Solving a polynomial with complex coefficients

[astrololo]

Homework Statement

z^6+(2i-1)z^3-1-i=0

Homework Equations

The Attempt at a Solution

I know that I must put k=z^3 and solve the quadratic. But I'm not able to simplify the quadratic. I get the square root of (-8i+1)

What am I supposed to do ?

 

[SteamKing]

Homework Statement

z^6+(2i-1)z^3-1-i=0

Homework Equations

The Attempt at a Solution

I know that I must put k=z^3 and solve the quadratic. But I'm not able to simplify the quadratic. I get the square root of (-8i+1)

What am I supposed to do ?

The square root of (1-8i) is, wait for it, another complex number!

If you know how to express (1-8i) in polar form using Euler's formula, then finding the square root should be a snap.

https://en.wikipedia.org/wiki/Complex_number

 

[SammyS]

Homework Statement

z^6+(2i-1)z^3-1-i=0

Homework Equations

The Attempt at a Solution

I know that I must put k=z^3 and solve the quadratic. But I'm not able to simplify the quadratic. I get the square root of (-8i+1)

What am I supposed to do ?

Check that result from quadratic formula again.

I get something much simpler under the square root.

 

[ogg]

FWIW, given any complex number z, then assume √z = a+bi and so z = (a+bi)² which means that Real(z²) = a²-b² and Im(z²) = 2abi; two equations, two unknowns. simple to solve, i hope.

 

[Ian Taylor]

ogg, I think you meant to write:
Re$(z)=a^2-b^2$ and Im$(z)=2ab$

 

[SammyS]

ogg, I think you meant to write:
Re$(z)=a^2-b^2$ and Im$(z)=2ab$

Right. The imaginary part doesn't include the imaginary unit, i , by the usual convention.

Hello, Ian Taylor. Welcome to PF !

If you notice, this thread is 1/2 year old.

 

[Ian Taylor]

Right. The imaginary part doesn't include the imaginary unit, i , by the usual convention.

Hello, Ian Taylor. Welcome to PF !

If you notice, this thread is 1/2 year old.

Thanks SammyS. He had also written $z^2$ rather than $z$. which contradicts his original definition.