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

[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 :)