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

[Dank2]

Homework Statement

(Z^3)+3i(conjugate z) = 0
find all solutions.

Homework Equations

The Attempt at a Solution

How can i isolate Z Tried factoring out z, didn't came out good: z(z^2+3i*(Conjugate z)/z) ==> Right part equal to zero. couldn't factor anymore.

 

[Samy_A]

Homework Statement

(Z^3)+3i(conjugate z) = 0
find all solutions.

Homework Equations

The Attempt at a Solution

How can i isolate Z Tried factoring out z, didn't came out good: z(z^2+3i*(Conjugate z)/z) ==> Right part equal to zero. couldn't factor anymore.

You could use $z=x+iy$, $\bar z = x-iy$, and see what equations you get when you plug that in into $z³+3i\bar z=0$.

 

[Dank2]

You could use $z=x+iy$, $\bar z = x-iy$, and see what equations you get when you plug that in into $z³+3i\bar z=0$.

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.

 

[ehild]

Write z in trigonometric form z=r(cosx+isinx) into the equation.

 

[Dank2]

Write z in trigonometric form z=r(cosx+isinx) into the equation.

r³(cos3x+isin3x) = 3ir(cosx + isin(-x))

==> r³ = 3ir ==> r = squareroot(3)

==> cos3x+isin3x = cosx + isin(-x)
==> cos3x = cosx (1)
==> isin3x = isin(-x) (2)

cos2x = 1 (1)
isin2x = -1 (2)

which x will satisfy both? or do i have a mistake?

 

[ehild]

r³(cos3x+isin3x) = 3ir(cosx + isin(-x))

==> r³ = 3ir ==> r = squareroot(3)

You can not do that! i can not be ignored.

==> cos3x+isin3x = cosx + isin(-x)

It is wrong again. You ignored r.

==> cos3x = cosx (1)
==> isin3x = isin(-x) (2)

cos2x = 1 (1)
isin2x = -1 (2)

which x will satisfy both? or do i have a mistake?

cos(3x) is not 3cos(x) and so on.

r³(cos3x+isin3x) = 3ir(cosx + isin(-x))
one solution can be that r=0. If r ≠ 0, simplify with r. Then collect the real and imaginary terms. You get two equations, one for the real terms, and one for the imaginary ones.

 

[Dank2]

You can not do that! i can not be ignored.

Ok so r^3= 3ir ==> r^2 =3i ==> r = sqrt(3i).
i just remember that r must be real, since it the size of the vector.

 

[ehild]

Ok so r^3= 3ir ==> r^2 =3i ==> r = sqrt(3i).
i just remember that r must be real, since it the size of the vector.

Yes, r is real, so r³ can not be equal to ir.
Let's start again. z = r (cos(θ) + i sin (θ)). You wrote z³ correctly, z³ = r³(cos (3θ) + i sin (3θ) ) and z* = r( cos(-θ)+isin(-θ).
The original equation is (Z^3)+3i(conjugate z) = 0, which means z³=-iz*. The magnitude of both sides has to be the same . What equation you get for r?
The phase angles also should be the same, or differ by 2kπ (k integer) . Write the right side as cos(φ)+ isin(φ) using that -i corresponds to -pi/2 phase angle.

 

[Dank2]

z3=-iz*

isn't it z^3 = -3iz* ?

then r^3 = -3r, if i put i into the parentesis of z*
==> r^2 = -3r, but then it's imaginary again

 

[ehild]

isn't it z^3 = -3iz* ?

yes, it is.

then r^3 = -3r, if i put i into the parentesis of z*

No. The magnitude can not be negative, and i z* ≠ -z. The magnitude of the product of complex numbers is equal to the product of the magnitudes. What is the magnitude of z? What is the magnitude of z*? What is the magnitude of i?
(Do you know the exponential form of complex numbers?)

 

[Dank2]

yes, it is.

No. The magnitude can not be negative, and i z* ≠ -z. The magnitude of the product of complex numbers is equal to the product of the magnitudes. What is the magnitude of z? What is the magnitude of z*? What is the magnitude of i?
(Do you know the exponential form of complex numbers?)

i think, yeah.

|z*| = |z|
|z^3| = |-3iz*| ==> |z|^3 = |-3i| * |z| ==? |z|^2 = |-3i| = 3 ==? |z| = sqrt(3)

 

[ehild]

i think, yeah.

|z*| = |z|
|z^3| = |-3iz*| ==> |z|^3 = |-3i| * |z| ==? |z|^2 = |-3i| = 3 ==? |z| = sqrt(3)

Correct! And r is the magnitude of z, so r=√3.
Now determine the phase angles at both sides. The angles add in case of a product. What are the phase angles of z* and -i?

 

[Dank2]

Correct! And r is the magnitude of z, so r=√3.
Now determine the phase angles at both sides. The angles add in case of a product. What are the phase angles of z* and -i?

-i*z = (cos(3/2pi) + isin(3/2pi))*sqrt(3)(cosx + isin-x). = sqrt(3)(cos(x+3/2pi)) + isin(3/2pi-x ))

 

[ehild]

Not quite.
Try to use the exponential form. z=re^iθ, z^*=re^-iθ, -i = e^-iπ/2. Remember you have the equation z³=-iz* to solve.

