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.