Hi everyone, Chris Tisdell here again.
In this presentation I am going to continue
my series of videos on complex numbers.
Now in previous videos I introduced the idea
of a complex number and the imaginary unit,
i, and I showed you how to do basic operations
with complex numbers: addition, subtraction,
multiplication, division, and I talked about
the geometry of these operations as well.
Now in this presentation I am going to show
you how to apply the quadratic formula when
quadratic equations have complex numbers as
solutions.
I am just going to work through an example
so let me share my screen with you and we
can get down to business.
So here is the example I am going to work
through.
we have got a quadratic equation here, z.
13z^2 – 6z + 1 = 0, and we are asked to
solve it and write the solutions in the Cartesian
form x + yi where x is real and y is real
and x is known as the real part of this and
y is known as the imaginary part.
let us just refresh our memory.
This imaginary unit i is defined to be the
√–1, so i^2 satisfies the equation i^2
= –1.
Now, to solve this kind of problem, it is
similar to what you did at school.
You would identify the coefficients, a, b
and c and you would plug them into the quadratic
formula and then you would try to simplify.
Now the challenging thing with this case is
that the b^2 – 4ac is going to be negative,
so I am just going to show you how to work
through that (it relies on this thing).
So, by the quadratic formula, let us call
this, say, (*), the solutions to (*) are the
following: Well, z = –b plus or minus the
√ (b^2 – 4ac) all over 2a.
All we need to do is plug in a = 13, b = –6,
c = 1 and see what we get here.
Now the tricky thing here is the √( b^2
– 4ac).
let us calculate the b^2 – 4ac and see what
happens.
This is really the only potential stumbling
problem on this.
So b^2 – 4ac is going to be (–6)^2 – 4
* 13 * 1 and we are going to get 36 here and
– 52 over here.
So if you simplify this you will get – 16
which is less than 0.
Now at school, you have been told that this
does not have a solution because b^2 – 4ac
is negative and you cannot take the negative
√ a negative number.
But with complex numbers you can take the
√ negative numbers.
So if we are looking for complex solutions,
we can definitely apply this formula (the
quadratic formula).
So, I can replace what is in here with -16
and then I should be able to simplify.
So plugging in now, b is –6 so –b is +6
and the square root sign will have -16 and
a is 13 so 2a will be 26.
So some of you may realise that this √–16
is 4i, but let me just write an extra line
down here.
–16 is 16 x -1 and I can write -1 as i^2.
So now I can write 16i^2 as 4i all squared.
So now what I have here is the square root
of a square, so the square root and the square
will cancel out and I will be left with (6
+ or – 4i)/26 and the I can basically divide
by 2 in top and bottom and that will be something
like (3 + or – 2i)/26.
Now it is not quite in the Cartesian form
yet (that should be a 13 there, let me fix
that up) so I can split that 13 into both
parts.
So let us just write the conclusion down.
So z = 3/13 + 2i/13 or 3/13 – 2i/13 , so
I have just squished it in there at the bottom.
So here are your solutions; there are two
of them and they are written in this Cartesian
form.
So that is how you solve the problem, let
us just analyse the kind of solution we have
here now.
These two solutions are the same, except you
have got a +2i/13 and a -2i/13, the real parts
are the same.
So if you were to plot these two solutions
on the complex plane or an Argand diagram,
to go from this to this you would just reflect
this point on the horizontal real axis.
The real parts are the same and the imaginary
parts differ by a sign.
This is kind of the idea known as the conjugate.
We saw the idea of a conjugate when we were
dividing one complex number by another.
It turns out that for all kinds of quadratic
equations where the coefficients are real.
If you have one complex number as a solution
then the conjugate of that complex number
will also be a solution.
So basically, if you know that you can go
‘okay just change the sign on that one and
that is also got to be a solution’.
So that idea relies on the conjugate and I
will speak about that in forthcoming videos.
So I hope you enjoyed this presentation; I
hope you found it useful.
Please join me for the next presentation on
the conjugate and its applications.
If you have any questions, any comments, you
can post them in the comments sections below.
If you have any suggestions or anything you
would like me to cover, please let me know.
So I hope you can join me in the next presentation;
thank you for watching.
