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

• Dank2
In summary: Remember you have the equation z3=-iz* to...z*=re-iθ, -i = e-iπ/2. Remember you have the equation z3=-iz* to...z*=re-iθ, -i = e-iπ/2. Remember you have the equation z3=-iz* to...z*=re-iθ, -i = e-iπ/2. Remember you have the equation z3=-iz* to...z*=re-iθ, -i = e-iπ/2. Remember
Dank2

## Homework Statement

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

## 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.

Dank2 said:

## Homework Statement

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

## 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##.

Samy_A said:
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.

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

Dank2
ehild said:
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?

Last edited:
Dank2 said:
r3(cos3x+isin3x) = 3ir(cosx + isin(-x))

==> r3 = 3ir ==> r = squareroot(3)
You can not do that! i can not be ignored.
Dank2 said:
==> cos3x+isin3x = cosx + isin(-x)
It is wrong again. You ignored r.
Dank2 said:
==> 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
ehild said:
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.

Dank2 said:
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.

Last edited:
Dank2
ehild said:
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

Dank2 said:
isn't it z^3 = -3iz* ?
yes, it is.
Dank2 said:
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
ehild said:
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)

Dank2 said:
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
ehild said:
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 ))

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

Dank2
ehild said:
Not quite.
Try to use the exponential form. z=re, 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

Dank2 said:
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) .

Dank2
Dank2 said:
-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
ehild said:
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 said:
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)

Dank2 said:
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.

Dank2
Dank2 said:
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 .

Last edited:
Dank2 said:
how come (sqrt(3))^3 = 3*sqrt(3) , the magnitude should be the same
√3*√3*√3= (√3)2√3=3*√3

Dank2
Dank2 said:
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
ehild said:
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 ;)

ehild
ehild said:
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)

Dank2 said:
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.

Last edited:
Dank2
ehild said:
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))

Dank2 said:
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
ehild said:
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))

Dank2 said:
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
ehild said:
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 said:
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)

Dank2 said:
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.

Dank2
Dank2 said:
0(cos0+isin0)
4[/SUP]√3(cos3/8pi + isin 3/8pi)
4√3(cos7/8pi + isin 7/8pi)

Dank2

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