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PreCalculus: Trig to find a Right Triangle angle 
PreCalculus: Simplifying Using Trig Identities 
PreCalculus: Trig Equations and Quadratic Formula 
PreCalculus: Solving Trigonometric Equations 
PreCalculus: Polar & Rectangular Coordinates 
PreCalculus: Trig to Find Right Triangle sides 
PreCalculus: Find Angle Complements & Supplements 
PreCalculus: Complex Numbers  Trig or Polar Form 
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PreCalculus: Solving Trig Equations by Factoring
About this Lesson
 Type: Video Tutorial
 Length: 10:43
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 115 MB
 Posted: 01/22/2009
This lesson is part of the following series:
Trigonometry: Full Course (152 lessons, $148.50)
PreCalculus Review (31 lessons, $61.38)
Trigonometry: Complex Numbers & Polar Coordinates (15 lessons, $26.73)
Trigonometry: Powers and Roots of Complex Numbers (4 lessons, $5.94)
Professor Burger explains how to find the powers and roots of complex numbers. The equation of a complex number is z= r(cosx + isinx). To raise the complex number to a power, n, the equation is z^n = r^n[cos(nx) + isin(nx)]. In general, if you are raising a complex number to the power of n or 1/n (taking the nth root), you will come up with n solutions, as you will always have one solution for each of the degrees of power. When taking the root of a complex number, you will find one solution for each degree of power. To find the nth root of a complex number the equation is n root of z = (n root r) *[cos ((x + 2 Pi K)/n) + 1 sin ((x + 2 Pi K)/n)] where k = 0, 1, 2,...n1.
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Precalculus. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/precalculus. The full course covers angles in degrees and radians, trigonometric functions, trigonometric expressions, trigonometric equations, vectors, complex numbers, and more.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
 Thinkwell
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11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
Thinkwell lessons feature a starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
Applications of Trigonometry
Powers and Roots of Complex Numbers
Roots of Complex Numbers Page [1 of 3]
So we saw that there’s a really easy way of taking a complex number written in trig form or in polar form and raising it
to higher and higher powers. You just use the following formula. If you have it in trig form like this,
r(cos(?) +isin(?)) , and if you want to raise that number to the n th power, where n is either 1, 2, 3, 4, 5, and so
forth, you just take r to that power, and then the angle you would have is the power times the angle. That’s all there
is to it. Real easy. But what if you want to go backwards? Suppose I give you now a complex number and I want you
to find the n th root? So I don’t want you to raise to a power; I want you to take a root now. What would you do?
Well, it’s going to be a similar albeit slightly different formulation of this kind of thing.
Let me begin with a very specific example where we can actually find the root and then see what the formula is.
Suppose I tell you that we have z4 =16 and I want you to find all the solutions to this. Well, z is a complex number
and it turns out there will always be four solutions. And in general, if I have a polynomial that’s of degree n , there will
always be n solutions. Even though we already saw when we took a look at finding real solutions to polynomials,
there may not be always n real solutions—like there could be actually a quadratic that has no real solutions, but there
will always be two complex solutions in that case. So in general when you’re working with complex numbers, you will
always have one solution for each of the degrees of power here. So what I’m actually asking for here is, I want to find
the 4th root of 16. There should be four of them. Let’s see if we can find them.
One way to find them in this easy example would be just to try to factor. I bring everything over to this side and then I
see this is the difference of two perfect squares, so I can factor that really easily. We can just try to factor it by hand.
So if you check, you’ll see that in fact that’s the factor of that. This is still a difference of two perfect squares, and so I
can factor this into (z?2)(z+2), and then I have this thing here. I’ll just keep that asis for a second. And if all
these things equal zero, therefore either this equals zero, which means z = 2 ; this equals zero, z = ?2 . And what
about this one? If this were to equal zero, that means that z2 would have to be ?4 . But if I take plus or minus the
square roots of both sides—I’m allowed to be complex now—then I would see ± ?4 , which is ±2i . So I see four
different solutions. I see 2, 2, 2 i , and 2 i . So in fact I found all the four solutions to this, and you can check. Take 2
and raise it to the 4th power and you get 16. Take 2 and raise it to the 4th power and you also get 16, because a
negative times a negative is a positive; times a negative is a negative; times a negative is a positive, +16. Try that
also with the 2 i and the 2 i . If you raise it to the 4th power, you’ll still get 16. So there is one where we can actually
find all the roots.
But what if you just can’t read them off that easily? Well, then you can use the following. So suppose that I give you
z in polar form, r(cos(?) +isin(?)) , and I want you to find the n th complex root of this number. I want you to take
the n th root and find all of them. There are going to be n different ones. Well, then it turns out that here’s how to
write them. So then the n complex numbers—I’m going to write down n complex numbers now—and here they are.
You take the n th root of r ; you take the n th root of that. And then for the angle, what we’re going to do is we’re going
to take that angle and we’re going to add 2? k and divide the whole thing by n . And I’m going to do the exact same
thing with the imaginary part, so I have
isin( ) 2k
n
? +?
