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Complex polynomial solutions

  • Thread starter Dank2
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  • #1
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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.
 

Answers and Replies

  • #2
Samy_A
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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##.
 
  • #3
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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:
  • #4
ehild
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Write z in trigonometric form z=r(cosx+isinx) into the equation.
 
  • #5
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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?
 
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  • #6
ehild
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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.
 
  • #7
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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.
 
  • #8
ehild
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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.
 
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  • #9
213
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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
 
  • #10
ehild
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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?)
 
  • #11
213
4
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)
 
  • #12
ehild
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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?
 
  • #13
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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 ))
 
  • #14
ehild
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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
213
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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
 
  • #16
ehild
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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) .
 
  • #17
ehild
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-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?
 
  • #18
213
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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)
 
  • #19
213
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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)
 
  • #20
ehild
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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.
 
  • #21
ehild
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Left hand-side, you mean by z^3 ? that would be 9(cos3x + isin3x)
No, it is (√3)3(cos3x + isin3x).
 
  • #22
213
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3x = 3/2pi-x
x = 3/8pi .
 
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  • #23
ehild
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how come (sqrt(3))^3 = 3*sqrt(3) , the magnitude should be the same
√3*√3*√3= (√3)2√3=3*√3
 
  • #24
ehild
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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
 
  • #25
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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 ;)
 

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