Complex Number, again

  • Thread starter JasonRox
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  • #1
JasonRox
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Complex Number, again... :(

This I'll give you the entire question and answer.

10. Solve:

[tex]z^4-2z^2+4=0[/tex]

That's all I got.

Answer:

[tex]+-1/2(\sqrt{6}+-\sqrt{2i}) (four combinations of signs).[/tex]

That is all.

I tried factoring, but I can't come up with anything. I also tried bisecting, but that is useless.

Knowing that the solution has complex numbers, I have no clue where they got this from.
 

Answers and Replies

  • #2
Dr Transport
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solve for [tex] z^{2} [/tex] then take the square root of those........
 
  • #3
JasonRox
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Ok then.

[tex]z^2=(\frac{z^2}{\sqrt{2}}-2i)(\frac{z^2}{\sqrt{2}}+2i)[/tex]

The square root of these?
 
  • #4
my results are
[tex]z=+-\sqrt{1+i\sqrt{3}}[/tex]
[tex]z=+-\sqrt{1-i\sqrt{3}}[/tex]

I dont know whether they are correct or not..
 
  • #5
HallsofIvy
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let x= z2 then the equation z4- 2z2+ 4= 0 becomes x2- 2x+ 4= 0. That is the same as x2- 2x= -4 or
x2- 2x+ 1= (x- 1)2= -3. From that x= [itex]1+- \sqrt{3}i[/itex].
Since x= z2, [itex]z^2= 1+- i\sqrt{3}[/itex] so [itex]x= +- \sqrt{1+- i\sqrt{3}}.
 
  • #6
JasonRox
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I can see that, but where would their solution come from?

Using the quadratic formula yield the same result as yours, too.
 

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