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

[Dank2]

How did you come up with that equation?

What is 0 + 4a²b² ?

That's what you should have on the right hand side. The left hand side should be a⁴-2a²b²-b⁴. Right?

a^4-2a^2b^2+b^4 =4a^2b^2 ==> (a^2-b^2)^2 = 4a^2b^2
taking root of both sides ==> a^2-b^2 = 2ab

 

[SammyS]

a^4-2a^2b^2+b^4 =4a^2b^2 ==> (a^2-b^2)^2 = 4a^2b^2
taking root of both sides ==> a^2-b^2 = 2ab

Oh! OK. That is correct.

Now solve for a in terms of b.

 

[Dank2]

Oh! OK. That is correct.

Now solve for a in terms of b.

How do i do that. a(a-2b) = b^2 ==> a = b^2 over (a-2b)

 

[SammyS]

How do i do that. a(a-2b) = b^2 ==> a = b^2 over (a-2b)

It's a quadratic equation.

Complete the square or use the quadratic formula.

To complete the square, add 2b² to $\ a^2-2ab -b^2 = 0 \ .$

Added in Edit:
Also notice that the equation should be

$\ a^2\pm 2ab -b^2 = 0 \ .$​

 

[Dank2]

It's a quadratic equation.

Complete the square or use the quadratic formula.

To complete the square, add 2b² to $\ a^2-2ab -b^2 = 0 \ .$

Added in Edit:
Also notice that the equation should be

$\ a^2\pm 2ab -b^2 = 0 \ .$​

So, a^2-2ab+b^2 = 2b^2, should i solve only left hand side?

 

[SammyS]

It's a quadratic equation.

Complete the square or use the quadratic formula.

To complete the square, add 2b² to $\ a^2-2ab -b^2 = 0 \ .$

No. (How is it even possible to solve only one side of an equation?)

Take the square root of bot sides.

Don't forget the ' ± ' .

​

 

[Dank2]

No. (How is it even possible to solve only one side of an equation?)

Take the square root of bot sides.

Don't forget the ' ± ' .

​

why +-

 

[SammyS]

why +-

Really?

You got it from taking the square root of both sides of some equation. Does that ring a bell?

 

[Sahil Kukreja]

i have got 6 solutions for z , what's the answer?

 

[Dank2]

i have got 6 solutions for z , what's the answer?

See message #41 4 solutions.

 

[ehild]

i have got 6 solutions for z , what's the answer?

There are 5 solutions. How did you get 6?
z= 0
√3(cos3/8pi + isin 3/8pi)
√3(cos7/8pi + isin 7/8pi)
√3(cos11/8pi+isin11/8pi)
√3(cos15/8pi+isin15/8pi)

 

[Sahil Kukreja]

There are 5 solutions. How did you get 6?
z= 0
√3(cos3/8pi + isin 3/8pi)
√3(cos7/8pi + isin 7/8pi)
√3(cos11/8pi+isin11/8pi)
√3(cos15/8pi+isin15/8pi)

i might have made a mistake, i will solve again

 

[Sahil Kukreja]

yes there are 5 solns
converting in principal form
z=0
z=√3e^(i3pi/8)
z=√3e^(i7pi/8)
z=√3e^(-i5pi/8)
z=√3e^(-ipi/8)

 

[Samy_A]

This is a very nice exercise, with, as seen in the thread, a number of ways to solve it.

One way to relatively easily see that there will be five solutions is as follows:

$z³+ 3i\bar z=0$
⇒ $z³ =-3i\bar z$
⇒ $|z|³ =3|z|$
$z=0$ is a solution, so from now on assume $z \neq 0$.
Dividing by $|z|$, we get $|z|²=3$, or $|z|=\sqrt 3$.

Now, multiplying the original equation by $z$ gives $z^4+3i\bar z z=0$
But $z^4+3i\bar z z= z^4 + 3i|z|²=z^4+9i$.
So finally, the non zero roots of the original equation are the (4) roots of $z^4=-9i$.

 

[Dank2]

This is a very nice exercise, with, as seen in the thread, a number of ways to solve it.

One way to relatively easily see that there will be five solutions is as follows:

$z³+ 3i\bar z=0$
⇒ $z³ =-3i\bar z$
⇒ $|z|³ =3|z|$
$z=0$ is a solution, so from now on assume $z \neq 0$.
Dividing by $|z|$, we get $|z|²=3$, or $|z|=\sqrt 3$.

Now, multiplying the original equation by $z$ gives $z^4+3i\bar z z=0$
But $z^4+3i\bar z z= z^4 + 3i|z|²=z^4+9i$.
So finally, the non zero roots of the original equation are the (4) roots of $z^4=-9i$.

That made it really simple :)