Hi again everyone!
Chris Tisdell here again.
In this presentation I am going to continue
my series of videos on complex Numbers.Now,
in particular I am going to show you how to
solve a quadratic equation that has complex
numbers as coefficients.
So we will work through an example and let
me share my screen with you and we can get
down to business.
Okay so square roots of complex numbers and
quadratic equations that have complex number
as its coefficients.
Okay so it is a two part problems and we are
going to solve the following problem first
and then we will use that information in the
second part.
We want to solve this z^2 = (x +yi)^2 = 15
+ 8i by computing the real and imaginary parts
of "z".
They are assumed to be integers or whole numbers.
Now this simplifies some of the calculations.
Okay and hence write down the square roots
of 15 + 8i.
We will use that in solving this and apply
the quadratic formula.
So θ forget about the z and θ compute x
and y.
Once we know x and y we have z!
How do we do it ? Well, for part 1, expand
this bracket and equate the real and imaginary
parts, okay so θ expand the brackets.
When I expand this I am going to get x^2 +
(yi)^2, remember i^2 = -1 ! So we get x^2
- y^2 . They are the real parts, and we are
going to get 2 * (x) * (yi).
So, the imaginary part would be 2xy.
This is going to be equal to 15 + 8i.
Okay, so two complex numbers are equal if
and only if their real parts are equal so
(x^2 -y^2) = 15 and their Imaginary parts
are equal so, 2xy = 8.
Okay so if we equate those, we set them equal
and we get two equations.
So θ equate the real and the imaginary parts
to obtain the following.
Okay so we are going to get (x^2 - y^2) = 15.
That is one equation. and 2 xy = 8.
Okay, so how do we solve this?
The nice thing now is that because we know
x and y are both integers, both whole numbers,
we can actually start with 2xy = 8.
Take the 2 to the other side 
and then I can come up with potential values
for x and y just by looking at the factors
of 4.
Okay, so 
factors of 4 are the following, (2 and 2 ) ,and
(-2 and -2), (4 and 1) and (-4,-1).
Now, we cannot have all of those! but we can
rule some of them out.
Okay let us put x = 2 and y = 2 in (x^2 - y^2)
= 15. you do not get 15.
Same with x = -2 and y = -2.
Put those in (x^2 - y^2) you dont get 15.
Those two can actually be scratched out because
they do not satisfy that equation.What about
these two.
well, if x = 4 and y = 1 you will get 15 and
same with -4 and -1.
Okay, so if we apply x ^2 - y^2 = 15, we see
that x and y is 4 and 1 or -4 and -1.
Okay so, there is going to be 2 complex numbers
that satisfy this 
and its going to be 4 + i and -4 - i.
These then are the roots of 15 + 8i.
Okay so that is the first part done.
So θ go and see if we can solve the second
part now, the second part is looking at the
quadratic equation.
quadratic equations that you have seen at
high school.
But this one has a twist.
You have got coefficients that are complex
numbers.
Okay so we still got in the form az^2 + bz
+ c = 0.
But b and c are both complex numbers here.
a is just 1.
What we are going to do is apply the quadratic
formula to solve this and actually the square
root sign we are going to get is going to
be square root of 15 + 8i.
We are going to use part 1 to solve part 2.
Let me show you how that works.
We are going to apply the quadratic formula
that we all know.
We know that z is going to be (-b (+/-) √(b^2
- 4ac))/2a. b in this case will be (-(2+3i)),
a would be 1 and c will be -5+i.
If you put those in here you get the following
. Negative will cancel out.
b ^2 its gonna be this -4*a and a = 1 and
c which is -5 + 1.
All over 2a.
Okay now this is a bit of a mess here but
if you expand this out you will get inside
that square root sign 15 + 8i.
Now we are not filling details in.
Now because we have computed the square root
of 15 + 8i in part 1, I can write an expression
down for this.
We got them to be this and this.
I have already got + and - sign so I will
replace square root with just 4 + i.
That is from part 1.
Now we all we need is to simplify.
If I use the + sign I am going to get (6 +
4i)/2 yielding 3 + 2i.
Negative sign yields -1 + i.
Solutions to that quadratic equations are
given here.
Couple of remarks now, if you do not know
that x and y are integers you can introduce
a third equation here.
You take modulus of both sides and use properties
of modulus to get a third equation involving
x^2 + y^2 and I have done those kind of examples
in other videos.
Now if you are asked to use part 1 in part
2 and you get down to part 2 and you do not
have that you want to check your calculations
quite carefully.
Also, there is an easier way to doing this
known as the polar form but we have not covered
that in this series of videos yet but we will.
Okay that is my presentation and hope you
enjoyed it.
if you want to post a comment or give me some
feedback or some suggestion I am happy to
receive them.
So please join me for next presentation where
I am going to continue my investigation into
complex numbers and hope to join you again!
Bye!