 

[Dank2]

Not quite.
Try to use the exponential form. z=re^iθ, z^*=re^-iθ, -i = e^-iπ/2.

We haven't learned this one yet, thought you meant somthing else. we have learned de moivre formula and it's properties

 

[ehild]

i haven't learned this one yet, though you meant somthing else. we have learned is de moivre formula and it's properties

OK, it is a bit more complicated. But you know that the angles add if you multiply two complex numbers. How can you write -i z*?
z*=r(cos(-θ)+isin(-θ)) and -i=isin(-π/2) .

 

[ehild]

-i = (cos(3/2pi) + isin(3/2pi))*sqrt(3)(cosx + isin-x). = sqrt(3)(cos(x+3/2pi)) + isin(3/2pi-x ))

Almost:)
-3i z* =3 (cos(3/2pi) + isin(3/2pi))*sqrt(3)(cos(-x) + isin(-x)) = 3sqrt(3)(cos(-x+3/2pi)) + isin(3/2pi-x )).
So the phase angle of the right-hand side is -x+3pi/2 +2pi*k.
And what is the phase of the left-hand side?

 

[Dank2]

OK, it is a bit more complicated. But you know that the angles add if you multiply two complex numbers. How can you write -i z*?

well, -i = 1(cos 3/2 π + isin 3/2 π)
z* = sqrt(3)( cosx + isin -x )
-i*z* = sqrt(3)(cos(3/2π+ x) + isin(3/2π -x))
so the product will be as written above, and we could compare it to left handside which is 9(cos 3x +isin3x)

 

[Dank2]

well, -i = 1(cos 3/2 π + isin 3/2 π)
z* = sqrt(3)( cosx + isin -x )
-i*z* = sqrt(3)(cos(3/2π+ x) + isin(3/2π -x))
so the product will be as written above, and we could compare it to left handside which is 9(cos 3x +isin3x)

Left hand-side, you mean by z^3 ? that would be 9(cos3x + isin3x)

 

[ehild]

well, -i = 1(cos 3/2 π + isin 3/2 π)
z* = sqrt(3)( cosx + isin -x )
-i*z* = sqrt(3)(cos(3/2π+ x) + isin(3/2π -x))
so the product will be as written above, and we could compare it to left handside which is 9(cos 3x +isin3x)

The argument of both the cosine and sine must be the same in Moivre formula. Write z* as
z* = sqrt(3)( cos(-x) + isin (-x )). Cos(-x) is the same as cos(-x)!
The left-hand side is (√3)³ (cos 3x +isin3x), the right-hand side is 3sqrt(3)(cos(3/2π- x) + isin(3/2π -x)), and you equated the magnitude of both sides already!
The phase of the left side is equal to the phase of the right side + 2πk.

 

[ehild]

Left hand-side, you mean by z^3 ? that would be 9(cos3x + isin3x)

No, it is (√3)³(cos3x + isin3x).

 

[Dank2]

3x = 3/2pi-x
x = 3/8pi .

 

[ehild]

how come (sqrt(3))^3 = 3*sqrt(3) , the magnitude should be the same

√3*√3*√3= (√3)²√3=3*√3

 

[ehild]

So, 3x = x+1/2pi
3x = -x +1/2pi

The angles both in cosine and sine must be the same.
And it was -x+3pi/2 why did you left out the 3? Equating the angles, you get 3x=-x+3pi/2 +2kpi

 

[Dank2]

The angles both in cosine and sine must be the same.
And it was -x+3pi/2 why did you left out the 3? You can assign also the angle -pi/2 to -i. So you can write the angle of the right-hand side as -x -pi/2 or -x+3pi/2, and add 2kpi to it.
So 3x=-x + 3pi/2 +2kpi. And what is x then?

i've noticed and it and fixed, was rushing ;)

 

[Dank2]

The angles both in cosine and sine must be the same.
And it was -x+3pi/2 why did you left out the 3? Equating the angles, you get 3x=-x+3pi/2 +2kpi

x1= 3/8π
x2= 3/8π+ 2π = 19/8π
r = sqrt(3)

 

[ehild]

x1= 3/8π
x2= 3/8π+ 2π = 19/8π
r = sqrt(3)

It was 4x=3pi/2 + 2kpi. The whole right-hand side has to be divided by 4. And include all angles between 0 and 2pi.

 

[Dank2]

The angles both in cosine and sine must be the same.
And it was -x+3pi/2 why did you left out the 3? Equating the angles, you get 3x=-x+3pi/2 +2kpi

So the first solution will be : 0(cos0+isin0)
second ³√3(cos(3/8pi)+isin(3/8pi)
third ³√3((cos(19/8pi)+isin(19/8pi))

 

[ehild]

So the first solution will be : 0(cos0+isin0)
second ³√3(cos(3/8pi)+isin(3/8pi)
third ³√3((cos(19/8pi)+isin(19/8pi))

The first two are correct the third is not. I edited my previous post.
And for the first solution, it is enough to write z=0.

 

[Dank2]

The first two are correct the third is not. I edited my previous post.
And for the first solution, it is enough to write z=0.

3√3((cos(7/8pi)+isin(7/8pi))

 

[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?