Complex polynomial solutions

  • Thread starter Dank2
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  • #51
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How did you come up with that equation?

What is 0 + 4a2b2 ?

That's what you should have on the right hand side. The left hand side should be a4-2a2b2-b4. Right?
a^4-2a^2b^2+b^4 =4a^2b^2 ==> (a^2-b^2)^2 = 4a^2b^2
taking root of both sides ==> a^2-b^2 = 2ab
 
  • #52
SammyS
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a^4-2a^2b^2+b^4 =4a^2b^2 ==> (a^2-b^2)^2 = 4a^2b^2
taking root of both sides ==> a^2-b^2 = 2ab
Oh! OK. That is correct.

Now solve for a in terms of b.
 
  • #53
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Oh! OK. That is correct.

Now solve for a in terms of b.
How do i do that. a(a-2b) = b^2 ==> a = b^2 over (a-2b)
 
  • #54
SammyS
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How do i do that. a(a-2b) = b^2 ==> a = b^2 over (a-2b)
It's a quadratic equation.

Complete the square or use the quadratic formula.

To complete the square, add 2b2 to ##\ a^2-2ab -b^2 = 0 \ . ##

Added in Edit:
Also notice that the equation should be
##\ a^2\pm 2ab -b^2 = 0 \ . ##​
 
  • #55
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It's a quadratic equation.

Complete the square or use the quadratic formula.

To complete the square, add 2b2 to ##\ a^2-2ab -b^2 = 0 \ . ##

Added in Edit:
Also notice that the equation should be
##\ a^2\pm 2ab -b^2 = 0 \ . ##​
So, a^2-2ab+b^2 = 2b^2, should i solve only left hand side?
 
  • #56
SammyS
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It's a quadratic equation.

Complete the square or use the quadratic formula.

To complete the square, add 2b2 to ##\ a^2-2ab -b^2 = 0 \ . ##
No. (How is it even possible to solve only one side of an equation?)

Take the square root of bot sides.

Don't forget the ' ± ' .

 
  • #57
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No. (How is it even possible to solve only one side of an equation?)

Take the square root of bot sides.

Don't forget the ' ± ' .

why +-
 
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  • #58
SammyS
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why +-
Really?

You got it from taking the square root of both sides of some equation. Does that ring a bell?
 
  • #59
i have got 6 solutions for z , whats the answer???
 
  • #61
ehild
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i have got 6 solutions for z , whats the answer???
There are 5 solutions. How did you get 6?
z= 0
√3(cos3/8pi + isin 3/8pi)
√3(cos7/8pi + isin 7/8pi)
√3(cos11/8pi+isin11/8pi)
√3(cos15/8pi+isin15/8pi)
 
  • #62
There are 5 solutions. How did you get 6?
z= 0
√3(cos3/8pi + isin 3/8pi)
√3(cos7/8pi + isin 7/8pi)
√3(cos11/8pi+isin11/8pi)
√3(cos15/8pi+isin15/8pi)
i might have made a mistake, i will solve again
 
  • #63
yes there are 5 solns
converting in principal form
z=0
z=√3e^(i3pi/8)
z=√3e^(i7pi/8)
z=√3e^(-i5pi/8)
z=√3e^(-ipi/8)
 
  • #64
Samy_A
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This is a very nice exercise, with, as seen in the thread, a number of ways to solve it.

One way to relatively easily see that there will be five solutions is as follows:

##z³+ 3i\bar z=0##
⇒ ##z³ =-3i\bar z##
⇒ ##|z|³ =3|z|##
##z=0## is a solution, so from now on assume ##z \neq 0##.
Dividing by ##|z|##, we get ##|z|²=3##, or ##|z|=\sqrt 3##.

Now, multiplying the original equation by ##z## gives ##z^4+3i\bar z z=0##
But ##z^4+3i\bar z z= z^4 + 3i|z|²=z^4+9i##.
So finally, the non zero roots of the original equation are the (4) roots of ##z^4=-9i##.
 
Last edited:
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  • #65
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This is a very nice exercise, with, as seen in the thread, a number of ways to solve it.

One way to relatively easily see that there will be five solutions is as follows:

##z³+ 3i\bar z=0##
⇒ ##z³ =-3i\bar z##
⇒ ##|z|³ =3|z|##
##z=0## is a solution, so from now on assume ##z \neq 0##.
Dividing by ##|z|##, we get ##|z|²=3##, or ##|z|=\sqrt 3##.

Now, multiplying the original equation by ##z## gives ##z^4+3i\bar z z=0##
But ##z^4+3i\bar z z= z^4 + 3i|z|²=z^4+9i##.
So finally, the non zero roots of the original equation are the (4) roots of ##z^4=-9i##.
That made it really simple :)
 

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