PROFESSOR: Hi.
Today we'll explore the complex numbers and Euler formula.
So the first part of the problem is
to write this complex number, minus 2 plus 3i, in polar form.
And at this point, it's written in rectangular coordinate form.
The second question asks you to do the reverse,
to write three exponential to the i*pi over 6 in rectangular
coordinate form.
Question c asks you to draw and label the triangle
relating the rectangular coordinates
to the polar coordinate form.
D asks you to compute 1 over this reverse complex number
that we already saw in question a.
And e asks you to find the cube root of 1.
And in all these questions, you'll be using Euler formula.
So why don't you pause the video,
take a few minutes to work out the problem,
and we'll come back.
Welcome back.
So we're asked throughout the problem
to go back and forth between coordinates in polar form
and in rectangular form.
So a key thing to remember is Euler formula from the start.
So we're just going to write it up here.
It allows us to express a complex exponential
into the sum of its cosine plus i sine theta.
So how do we tackle question a?
Question a gives us a complex number in rectangular form.
So in this form, a plus i*b.
And we're asked to write it in polar form, which
introduces the modulus of the complex number r and its phase
theta.
So r, the modulus of the complex number
that we can compute when we know its rectangular form,
with its real form squared plus imaginary part squared,
the whole thing under the root.
So in this case, we have 4 plus 9.
So we end up with root of 13 for the modulus
of the complex number z.
So now for the phase.
Using Euler formula, we can see that we
can relate the rectangular form to the polar form
by just introducing-- I'm going to keep r.
And you can see now that we can extract
the sine and the cosine of the angle theta
and relate that to a ratio of a.
And the modulus r that we just found,
b modulus r that we just found.
Or in one move, just express it as the tangent of the angle
theta, just sine over the cosine,
just becomes basically b over a, which we have here.
So we have the modulus r, which is now root of 13,
and the angle theta that we can now
extract by using the reverse of the function tan.
So just before we move to the next question,
this is not one of the classical angles that you learned.
So just to have an idea of where this angle lies
on the trigonometric unit circle,
just recall here that the sine is positive
and the cosine of this angle is negative.
So we're bound to be in this region, where basically theta
is between pi and pi over 2.
And that's the answer to question a.
So now for question b, we're asked to do the reverse,
expressing the polar number 3 i*pi over 6 in rectangular
coordinates.
So now this is just a straightforward application
of the Euler formula that we just saw.
By just expanding the exponential,
as I already wrote there.
Plus i 3 sine pi over 6.
And on the same trigonometric circle
here, that's roughly where pi over 6 lie.
And you can just re-express this as 3 root of 3 over 2, plus i 3
over 2.
So that ends the solution for question b.
So now question c.
Let me just add a line here.
Question c we're asked to draw and label
the triangle relating rectangular to polar
coordinates.
So that's what we already had a sense with when we wrote this
formula going from a plus i*b to r exponential of i*theta.
So in the complex plane, we have the real axis,
an imaginary axis, and a complex number
lying on this plane written in this form
in a rectangular coordinate.
You'd have a projection of a in the real axis, projection
of value b on the imaginary axis.
And in polar form this would be its modulus
or distance from the origin, and its phase theta that
would come in the polar form.
So that's the triangle that allows
us to go back and forth between the rectangular and polar
coordinates.
So to almost finish, question d now
asks us to compute the reverse of the original complex number
that we used.
So 1 over minus 2 plus 3i.
So to do this, we can stay in rectangular form
and basically multiply the numerator and denominator
by the complex conjugate of the number.
But clearly now that we learned how
to use polar coordinate expressions of this number,
it's much easier to just write it directly in this form.
In one step we basically arrived to the results,
where we express that the angle was the reverse tan of minus 3
over 2.
And that's done.
So now are for the last question,
we were asked to compute the one third root of 1.
So basically, 1 to the 1/3.
So here, obviously, we're treating 1 as a complex number.
And if we go in the complex plane
and I just introduce here the number 1,
we see that in polar form 1 is just basically
a complex number with modulus 1, and angle 0, modulo 2pi.
So we can write 1 as exponential 2n*pi,
because it's basically angle 0 modulus 2pi.
And from here we know that we're looking at third roots,
so we're going to have three roots.
And these roots are going to be expressed
by changing the value of n.
First one, n equals to 0 is just going
to give us back root of 1, because we're
going to have exponential to 0 is just 1.
Power of 1/3 is just 1.
n equals to 1.
We are going to have exponential of 2pi
over 3, which we can express, again using the Euler formula,
also in coordinate form.
And then just write down the values.
And for the third root we take the value n equals to 2,
so we have i*4pi over 3, which again we can express
as the cosine plus the sine of 4pi over 3.
So where do these roots lie?
So we have root 1 for n equals to 0.
The second root, exponential 2pi over 3, basically in polar form
would be here, where we would have the angle 2pi over 3.
So 1pi over 3 would here.
2pi over 3 would be here.
3pi over 3 would be here.
And 4pi over 3 is our third root, it would be here.
So then we can just write down the values.
And you can do that when you know the angles,
or just keep it in either form when you don't know directly
the expression for the angles.
So this completes the problems.
In all of these problems what we kept using
is Euler formula to go back and forth between coordinate
in rectangular form to expression
of complex number in polar form.
And that's the key formula that we kept using.
And you'll be using this repeatedly when
we will be solving other ODEs for which we
can use complex number as a trick for solutions.
And this ends this session.