PDA

View Full Version : Complex Zeros and the Fundamental Theorem of Algebra



Ihatenotunderstandingmath
08-08-2011, 06:56 PM
I have tried for many hours to grasp the concept of complex zeros and its use, but I have failed or it's just that my pre-calculus textbook is just written confusingly. If you don't mind, I have a few questions that I can't figure out:

1) My pre-calculus textbook tells me that the degree of a polynomial is the number of complex zeros it will have. That doesn't make sense to me because what if the equation was 3x^8 - 3? It would have -1 and 1 as it's real solutions; where does the complex number come in as a solution? (is it supposed to be 'the degree of a polynomial is the number of either real or complex zeros it will have'?)

2) What exactly IS the fundamental theorem of Algebra? My textbook says: Every complex polynomial function f(x) of degree n is greater than or equal to one has at least one complex zero.

3) What IS a complex polynomial? My textbook keeps using that but it never gives any examples of a complex polynomial. I'm assuming it's a polynomial with a complex coefficient like for example, 3i(x^2) for example?

4) I can't figure out the zeros of this problem: 2(x^4) + 5(x^3) + 5(x^2) + 20x -12 ; zero: -2i
I was thinking that since -2i is a zero and it's conjugate 2i must be a zero, we should multiply the conjugates and then divide the expression by the product. But then -2i * 2i = 4 and how are you supposed to divide the polynomial by 4? Even if you tried synethic division (assuming x - 4 as the divisor), you come up with some gigantic number. How would you solve this problem?

Thank you very much!

mmm4444bot
08-08-2011, 07:36 PM
1) My pre-calculus textbook tells me that the degree of a polynomial is the number of complex zeros it will have.

Yes, but that definition of degree could be better presented. I would first say that the degree of a polynomial is the same as the largest exponent that appears.

I would say second that the number of Complex zeros is the same as the degree.

That doesn't make sense to me because what if the equation was 3x^8 - 3? It would have -1 and 1 as it's real solutions; where does the complex number come in as a solution?

Remember that the set of Real numbers is a subset of the set of Complex numbers. In other words, all Real numbers are also Complex numbers.

Hence, 1 and -1 are Complex zeros, as well as Real zeros.

The eight Complex zeros of 3x^8 - 3 are 1, -1, i, -i, 1/2 sqrt(2) [1 + i], 1/2 sqrt(2) [1 - i], -1/2 sqrt(2) [1 + i], -1/2 sqrt(2) [1 - i]


2) What exactly IS the fundamental theorem of Algebra? My textbook says: Every complex polynomial function f(x) of degree n is greater than or equal to one has at least one complex zero.

That is correct. Another interpretation is: "For any polynomial in x that you can create (with imaginary or strictly Real coefficients, or even a combination of the two), there will always be at least one value of x that causes the polynomial to evaluate to zero." Pretty amazing, actually.

3) What IS a complex polynomial?

You're correct; it's a polynomial with Complex coefficients.

Again, this can mean any of the following examples because Real coefficients are also Complex coefficients.

3x^8 - 3

3i x^2 + 1

x^2 + i

4) I can't figure out the zeros of this problem: 2(x^4) + 5(x^3) + 5(x^2) + 20x -12 ; zero: -2i

We do not need to write grouping symbols around the powers of x:

2x^4 + 5x^3 + 5x^2 + 20x - 12

I was thinking that since -2i is a zero and it's conjugate 2i must be a zero, we should multiply the conjugates and then divide the expression by the product.

That's almost correct.

We multiply the factors, not the zeros.

In other words, since we know that 2i and -2i are both zeros, we know that (x+2i) and (x-2i) are factors of the polynomial.

Multiply those factors (think "difference of squares"), and divide the polynomial by the resulting product (which is also a factor of the polynomial, so the remainder of the division will be zero).

You should get a quadratic polynomial for the quotient. Find the remaining two zeros from that.

Thank you very much!

You are most welcome. Thank you for asking direct questions about specific parts of your exercise. Most people here don't ask diddly.



