Complex algebra problem (roots)

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SUMMARY

The discussion centers on finding the roots of the expression z4 + 4 = 0 and factorizing it into quadratic factors with real coefficients. The correct factorization is (z2 + 2z + 2)(z2 - 2z + 2). The roots are derived using Euler's identity and the magnitude of z, leading to the four solutions: (+/-)1 and (+/-)i. The participants emphasize the importance of complex analysis and polynomial algebra in solving such problems.

PREREQUISITES
  • Complex analysis fundamentals
  • Euler's identity and its applications
  • Polynomial algebra techniques
  • Understanding of complex roots and their geometric representation
NEXT STEPS
  • Study the derivation of roots using Euler's formula
  • Learn about polynomial factorization techniques in complex numbers
  • Explore the geometric interpretation of complex numbers
  • Practice solving higher-degree polynomial equations
USEFUL FOR

Students in advanced mathematics, particularly those studying complex analysis, algebra, and anyone seeking to improve their problem-solving skills in polynomial equations.

Remixex
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Homework Statement


First off i wasn't sure if i should put this in precalc or here so i just tossed a coin[/B]
I must find the roots of the expression z^4 +4=0 (which I've seen repeatedly on the internet)
Use it to factorize z^4 +4 into quadratic factors with real coefficients
The answer is (z^2 + 2z + 2)(z^2 - 2z + 2)

Homework Equations


Just the one given on the top

The Attempt at a Solution


Ok so, the question is quite basic as I've seen people pull it off in 10 mins, BUT after 3 semesters of dealing with only real numbers i forgot about complex ones, and now that i took complex analysis I'm in trouble D:
First off i had to search Euler's identity, that's how lost i was
I managed to establish that the magnitude of Z (r) is sqrt(2), i also got to the point where phi is -pi/4 +k*pi/2
so the 4 roots would be given by the sqrt(2) * e^i*phi given k takes the values within (0,1,2,3)
This is where i die i have no idea how to turn those into the final equation, i know the 4 solutions are
(+/-)1 (+/-)i, and from there is just polynomial algebra, but i don't know how to get there, Am i supposed to be able to deduct this as obvious from my last step?
Thanks in advance, i firmly believe there are no such thing as dumb questions, but given this is something I'm supposed to manage, it's quite the candidate :D
 
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Remixex said:

Homework Statement


First off i wasn't sure if i should put this in precalc or here so i just tossed a coin[/B]
I must find the roots of the expression z^4 +4=0 (which I've seen repeatedly on the internet)
Use it to factorize z^4 +4 into quadratic factors with real coefficients
The answer is (z^2 + 2z + 2)(z^2 - 2z + 2)

Homework Equations


Just the one given on the top

The Attempt at a Solution


Ok so, the question is quite basic as I've seen people pull it off in 10 mins, BUT after 3 semesters of dealing with only real numbers i forgot about complex ones, and now that i took complex analysis I'm in trouble D:
First off i had to search Euler's identity, that's how lost i was
I managed to establish that the magnitude of Z (r) is sqrt(2), i also got to the point where phi is -pi/4 +k*pi/2
so the 4 roots would be given by the sqrt(2) * e^i*phi given k takes the values within (0,1,2,3)
This is where i die i have no idea how to turn those into the final equation, i know the 4 solutions are
(+/-)1 (+/-)i, and from there is just polynomial algebra, but i don't know how to get there, Am i supposed to be able to deduct this as obvious from my last step?
Thanks in advance, i firmly believe there are no such thing as dumb questions, but given this is something I'm supposed to manage, it's quite the candidate :D

You can write ##-1## as
-1 = e^{i \pi} = e^{3 i \pi} = e^{-i \pi} = e^{-3 i \pi},
so the four 4th roots of ##-1## are obtained by taking 1/4 of the argument of each of the above. That is, the four roots ##r_1,r_2,r_3,r_4## are
<br /> r_1 = e^{i \pi/4} = (1+i)/ \sqrt{2}, \; r_2 = e^{i 3 \pi/4} = (-1 + i)/ \sqrt{2} \\ <br /> r_3 = e^{-i \pi/4} = (1-i)/ \sqrt{2}, \; r_4 = e^{-i 3 \pi/4} = (-1-i)/ \sqrt{2}
From these you can determine the four 4th roots of ##-4##.

Note that ##e^{i \theta} = \cos(\theta) + i \sin(\theta)##, so we can easily write the four roots in ##a + ib## form, just using elementary geometry/trigonometry.
 

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