# Complex polynomial solutions

## 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.

Samy_A
Homework Helper

## 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##.

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:
ehild
Homework Helper
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?

Last edited:
ehild
Homework Helper
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
Homework Helper
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
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
Homework Helper
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
Homework Helper
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
Homework Helper
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
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

ehild
Homework Helper
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
ehild
Homework Helper
-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)

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
Homework Helper
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
ehild
Homework Helper
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:
ehild
Homework Helper
how come (sqrt(3))^3 = 3*sqrt(3) , the magnitude should be the same
√3*√3*√3= (√3)2√3=3*√3

Dank2
ehild
Homework Helper
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 ;)

ehild
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
Homework Helper
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
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
Homework Helper
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
Homework Helper
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

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
Homework Helper
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
ehild
Homework Helper
0(cos0+isin0)
4[/SUP]√3(cos3/8pi + isin 3/8pi)
4√3(cos7/8pi + isin 7/8pi)

Dank2