Once again, it's really difficult to see how this factor is related to the two initial factors, and
we will find out now how they are related.
Before we can do this, we need another trick.
We need to first understand how we can calculate exponential growth.
So exponential of complex numbers.
And I'm not sure if you know this, but the exponential function is actually defined as a role, so
it's not really clear how the exponential function is defined.
And in very simple terms, you really have to consider this role, which is a sum of infinite terms.
So you start by using this expression here and set an equal to zero.
So you get here the number to the power of zero, which is one, and then you get to zero facultative,
which is also one.
So you get one and then you go to the next term where you use an equal one, then another one where
you use an equal two and so on, and you add up an infinite number of terms.
But due to this denominator here, these individual terms will decay and they will decrease to zero.
So this means you add up an infinite number of terms, but the terms will get smaller and smaller.
So the whole row converges to a finite value.
And this is really the definition of the exponential function.
And this is how it is defined in in real space for real numbers.
But we can just use the same definition also for complex numbers as as is written down here.
So let's use this for purely imaginary numbers, so instead of writing down X plus I y, I just used
to I y this was just a special case for imaginary numbers.
And this is actually all we need to understand how we can geometrically multiply and divide complex
numbers.
So let's do this.
Let's use Iwai instead of Z so we get Y to the power of end and I to the power of an ant.
Now just to get a feeling, let's write down the first six terms so we get one, as I already mentioned.
Then we get y times I divide it by one faculty and then we get here minus we get first of all, why
square and then we get here I square which is minus one.
This is why we have this one here.
So it's really simple.
The only thing that you have to take care of this is I.
And here we find that I to the power of three is AI times AI to the power of two which is minus AI.
And then when we have eye to depart for, we have I squared Times Square, which is minus one times,
minus one, which is plus one.
So the exponential of this, I hear they go in circles.
You start with one, then I and minus one minus I.
And then again one this you can see here one eye minus one minus I, one eye and so on and so on.
So now we have this terms and now we can all of them, we can order them by the real part and by the
imaginary part, and you will see that every second term will be real and every second term will be
imaginary.
So here we have all the real parts, which are the terms for N equals zero for two, for four and so
on.
So for even numbers and here we have two imaginary parts for art.
And so just for one, three, five and so on.
And now our next step is to use these real parts and these imaginary parts and we will re express them
in terms of rows, because as you can see here, we are still adding up an infinite number of terms
and here as well.
So we can once again write it as a row, which is a sum that goes to infinity.
Now, here, since we have only even numbers in the exponent and in the denominator, we write down
to end.
And he has a prefect to account for this plus minus, plus minus.
We write minus one to the power of end because this gives us exactly a positive sign, a negative sign
and so on.
And for the other row we have the same thing.
But just instead of writing down to the end, we write down to and plus one so that we get the odd numbers.
So this if you know about trigonometric functions, this is the real expression of the cosine function
of Y.
And this here is the sign expression or the the real expression of the sign of Y.
So what we have here is that the exponent of Iwai is the cosine of Y plus the sign of Y times eye.
And this is a very, very famous equation.
This is called the Solar Equation, named after Oilor, which is the same guy after which the E constant
is named this two point seven something.
And it allows us essentially to express a complex number in terms of the length times some E to the
I find.
So here we know now that this each of the five corresponds to this term and we have here a real part
and an imaginary part.
And the ratio of these two parts can take any value.
So the ratio of cosine divided by sine can take any value, positive and negative.
So this means we can express any complex number in terms of such a relation here.
And then we just have to scale the length because the absolute value of this expression here is always
one, because the absolute value is cosine squared plus sine square, which is always one.
So this is our main result here.
We have expressed our Z, our complex number in terms of its length times, some exponential function
with a purely imaginary argument.
And I go I go one step back.
So here you can see this angle.
Phi is determined by the ratio of these two terms, as I mentioned.
And the ratio of these two terms is the Arcus tangency function Y divided by X.
And now I will show you this later.
In an example, the across tangent function is not really well defined for defining this angle here
because it gives the same value for, let's say, positive and the positive Y and X and also for a negative
Y index because these two negative signs would cancel out.
So by using just this across Tongans, we would get the same angle for these two examples, even though
the real angle would be different by one hundred and eighty.
So there is something missing here.
And a better way to define this angle is to use this across tokens to function, which is often defined
in programming.
For example, if you use Python, you have to turn to function already available.
You can just go ahead and use it.
All right.
So this is a very important result, as you will see now.
So we are now able to express any complex number in terms of this relation here.
Now we can revisit our multiplication from previously.
So let us calculate once again the product of C1 and C2.
So we have already done this previously using this standard equation here and we got the result minus
five, minus five.
So let's use our new Arless relation where we write down the one as the absolute value of C1 times this
yeah.
This complex E function.
And for Z2 two we do the same thing.
So we have now this relation here.
So we have to do is we have to multiply the absolute values and we have to add the the angles five one
and five two.
So we must check now that this term here gives the same thing as previously.
So we need to arrive at minus five, minus five.
So of course, we need, first of all, the absolute values of both numbers.
So the absolute value of C one is the square root of.
The square of the real part, which is one square, which is one and square of the imaginary part,
which is three square, which is nine, so we get a square root of ten for the two, we get a square
root of the square of minus two, which is three apart, which is four plus the square of the imaginary
part, which is one square, which is one.
So we have to square root of five.
So our product C one times two is the square root of the 50.
And we can write 50 as the product of 25 times to and this is good because we can calculate the square
root of 25, which is five.
So we have a square root of two times five.