. Now I haven’t told you what k is yet. You know what n is.
That’s the n th root I’m taking. But what’s k ? Where k is going to take on the following values: 0, 1, 2, all the way
out to n ?1. If I start with zero and count all the way up to n ?1, you’ll notice that k will take on exactly n values.
You see n ?1 value here and then plus the zero would be a total of n values. So in fact there are n different
numbers and you get them by first letting k be zero and plugging that in. Then let k =1 and plug that in. Then let
k = 2 and plug that in. And if you keep plugging in, you’re going to get n different numbers. And it turns out then,
the n complex numbers—and here they all are—are the n th roots of z .
So if you want to find the n th roots of a complex number, there are going to be n of them and this is exactly how you
find them by using this formula. So it’s a little bit complicated. The beginning is pretty straightforward. You take the
n th root. But here with the angle it’s a little tricky. You take ? and then you add it; you shift it by 2? k , where the k
is going to vary, and then divide by n . It’s the opposite of the DeMoivre Formula.
Applications of Trigonometry
Powers and Roots of Complex Numbers
Roots of Complex Numbers Page [2 of 3]
Let’s do an example of this and let’s in fact look at the easy example we just looked at. The easy example we just
looked at, let me remind you, was to find the 4th roots of 16. So let’s find the 4th roots of 16 using the method I just
showed you and hope we come up with the same answers. So what do you do with the 4th roots of 16? Well, let’s
see what 16 equals. Well, 16, how do you write that in polar form? That’s 16(cos 0 + i sin 0) . We wanted that
because look at where 16 is. Sixteen is right here. So its length is 16, and the angle it makes here is zero, so it’s
pretty easy to write 16. It’s just 16(cos 0 + i sin 0) .
So now what does this formula say to do? Well, this formula says that what we’ve got to do is you want to find all the
4th roots because you’ve got to make this number where you put in 4 here and 16 here. And then what you’ve got to
do is
cos 0 2
4
+ ? k
, and the same thing here. And if you do that, let’s see what you get. So we consider the following.
We consider the 4th root of 16 times cosine of—and now what do we do? We take ? , which in this case is zero, plus
2
4
? k
plus i times the sine of that same thing. So zero plus—oops, I’m running out of room here. Let me squeeze
myself back. This is so important I’m going to squeeze myself. Here we go. It’s always great to squeeze yourself out
of the picture a little bit. Ready? Okay, that should do it right there. And so
2
4
? k
.
Now, what does k do? Well, k is going to range 0, 1, 2, 3. Notice there are four values there. Each of those should
be a 4th root. What’s the 4th root of 16? Well, that’s just 2, so that’s pretty easy. And what do I see here? Cosine
of—let’s let k be zero. If k is zero, I put in a zero here and this whole thing becomes zero. So I see 0+ isin0.
And cos0 =1; sin 0 = 0 ; so this just equals 2. So one of the roots should be 2.
What about another root? Well, let’s start with k now being 1. So I still have this 2 out in front. Cosine of—and if I let
k be 1, I see just a
2
4
?
, which is
2
?
. So I see sin( )
2 2
?+ i ?. And what’s the cosine of
2
?
? Well, the cosine of
2
?
is
zero. The sine of
2
?
is 1. So I just see an i here. So 2 times i equals 2i .
What if I let k be the next number, k be 2? If k = 2 , then what do we get? Well then I get 2 times cosine of—now
k is now 2, so 2 x 2 is 4? , divided by 4 is just ? . So I see cos(?) + i sin(?) , and what does that equal. Well,
cos(? ) is what? Well, that’s 1. Sine of ? is zero, and so here I just see (?1)(2)=?2.
There should be one last solution, and that’s what I get when I plug k and evaluate that at 3. So I see cosine of—
now I put a 3 in here. Here I’ll see
(2)(3 )
4
?
which is
6
4
?
. But
6 3
4 2
?= ?, so I see
3 sin(3)
2 2
?+ i ?. And what do I
see? Well,
cos(3 ) 0
2
? = ;
sin(3 ) 1
2
? = ? , so I see (?i)(2)=?2i.
So using this formula I see four roots: 2, 2i,?2,?2i. But remember that when we did it by hand, that’s exactly what
we got. We got 2,?2 , and then 2i and ?2i . So in fact we’re seeing the exact same answers using this formula.
So that shows that the formula holds, at least in this case, and in fact the formula holds in general. So the recipe is—
it’s a little bit elaborate, but all you do if you want to take the n th roots of a complex number written in polar form, you
take the n th root of the modulus, and then with the angle you take ? and you add to it 2? k , divide the whole thing by
n , the root you’re taking, and let k range from zero up to n ?1.
Applications of Trigonometry
Powers and Roots of Complex Numbers
Roots of Complex Numbers Page [3 of 3]
Up next we’ll take a look at one more example of using this formula in order to take roots, just so you get a little more
practice. I’ll see you there.
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Great online video tutorial. This totally clears up any problems I had trying to find the root of a polynomial solution. Thanks for posting!