please help me

**flower3** · July 10th 2009, 10:14 AM

use complex analysis to factor the following polynomials into a product of two quadratic polynomials:
$x{^4} + 1$
---------------------------------------------------------------------------------------------------------------
sketch :
$\mid Arg z \mid \leq \frac{\pi}{4}$ and determine if the set is :
open,closed,bounded,connected ,simply connected , domain,compact or region
answer :
the given set is :
closed,connected ,simply connected ,and region
but not :
open,bounded,domain,and compact
my question : if the question : arg z not Arg z"principle argument" the answer is the same or not ???

**AlephZero** · July 10th 2009, 01:55 PM

From the law $a^2-b^2=(a+b)(a-b),$ the fact that $i^2=-1,$ and the fact that $x^4+1=x^4-(-1),$ you should be able to prove the first part easily.

**Soroban** · July 11th 2009, 07:05 PM

Hello, flower3!

Use complex analysis to factor the following polynomial
into a product of two quadratic polynomials: . $x{^4} + 1$

Complex analysis? . . . It requires only Algebra 1.

Add and subtract $2x^2\!:\quad x^4 \:{\color{red}+ \:2x^2} + 1 \:{\color{red}- \:2x^2} \;=\;\left(x^2+1\right)^2 - \left(\sqrt{2}\:\!x\right)^2$

We have a difference of squares:

. . $\left[(x^2+1) - \sqrt{2}\:\!x\right]\,\left[(x^2+1) + \sqrt{2}\:\!x\right]$

. . . $=\;\left(x^2 - \sqrt{2}\:\!x + 1\right)\,\left(x^2 + \sqrt{2}\:\!x + 1\right)$

**malaygoel** · July 11th 2009, 08:49 PM

Originally Posted by flower3

use complex analysis to factor the following polynomials into a product of two quadratic polynomials:
$x{^4} + 1$

$x^4 + 1$

= $x^4-i^2$

= $(x^2+i)(x^2-i)$

**mr fantastic** · July 12th 2009, 03:59 AM

Originally Posted by malaygoel

$x^4 + 1$

= $x^4-i^2$

= $(x^2+i)(x^2-i)$

I suspect the OP meant two quadratic polynomials with real coefficients.

Originally Posted by Soroban

Hello, flower3!

Complex analysis? . . . It requires only Algebra 1.

Add and subtract $2x^2\!:\quad x^4 \:{\color{red}+ \:2x^2} + 1 \:{\color{red}- \:2x^2} \;=\;\left(x^2+1\right)^2 - \left(\sqrt{2}\:\!x\right)^2$

We have a difference of squares:

. . $\left[(x^2+1) - \sqrt{2}\:\!x\right]\,\left[(x^2+1) + \sqrt{2}\:\!x\right]$

. . . $=\;\left(x^2 - \sqrt{2}\:\!x + 1\right)\,\left(x^2 + \sqrt{2}\:\!x + 1\right)$

Unfortunately, in an exam setting, if the question gives the instruction to use complex analysis then a correct answer using an alternative technique will not score any marks. Not even an answer mark.

Originally Posted by flower3

use complex analysis to factor the following polynomials into a product of two quadratic polynomials:
$x{^4} + 1$
[snip]

Start by finding the four complex roots of -1. Can you do this? (If not, see http://www.mathhelpforum.com/math-he...ex-number.html, for example).

**mr fantastic** · July 12th 2009, 04:19 AM

Originally Posted by flower3

[snip]
sketch :
$\mid Arg z \mid \leq \frac{\pi}{4}$ and determine if the set is :
open,closed,bounded,connected ,simply connected , domain,compact or region
answer :
the given set is :
closed,connected ,simply connected ,and region
but not :
open,bounded,domain,and compact
my question : if the question : arg z not Arg z"principle argument" the answer is the same or not ???

Your question can be easily answered by noting that $\mid \text{Arg} (z) \mid \leq \frac{\pi}{4}$ is equivalent to $- \frac{\pi}{4} \leq \text{Arg} (z) \leq \frac{\pi}{4}$ ....

**flower3** · July 12th 2009, 10:56 AM

Originally Posted by mr fantastic

Your question can be easily answered by noting that $\mid \text{Arg} (z) \mid \leq \frac{\pi}{4}$ is equivalent to $- \frac{\pi}{4} \leq \text{Arg} (z) \leq \frac{\pi}{4}$ ....

i know that , but i mean if the question $\mid arg z \mid \leq \frac{\pi}{4}$ is the answer the same or not ???
closed,connected ,simply connected ,and region
but not :
open,bounded,domain,and compact

**flower3** · July 12th 2009, 11:00 AM

Originally Posted by mr fantastic

I suspect the OP meant two quadratic polynomials with real coefficients.

Unfortunately, in an exam setting, if the question gives the instruction to use complex analysis then a correct answer using an alternative technique will not score any marks. Not even an answer mark.

Start by finding the four complex roots of -1. Can you do this? (If not, see http://www.mathhelpforum.com/math-he...ex-number.html, for example).

Thanks !
if i find one of theses roots and then multiply it by its conjugate then by using long division i find the 2nd quadratic polynomial "true or not ???"

**mr fantastic** · July 12th 2009, 05:23 PM

Originally Posted by flower3

Thanks !
if i find one of theses roots and then multiply it by its conjugate then by using long division i find the 2nd quadratic polynomial "true or not ???"

The roots will have the form $z = \alpha_1$ , $z = \overline{\alpha_1}$ , $z = \alpha_2$ and $z = \overline{\alpha_2}$ .

Therefore the two quadratic factors you require will obviously be $(z - \alpha_1)(z - \overline{\alpha_1})$ and $(z - \alpha_2)(z - \overline{\alpha_2})$ .

Originally Posted by flower3

i know that , but i mean if the question $\mid arg z \mid \leq \frac{\pi}{4}$ is the answer the same or not ???
closed,connected ,simply connected ,and region
but not :
open,bounded,domain,and compact

The answer is contained in post #6.

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