use complex analysis to factor the following polynomials into a product of two quadratic polynomials:
$\displaystyle x{^4} + 1$
---------------------------------------------------------------------------------------------------------------
sketch :
$\displaystyle \mid Arg z \mid \leq \frac{\pi}{4}$ and determine if the set is :
open,closed,bounded,connected ,simply connected , domain,compact or region
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 ???

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

3. Hello, flower3!

Use complex analysis to factor the following polynomial
into a product of two quadratic polynomials: .$\displaystyle x{^4} + 1$
Complex analysis? . . . It requires only Algebra 1.

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

We have a difference of squares:

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

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

4. Originally Posted by flower3
use complex analysis to factor the following polynomials into a product of two quadratic polynomials:
$\displaystyle x{^4} + 1$
$\displaystyle x^4 + 1$

=$\displaystyle x^4-i^2$

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

5. Originally Posted by malaygoel
$\displaystyle x^4 + 1$

=$\displaystyle x^4-i^2$

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

We have a difference of squares:

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

. . . $\displaystyle =\;\left(x^2 - \sqrt{2}\:\!x + 1\right)\,\left(x^2 + \sqrt{2}\:\!x + 1\right)$
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:
$\displaystyle x{^4} + 1$
[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).

6. Originally Posted by flower3
[snip]
sketch :
$\displaystyle \mid Arg z \mid \leq \frac{\pi}{4}$ and determine if the set is :
open,closed,bounded,connected ,simply connected , domain,compact or region
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 $\displaystyle \mid \text{Arg} (z) \mid \leq \frac{\pi}{4}$ is equivalent to $\displaystyle - \frac{\pi}{4} \leq \text{Arg} (z) \leq \frac{\pi}{4}$ ....

7. Originally Posted by mr fantastic
Your question can be easily answered by noting that $\displaystyle \mid \text{Arg} (z) \mid \leq \frac{\pi}{4}$ is equivalent to $\displaystyle - \frac{\pi}{4} \leq \text{Arg} (z) \leq \frac{\pi}{4}$ ....
i know that , but i mean if the question $\displaystyle \mid arg z \mid \leq \frac{\pi}{4}$ is the answer the same or not ???
closed,connected ,simply connected ,and region
but not :
open,bounded,domain,and compact

8. 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 ???"

9. 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 $\displaystyle z = \alpha_1$, $\displaystyle z = \overline{\alpha_1}$, $\displaystyle z = \alpha_2$ and $\displaystyle z = \overline{\alpha_2}$.

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

Originally Posted by flower3
i know that , but i mean if the question $\displaystyle \mid arg z \mid \leq \frac{\pi}{4}$ 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.