# Solving a polynomial with complex coefficients

1. Nov 1, 2015

### astrololo

1. The problem statement, all variables and given/known data
z^6+(2i-1)z^3-1-i=0

2. Relevant equations

3. 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 ?

2. Nov 1, 2015

### SteamKing

Staff Emeritus
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

3. Nov 1, 2015

### SammyS

Staff Emeritus
Check that result from quadratic formula again.

I get something much simpler under the square root.

4. Nov 2, 2015

### 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.

5. Apr 28, 2016

### Ian Taylor

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

6. Apr 28, 2016

### SammyS

Staff Emeritus
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.

7. Apr 29, 2016

### Ian Taylor

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