# complex roots confusion

Show 40 post(s) from this thread on one page
Page 1 of 3 123 Last
• Aug 18th 2012, 11:04 AM
anthonye
complex roots confusion
Hi;
If I have a 3rd degree polynomial it has three roots(zeros) of which three can be complex.

Why a complex number has two parts real and imaginary and I thought these came in conjugate pairs?

Thanks.
• Aug 18th 2012, 11:43 AM
Plato
Re: complex roots confusion
Quote:

Originally Posted by anthonye
Hi;
If I have a 3rd degree polynomial it has three roots(zeros) of which three can be complex.
Why a complex number has two parts real and imaginary and I thought these came in conjugate pairs?

The polynomial $(z-i)(z+i)(z-1-i)=z^3-(1+i)z^2+z-(1+i)$ has three complex roots.

But note the coefficients are not all real.
• Aug 18th 2012, 11:49 AM
anthonye
Re: complex roots confusion
What still confised.
• Aug 18th 2012, 11:54 AM
HallsofIvy
Re: complex roots confusion
If a polynomial equation has only real coefficients, then non-real roots must come in conjugate pairs. If you had a third degree equation with real coefficients, it must have either three or one real root.

However notice that I said "non-real", not "complex". The set of all complex number includes the real numbers. That is why we can say that any nth degree polynomial equation has n (counting multiplicity) complex roots. The equation $x^3- 6x^2+ 11x- 6= (x- 1)(x- 2)(x- 3)= 0$ has three "complex" roots- they are 1, 2, and 3.

The equation $x^3- 4x^2+ 5x-2= (x- 1)^2(x- 2)= 0$ has three "complex" roots- they are 1 (of multiplicity 2) and 2.
• Aug 18th 2012, 12:00 PM
anthonye
Re: complex roots confusion
So a true complex number would be a - bi or a + bi.

But a complex number can also be say 1,2,3...ect it's just that the imaginary part is zero?

I could also have bi as a complex number where the real part is zero?
• Aug 18th 2012, 12:16 PM
Soroban
Re: complex roots confusion
Hello, anthonye!

Quote:

If I have a 3rd degree polynomial, it has three roots (zeros) of which three can be complex.
. . This is impossible.

Consider the graph of a cubic function.
The zeros of the polynomial are the (real) x-intercepts.
Code:

                          *             *            *           *    *         *      *      *                   *    *         *            *       *

It can have one x-intercept.
(One real zero, two complex zeros.)
Code:

          |           |                *           |           |           |  *            *           *    *         *|      *      *           |      *    *         * |          *           |           |     ---o--+----------------           |           |           |       *  |           |

It can have two x-intercepts.
(Two real zeros, one is repeated.)
Code:

              |               |          *               |             *|          *           *  | *         *    |  *      *               |  *    *     ----o-----+------o---               |               |       *      |               |

It can have three x-intercepts.
(Three real zeros.)
Code:

            |             |              *             |             |             |*            *           * |  *       ---o--+----o-------o----             |    *    *         *  |        *             |             |       *    |             |

But it cannot have no x-intercepts (three complex zeros).
• Aug 18th 2012, 12:19 PM
anthonye
Re: complex roots confusion
Ok thanks for taking the time Soroban I'll look later.
• Aug 18th 2012, 05:53 PM
HallsofIvy
Re: complex roots confusion
Quote:

Originally Posted by anthonye
So a true complex number would be a - bi or a + bi.

But a complex number can also be say 1,2,3...ect it's just that the imaginary part is zero?

I could also have bi as a complex number where the real part is zero?

Yes, to each of those. "bi as a complex number where the real part is zero" would be a pure imaginary number. We can say that "every complex number is the sum of a real number and a pure imaginary number" but we have to count 0 as both real and complex. That's not such an unreasonable thing to do. In the "complex plane", the real and imaginary axes intersect at 0: 0+ 0i.
• Aug 19th 2012, 11:41 AM
anthonye
Re: complex roots confusion
Ok getting there now.

This is what I have:

For the equation x^4 - 3x^3 - x^2 - x + 3 = 0

Find the number of complex roots,The possible number of real roots
and the possible number of rational roots.

complex roots = 4 (The reason for my answer is complex numbers are almost any numbers)

real roots = 0,2,4 ( the reason for my answers are 0 the answer may all be complex) and this is where i'm confused

real numbers are on the number line -3,-2,-1,0,1,2,3ect But if a complex number is say 3(imaginary part=0) is 3 complex or
real?

Thanks.
• Aug 19th 2012, 11:52 AM
MaxJasper
Re: complex roots confusion
Quote:

Originally Posted by anthonye
Hi;
If I have a 3rd degree polynomial it has three roots(zeros) of which three can be complex.

Why a complex number has two parts real and imaginary and I thought these came in conjugate pairs?

Thanks.

I recommend you start reading: $\text{The } \text{Story } \text{of } \sqrt{-1}$
• Aug 19th 2012, 11:56 AM
anthonye
Re: complex roots confusion
Yes I get that thats a imaginary number but it must also be complex with the real part being zero yes/no?
• Aug 19th 2012, 12:09 PM
Plato
Re: complex roots confusion
Quote:

Originally Posted by anthonye
For the equation x^4 - 3x^3 - x^2 - x + 3 = 0
Find the number of complex roots,The possible number of real roots
and the possible number of rational roots.
complex roots = 4 (The reason for my answer is complex numbers are almost any numbers)
real numbers are on the number line -3,-2,-1,0,1,2,3ect But if a complex number is say 3(imaginary part=0) is 3 complex or real?

I think finally I understand your confusion.
Every real number is a complex number.
Every complex number has the form $a+bi$ where $a~\&~b$ are real numbers and $i^2=-1$.
If $b=0$ we say that $a+bi$ is a real number.

So in your question in the quote $3=3+0i$ is said to be real.

In the polynomial $\sum\limits_{k = 0}^n {{\alpha _k}{z^k}}$ if $\forall k$ we have $\text{Im}(\alpha_k)=0$ [the imaginary part is zero] we say that polynomial has real coefficients. Therefore, any complex roots (meaning the imaginary part is not zero) must occur in conjugate pairs.
• Aug 19th 2012, 12:10 PM
MaxJasper
Re: complex roots confusion
Quote:

Originally Posted by anthonye
Yes I get that thats a imaginary number but it must also be complex with the real part being zero yes/no?

$\text{Complex } \text{number } = \left\{\text{real } \text{part} 1 + \text{real } \text{part} 2 * \sqrt{-1}\right\}$
• Aug 19th 2012, 12:14 PM
anthonye
Re: complex roots confusion
Ok thanks again all have to go now back later(I'm starvin).
• Aug 20th 2012, 10:24 AM
anthonye
Re: complex roots confusion
So going back to the equation x^4 - 3x^3 - x^2 - x + 3 = 0.

How can I tell from just the equation how many roots of each type it will have?

I know I can use the rational root theorem and synthetic division to find the first two roots
and then the quadratic formula to find the last two that will tell me what they are and how many of each type
but can I tell what type and how many before I know what their values will be?

Thanks.
Show 40 post(s) from this thread on one page
Page 1 of 3 123 Last