OK, now we need the angles.
This is something you can calculate when you take your calculator.
You can of course, not do it by hand.
It's it's difficult because you have Tsakos tangents.
And here I will use the old relation that I had previously to show you where the problem is.
So we take four or five one, we take Y one divided by X one.
And so this is equal to Archos Tongans.
Of three, divided by one, which is three, and if we put it in the calculator, we get a value of
one point two for nine, which is equal to if you transform it to an angle, you have to multiply it
by 180 degree and divide by PI.
And what you get is seventy one point six degree.
And that is, of course, the angle here for this blue vector.
Now, four, five, two, we can do the same thing and this time we get our cost tangency of Y two divided
by X two.
So we get the Arcus tangents of um, one divided by minus two, which is minus one half.
And here we get the value minus zero point four six four, which gives us an angle of minus twenty six
point six degrees.
And here you see, this is actually wrong because our angle, as we can see, is somewhere between 90
degree and 180 degree.
So we have to calculate the angle from the positive real access to this green vector and minus twenty
six would be exactly the opposite.
So this is the problem of the Yarkas tanginess function because it is not sine sensitive here for these
two arguments.
So if we would have used the acquis tannin's two, we would have gotten the correct result and we can
get the correct result here by adding PI or by adding 180 degree.
So the correct answer is two point six, seven, eight, which is this value here.
But adding PI and this is equal to this, this corresponds to one hundred fifty three point four degrees.
OK, so now we can just we already have already calculated the product of the absolute values of the
One and Z two.
Now we calculate the sum of five one and five two.
So let's calculate five one plus five two.
Um, this is equal to.
So please you have to take the calculator.
You shouldn't do it by hand.
Would be very difficult.
So you have here the, the, you know, the value and radials and.
This corresponds to an angle of exactly 225 degree, if you would take here all the following numbers
into account, and this is equal to.
The stun guns or this tend to function of minus five, minus five, which is exactly what we wanted,
because we have here minus five, minus five.
So this means our numbers, the one Z two is equal to square root two times five times E to the power
of EI.
And let's just write it down in terms of angles.
And I like this a bit more.
It's a bit easier.
And if you calculate this, you get minus five, minus five EI, which is exactly what we wanted.
So now let's look at what happens here once again.
So I take the laser pointer to show you.
So here we have our two numbers, C one and Z two in blue and green.
And the length is, of course, the absolute value.
And then you have here these angles that have to be measured from the positive, real access to the
vector itself.
So you have five one five two.
And you can see here, if you add these two angles up, you get this one.
And for the length, you just have to multiply these length.
So C one time, Z two and you get this vector.
So now we finally have found out how we can really visualize multiplication of two reciprocal, two
complex numbers, and for this we need the oil gas relation.
We need to express the numbers in terms of the absolute value and then this argument here.
So while this may look a bit difficult, a bit of a long calculation because of all of these Arkless
tangents, it is actually really, really easy if you just do it.
Yeah.
By looking at these vectors.
So if you would not have known what the product is and you don't want to calculate this, you can just
draw these two vectors in the complex plane and roughly estimate the length.
You say, OK, this is a bit longer than three.
This is a bit longer than two.
So yeah, we have here the product of these two, so it has to be a bit longer than six.
OK, this is actually a bit even longer, but yeah, approximately six, maybe seven.
And then here for the angle, you say, OK, this is maybe approximately 70 degrees.
This is approximately I don't know, maybe.
Yeah, 150.
So you take to some one probably 220 degrees.
So yeah.
Goes along this direction.
So you have already a very good feeling what the product of these two numbers will look like.
And you see, if you really do it carefully, it turns out correct.
Now we can do the same thing for the division.
So here we already had our result.
It was one of five minus seven over five times I.
And here we again can use the only representation where we express our number as the absolute value
times this complex function.
So this time we need to calculate the quotient of those two lengths and we need to calculate the difference
in these two vectors.
So first, let's start with calculating the quotient of the two length.
So we have C1 divided by Z to see one more square root of ten.
This one was the square root of five.
So we get the square root of two as our length and for the angles we have five one minus five two and
this is equal to minus one point four to nine, which corresponds to an angle of three minus eighty
one times nine degree.
So it looks a bit ugly, minus eighty one point nine degree, or this is also the same as two hundred
seventy eight point one degree.
And this one is equal to the Archos tangent to function of.
Minus seven over five and one over five.
And now if we if we if we write down our quotients, the one divided by Z two, we get a square root
of two three to the power of IE two hundred seventy eight point one degree, which is exactly one of
a five minus seven over five high, which is what we want.
So now we can yeah, we can do the same argumentation as before.
I will once again switch to the laser pointer and here we can see that we have these two A length C
one, zoo two and these two angles.
And what we have to do is we have to subtract the angles.
So you see, since FI two is larger than five one, we get into negative territories and since these
have almost the same length, but this one is a bit larger, we get the length that is just a bit larger
than one.
But in fact, it's a square root of two, which is one point for approximately.
So once again, you do not have to do this difficult looking calculation.
It would probably be easier to just use this calculation here.
But as you can see, this is also correct.
And this allows us to just.
Yeah, make these considerations so you can really estimate what the quotient of these two numbers will
look like without having to actually calculate it.
Because I think that's really, really nice, you have now learned what complex numbers are, we have
understood how we can do basic mathematical operations and we have learned about this very important
Euler's relation and how you can multiply and divide two numbers in a graphical way.
So this is something that would not have been possible by using vectors.
So now we are good to go to use complex numbers in our project that follows up.