What are the solutions to the complex polynomial equation ##z^3+3i\bar z=0##?

[ehild]

3√3((cos(11/8pi)+isin(11/8pi))

z=√3(cos(x)+isin(x)), why did you write 3√3?
And there are more solutions. What do you get for k=1 and k=3?

 

[Dank2]

z=√3(cos(x)+isin(x)), why did you write 3√3?
And there are more solutions. What do you get for k=2?

i meant superscript(3)
4x = 3pi/2 + 0 pi = > x =3/8 pi
4x = 3pi/2 + 2pi = x = 7/8 pi
4 x = 3pi/2 +4pi = > x = 11/8 pi = 3/8pi

 

[Dank2]

i meant superscript(3)
4x = 3pi/2 + 0 pi = > x =3/8 pi
4x = 3pi/2 + 2pi = x = 7/8 pi
4 x = 3pi/2 +4pi = > x = 11/8 pi = 3/8pi

0(cos0+isin0)
⁴√3(cos3/8pi + isin 3/8pi)
⁴√3(cos7/8pi + isin 7/8pi)

 

[ehild]

i meant superscript(3)
4x = 3pi/2 + 0 pi = > x =3/8 pi
4x = 3pi/2 + 2pi = x = 7/8 pi
4 x = 3pi/2 +4pi = > x = 11/8 pi = 3/8pi

11/8 pi is not the same as 3/8 pi . The period is 2pi, not pi. So x= 3/8 pi + pi is different, and you have one more angle, less than 2pi.
And do not forget that r=√3. Do not change it to everything else.

 

[ehild]

0(cos0+isin0)
^{4[/SUP]√3(cos3/8pi + isin 3/8pi)
4√3(cos7/8pi + isin 7/8pi)}

 

[Dank2]

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

 

[Dank2]

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

or is it a complex polynomial

 

[ehild]

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.

 

[ehild]

hehe, that's because r = sqrt (3)

You mean a square, with 4 sides ?

 

[ehild]

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.

 

[Dank2]

You mean a square, with 4 sides ?

i remember t

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.

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

 

[Dank2]

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

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 * ?

 

[ehild]

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 * ?

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]

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).

that got me a bit of confused, since I've seen z^3.

thanks for the help ;)

 

[ehild]

that got me a bit of confused, since I've seen z^3.

thanks for the help ;)

But there was also z*, and it is not power of z.

 

[Dank2]

But there was also z*, and it is not power of z.

yep, that also, and other algebra mistakes, i think i will use this forum from time to time ;)

 

[ehild]

I am looking forward to see you soon again

 

[SammyS]

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.

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\$
Multiply by $\ {(-b)}\$: $\ \ 3(a^2)b-( b^3)+3a = 0\$​

Add the equations.

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.

 

[Dank2]

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\$
Multiply by $\ {(-b)}\$: $\ \ 3(a^2)b-( b^3)+3a = 0\$​

Add the equations.

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.

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?

 

[SammyS]

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?

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?

 

[Dank2]

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

 

[SammyS]

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.

 

[Dank2]

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)

 

[SammyS]

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

$\ a^2\pm 2ab -b^2 = 0 \ .$​

 

[Dank2]

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?

 

[SammyS]

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 ' ± ' .

​

 

[Dank2]

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 +-

 

[SammyS]

why +-

Really?

You got it from taking the square root of both sides of some equation. Does that ring a bell?

 

[Sahil Kukreja]

i have got 6 solutions for z , what's the answer?

 

[Dank2]

i have got 6 solutions for z , what's the answer?

See message #41 4 solutions.