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

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
  • #1
Dank2
213
4

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.
 
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  • #2
Dank2 said:

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
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.
 
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  • #4
Write z in trigonometric form z=r(cosx+isinx) into the equation.
 
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  • #5
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?
 
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  • #6
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.
 
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  • #7
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.
 
  • #8
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.
 
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  • #9
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
 
  • #10
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?)
 
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  • #11
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)
 
  • #12
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?
 
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  • #13
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 ))
 
  • #14
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.
 
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  • #15
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
 
  • #16
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) .
 
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  • #17
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?
 
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  • #18
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)
 
  • #19
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)
 
  • #20
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.
 
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  • #21
Dank2 said:
Left hand-side, you mean by z^3 ? that would be 9(cos3x + isin3x)
No, it is (√3)3(cos3x + isin3x).
 
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  • #22
3x = 3/2pi-x
x = 3/8pi .
 
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  • #23
Dank2 said:
how come (sqrt(3))^3 = 3*sqrt(3) , the magnitude should be the same
√3*√3*√3= (√3)2√3=3*√3
 
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  • #24
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
 
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  • #25
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 ;)
 
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  • #26
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)
 
  • #27
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.
 
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  • #28
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))
 
  • #29
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.
 
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  • #30
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))
 
  • #31
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?
 
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  • #32
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
 
  • #33
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)
 
  • #34
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.
 
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  • #35
Dank2 said:
0(cos0+isin0)
4[/SUP]√3(cos3/8pi + isin 3/8pi)
4√3(cos7/8pi + isin 7/8pi)
:nb)
 
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<h2>1. What is a complex polynomial equation?</h2><p>A complex polynomial equation is an equation that involves complex numbers, which are numbers that can be written in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).</p><h2>2. What does the notation ##z^3+3i\bar z=0## mean?</h2><p>The notation ##z^3+3i\bar z=0## represents a complex polynomial equation with a degree of 3, where the variable z is raised to the third power and is multiplied by 3i times the complex conjugate of z. The equation is set equal to 0, indicating that the goal is to find the values of z that make the equation true.</p><h2>3. How do you solve a complex polynomial equation?</h2><p>To solve a complex polynomial equation, you can use a variety of methods such as factoring, the quadratic formula, or the cubic formula. Additionally, you can also use graphing techniques or numerical methods like Newton's method to approximate the solutions.</p><h2>4. What is the difference between a real and complex solution to a polynomial equation?</h2><p>A real solution to a polynomial equation is a value that satisfies the equation and is a real number. A complex solution, on the other hand, is a value that satisfies the equation but involves complex numbers. Complex solutions often come in the form of complex conjugate pairs, where the real parts are the same and the imaginary parts have opposite signs.</p><h2>5. How many solutions can a complex polynomial equation have?</h2><p>A complex polynomial equation of degree n can have up to n complex solutions, including repeated solutions. However, not all of these solutions may be distinct, as some may be complex conjugate pairs. For example, a cubic equation can have up to 3 complex solutions, but they could all be real numbers or two could be complex conjugate pairs.</p>

1. What is a complex polynomial equation?

A complex polynomial equation is an equation that involves complex numbers, which are numbers that can be written in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).

2. What does the notation ##z^3+3i\bar z=0## mean?

The notation ##z^3+3i\bar z=0## represents a complex polynomial equation with a degree of 3, where the variable z is raised to the third power and is multiplied by 3i times the complex conjugate of z. The equation is set equal to 0, indicating that the goal is to find the values of z that make the equation true.

3. How do you solve a complex polynomial equation?

To solve a complex polynomial equation, you can use a variety of methods such as factoring, the quadratic formula, or the cubic formula. Additionally, you can also use graphing techniques or numerical methods like Newton's method to approximate the solutions.

4. What is the difference between a real and complex solution to a polynomial equation?

A real solution to a polynomial equation is a value that satisfies the equation and is a real number. A complex solution, on the other hand, is a value that satisfies the equation but involves complex numbers. Complex solutions often come in the form of complex conjugate pairs, where the real parts are the same and the imaginary parts have opposite signs.

5. How many solutions can a complex polynomial equation have?

A complex polynomial equation of degree n can have up to n complex solutions, including repeated solutions. However, not all of these solutions may be distinct, as some may be complex conjugate pairs. For example, a cubic equation can have up to 3 complex solutions, but they could all be real numbers or two could be complex conjugate pairs.

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