Complex polynomial solutions

1. May 10, 2016

Dank2

1. The problem statement, all variables and given/known data
(Z^3)+3i(conjugate z) = 0
find all solutions.

2. Relevant equations

3. 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.

2. May 10, 2016

Samy_A

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

3. May 10, 2016

Dank2

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: May 10, 2016
4. May 11, 2016

ehild

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

5. May 11, 2016

Dank2

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: May 11, 2016
6. May 11, 2016

ehild

You can not do that! i can not be ignored.
It is wrong again. You ignored r.
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.

7. May 11, 2016

Dank2

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.

8. May 11, 2016

ehild

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: May 11, 2016
9. May 11, 2016

Dank2

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

10. May 11, 2016

ehild

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?)

11. May 11, 2016

Dank2

i think, yeah.

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

12. May 11, 2016

ehild

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?

13. May 11, 2016

Dank2

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

14. May 11, 2016

ehild

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.

15. May 11, 2016

Dank2

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

16. May 11, 2016

ehild

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) .

17. May 11, 2016

ehild

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?

18. May 11, 2016

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)

19. May 11, 2016

Dank2

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

20. May 11, 2016

ehild

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.