squared times π squared. And then since π squared is
negative one, we say that π to the power of four is negative one multiplied by
negative one and that simply one. And at this point, we can begin to
generalise.

Weβre going to raise this entire
equation to the πth power. And this works for integer values
of π. When we do, we see that π to the
power of four π is equal to one to the power of π. But actually, one to the power of
anything is just one. So we can see that π to the power
of four π equals one. We might then choose to multiply
both sides of this equation by π or π to the power of one.

Remember when we multiply two
numbers with the same base, here thatβs π, we add the powers. So π times π to the power of four
π is π to the power of four π plus one and π to the power four π plus one is
equal to π. Letβs do this again. When we do, we see that π to the
power of four π plus two is equal to π squared. But π squared is simply negative
one. So π to the power of four π plus
two is equal to negative one. Weβll repeat this process one more
time. And we see that π to the power of
four π plus three is negative π.

And now we stop. Why? Well, if we were to multiply by π
again, weβd have π to the power of four π plus four. But four is a multiple of four. So this will have the same result
as π to the power of four π. And this cycle repeats
infinitely. Thereβs a nice graphic we can use
to help us to find any power of π. For integer values of π, we can
use this cycle to define any power of π. Letβs look at the potential of
these results by simplifying an expression in terms of powers of π.

Simplify π to the power of 30.

To simplify this expression, we
really donβt want to write π out 30 times and evaluate each pair. Instead, weβll recall the cycle
that helps us remember the identities for various powers of π. Letβs compare our number π to the
power of 30 to this cycle. We need to represent the power 30
in the form four π plus π. And to correspond to the powers of
π in our cycle, π would be zero, one, two, or three.

Now in fact, 30 can be written as
four multiplied by seven plus two. So π to the power of 30
corresponds to the part of the cycle where π is to the power of four π plus
two. According to this, π to the power
of four π plus two is equal to negative one. And this means that π to the power
of 30 is negative one.

Now another method we could have
chosen would still have been to write π to the power of 30 as π to the power of
four times seven plus two. And we know by the rules of
exponents that this is the same as π to the power of four to the power of seven
times π to the power of two. π to the power of four is one and
π squared is negative one. So our expression becomes one to
the power of seven multiplied by negative one which is once again negative one.

So weβve seen how this cycle can
save us time when working with positive powers of π. And in fact, itβs important to
remember that these sets of rules for simplifying powers of π do actually work for
negative powers too.

Letβs look at a more in-depth
example of this.

Given that π is an integer,
simplify π to the power of 16π minus 35.

Remember the cycle that helps us
remember the identities for various powers of π is as shown. So we can do one of two things. Our first method is to use the laws
of exponents to essentially unsimplify our expression a little. We know that π₯ to the power of π
times π₯ to the power of π is the same as π₯ to the power of π plus π. So we can reverse this and say that
π to the power of 16π minus 35 is equal to π to the power of 16π times π to the
power of negative 35.

π to the power of 16π can
actually be further written as π to the power of four to the power of four π. This corresponds to the part of our
cycle π to the power of four π. So we can see that π to the power
of 16π can be written as one. And what about π to the power of
negative 35? This one is a little more
complicated. Weβre going to write negative 35 in
the form four π plus π, where π can take the values zero, one, two, or three to
correspond with the values in our cycle. Itβs the same as four times
negative nine plus one.

Remember four times negative nine
is negative 36 and adding one gets us negative 35. And we chose negative nine instead
of negative eight as we needed π to be zero, one, two, or three. And we certainly donβt want it to
be a negative value. So π to the power of negative 35
will have the same result as π to the power of four π plus one in our cycle;
thatβs π. So π to the power of 16π minus 35
is one multiplied by π which is π.

Letβs have a look at the
alternative method. Here we would have jumped straight
into writing the power β thatβs 16 π minus 35 β in the form four π plus π, where
π again is zero, one, two, or three. We can write 16π as four times
four π and negative 35 as four times negative nine plus one. We can factor this expression and
we see that 16 π minus 35 is the same as four multiplied by four π minus nine plus
one. So we can see that once again π to
the power of 16π minus 35 will have the same result as π to the power of four π
plus one in our cycle; thatβs π.

Our very final example involves one
of the laws of radicals weβve briefly looked at in this lesson. Thatβs the square root of π times
π is equal to the square root of π times the square root of π. We need to be extremely careful
with this rule. As whilst it works for all positive
real numbers, the same cannot be said for negatives.

Simplify the square root of
negative 10 times the square root of negative six.

Weβll begin by expressing each
radical in terms of π. Remember π squared is equal to
negative one. So we can say the square root of
negative 10 is the same as the square root of 10π squared. And similarly, the square root of
negative six is the same as the square root of six π squared. And at this point, we can split
this up. We get the square root of 10 times
the square root of π squared. And since the square root of π
squared is π, we can see that the square root of negative 10 is the same as root 10
π. And similarly, the square root of
negative six is root six π.

Next, we multiply these
together. Multiplication is commutative. So we can rearrange this a little
and say itβs equal to the square root of 10 times the square root of six which is
root 60 times π squared. And since π squared is negative
one, we see that the square root of negative 10 times the square root of negative
six is negative root 60. And in fact, we need to simplify
this as far as possible.

There are a number of ways to do
this. We could consider 60 as a product
of its prime factors. Alternatively, we find the largest
factor of 60 which is also a square number. In fact, that factor is four. So this means that the square of 60
is the same as the square root of four times the square root of 15 which is equal to
two root 15. And we fully simplified our
expression. We get negative two root 15.

Letβs look at what would have
happened had we applied the laws of radicals. We would have said that the square
root of negative 10 times the square root of negative six is equal to the square of
negative 10 times negative six which is equal to the square root of positive 60 or
two root 15 and thatβs patently different to our other solution.

In this video, weβve learned that
many of the rules of arithmetic and algebra that weβre so confident with can be
extended into the world of imaginary and complex numbers. Weβve also seen that some of the
rules requires to be a little bit more careful, such as generalising the law for
multiplying radicals when these radicals include negative numbers. We also saw how the integer powers
of π form a cycle and that allows us to simplify any real power of π fairly
quickly.