- #36
Dank2
- 213
- 4
hehe, that's because r = sqrt (3)ehild said:
i'm confused now, how many roots does a complex polynomial have, i thought it was 3, as the highest exponent of that polynomial
hehe, that's because r = sqrt (3)ehild said:
or is it a complex polynomialDank2 said:hehe, that's because r = sqrt (3)
i'm confused now, how many roots does a complex polynomial have, i thought it was 3, as the highest exponent of that polynomial
It is not a polynomial of z if it includes the complex conjugate.Dank2 said:hehe, that's because r = sqrt (3)
i'm confused now, how many roots does a complex polynomial have, i thought it was 3, as the highest exponent of that polynomial
You mean a square, with 4 sides ?Dank2 said:hehe, that's because r = sqrt (3)
i remember tehild said:You mean a square, with 4 sides ?
0(cos0+isin0)ehild said:Remember your first attempt with that two third-order equations for a and b. If you eliminate one of them you get a polynomial of degree higher than 3.
is there any thumb rule as to how much many roots should i expect? like is it the number of roots of z + number of roots of * ?Dank2 said:i remember t
0(cos0+isin0)
√3(cos3/8pi + isin 3/8pi)
√3(cos7/8pi + isin 7/8pi)
√3(cos15/8pi+isin15/8pi)
4x = 3pi + 8pi ==> 19/8 pi ==> 2pi + 3/8pi = 3/8pi
I do not know any thumb rules. You need all angles possible between zero and 2pi. Remember, the 3rd power brought in 3x and the conjugate on the other side brought in -x. So there were 4x=something+2pik. That is four roots, except for the trivial one (z=0).Dank2 said:is there any thumb rule as to how much many roots should i expect? like is it the number of roots of z + number of roots of * ?
that got me a bit of confused, since I've seen z^3.ehild said:I do not know any thumb rules. You need all angles possible between zero and 2pi. Remember, the 3rd power brought in 3x and the conjugate on the other side brought in -x. So there were 4x=something+2pik. That is four roots, except for the trivial one (z=0).
But there was also z*, and it is not power of z.Dank2 said:that got me a bit of confused, since I've seen z^3.
thanks for the help ;)
yep, that also, and other algebra mistakes, i think i will use this forum from time to time ;)ehild said:But there was also z*, and it is not power of z.
I see that my friend, ehild, has guided you to the solution.Dank2 said:so i opened it and came out something hairy with a^3 - b^3i -3ab(b-ai) +3(b+ai) = 0
(a^3)-3(b^2)a+3b+(3(a^2)b-b^3+3a)i ==>
(a^3)-3(b^2)a+3b = 0
3(a^2)b-b^3+3a = 0
What can i try to do here?
btw the solutions can be shown as trigonometric (cosx+isinx) aswell.
SammyS said:I see that my friend, ehild, has guided you to the solution.
I considered solving this an alternate way, by using your results in the attached quote.
Multiply by ##\ (a)\ ##: ##\quad \ (a^3)-3(b^2)a+3b = 0\ ##Add the equations.
Multiply by ##\ {(-b)}\ ##: ##\ \ 3(a^2)b-(
b^3)+3a = 0\ ##
You get
##a^4-6a^2 b^2+b^4 = 0\ ##
Adding ##\ 4a^2b^2\ ## or ##\ 8a^2b^2\ ## will give easily solved results.
I haven't tried anything beyond this.
How did you come up with that equation?Dank2 said:I've added 4a^2b^2, came up with a^2-2ab-b^2 = 0
How can i find the roots from here? i can see how i can get a=0 and b = 0, which is the first soltution, what about the rest?
a^4-2a^2b^2+b^4 =4a^2b^2 ==> (a^2-b^2)^2 = 4a^2b^2SammyS said: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?
Oh! OK. That is correct.Dank2 said: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
How do i do that. a(a-2b) = b^2 ==> a = b^2 over (a-2b)SammyS said:Oh! OK. That is correct.
Now solve for a in terms of b.
It's a quadratic equation.Dank2 said:How do i do that. a(a-2b) = b^2 ==> a = b^2 over (a-2b)
So, a^2-2ab+b^2 = 2b^2, should i solve only left hand side?SammyS said: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 \ . ##
No. (How is it even possible to solve only one side of an equation?)SammyS said: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 \ . ##
why +-SammyS said: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 ' ± ' .
Really?Dank2 said:why +-
See message #41 4 solutions.Sahil Kukreja said:i have got 6 solutions for z , what's the answer?
There are 5 solutions. How did you get 6?Sahil Kukreja said:i have got 6 solutions for z , what's the answer?
i might have made a mistake, i will solve againehild said: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)
That made it really simple :)Samy_A said: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##.