Please show your work, if you would like more assistance with this exercise..

Finally, be aware that people are not always totally clear when they talk about Complex numbers. I try to use phrases like "imaginary number" or "Complex number with an imaginary part" to mean exactly that. I try to use phrases like "strictly Real" or "Complex number with no imaginary part" to mean a Real number.

"Complex number" means a number of the form A+B*i, where symbols A and B are any strictly-Real numbers. In other words, "Complex number" means A+B*i regardless of whether A, or B, or both A and B are zero.

Cheers ~ Mark

lookagain
08-08-2011, 08:40 PM
I was unable to.

----------------------------------------------------------------------

"1) My pre-calculus textbook tells me that the degree of a polynomial is
the number of complex zeros it will have. That doesn't make sense to
me because what if the equation was 3x^8 - 3?"

----------------------------------------------------------------------

That is not an equation; it is an expression. There is
no equals sign.

The corresponding equation is

3x^8 - 3 = 0.

What you typed, the expression, is specifically a polynomial.

Ihatenotunderstandingmath
08-09-2011, 12:02 AM
Wow. Thank you so much, Mark! I didnít know complex zero included real zeros and that a real coefficient is also a complex coefficient. Now things are starting to make sense. J I have two questions though:

How did you figure out The eight Complex zeros of 3x^8 - 3 are 1, -1, i, -i, 1/2 sqrt(2) [1 + i], 1/2 sqrt(2) [1 - i], -1/2 sqrt(2) [1 + i], -1/2 sqrt(2) [1 - i] ? How does Ĺ sqrt(2)[1+i] even make sense when you try to plug it in? (or is that beyond the scope of pre-calculus?)
Iím confused about the last part you wrote ď"Complex number" means a number of the form A+B*i, where symbols A and B are any strictly-Real numbers. In other words, "Complex number" means A+B*i regardless of whether A, or B, or both A and B are zero.Ē Does that mean a complex number MUST include i? (whereas a complex zero and a complex coefficient donít have to include the i part?) And I donít understannd how a complex number means A+B*i if A or B or both A and B can be zero. Doesnít that mean EVERYTHING is a complex number? For example: 3 can be a complex number (assuming B is 0) because itís 3+0*i ? And then an equation where A and B arenít zero would also be written as a ďcomplex numberĒ. For example: 4+4 can be written as 4+-4i^2

Thank you so much for your answer!


Lookagain, thank you for answering! Youíre right Ė I forgot to add the = 0 part at the end.

wjm11
08-09-2011, 12:52 AM
Doesnít that mean EVERYTHING is a complex number? For example: 3 can be a complex number (assuming B is 0) because itís 3+0*i ?

Yes, the set of Real numbers is just a subset of Complex Numbers.

mmm4444bot
08-09-2011, 01:42 AM
How did you figure out The eight Complex zeros

I asked a computer program to do it for me, but one could also use the same methods in your exercise to do it by hand.


How does Ĺ sqrt(2)[1+i] even make sense when you try to plug it in?

Expand this factored imaginary number, and the resulting form is A + B*i, where A & B both equal 1/2 sqrt(2).

It turns out that the power [1/2 sqrt(2) + 1/2 sqrt(2) i]^8 equals 1, and it is good algebra practice to do this kind of multiplication by hand (at least once).

So, we have:

3x^8 - 3

3[1/2 sqrt(2) + 1/2 sqrt(2) i]^8 - 3

3(1) - 3

0


Does that mean a complex number MUST include i?

No. Only the Complex numbers that have an imaginary part contain i. The Complex numbers that do not have an imaginary part (i.e., B = 0) do not contain i, and these are also called Real numbers.


And then an equation where A and B arenít zero would also be written as a ďcomplex numberĒ. For example: 4+4 can be written as 4+-4i^2

Your example is not an equation, so I might be misunderstanding your intent. 4 + 4 is an expression.

The expression 4 + (-4)(i^2) is not of the form A + B*i because i is squared; however, the value of this expression can be written as 8 + 0i. This Complex number has no imaginary part, so it's also a Real number.

.