# Roots

• Apr 17th 2008, 04:21 AM
NeloAngelo
Roots
Hi

Got problem with these questions any suggestions?
Show that 2-i is a root of z^4 - 8z^3+28z^2-48z+35. Hence find all roots
and
Find all 5th roots of -32 and all 4th rots of 81i
In form x+iy where x and y in 4 DP
Thx
• Apr 17th 2008, 04:30 AM
bobak
Quote:

Originally Posted by NeloAngelo
Hi

Got problem with these questions any suggestions?
Show that 2-i is a root of z^4 - 8z^3+28z^2-48z+35. Hence find all roots

all the coefficients of the quartic are real so the complex conjugate is a root.
so we have two factors $\displaystyle (x-(2-i))$ and $\displaystyle (x-(2+i))$.
so $\displaystyle (x-(2-i))(x-(2+i))$ is a factor of your quartic, divide your quartic by this factor and then solve the quadratic to get the two remaining roots.

Quote:

Find all 5th roots of -32 and all 4th rots of 81i
you want the solve the equation $\displaystyle z^5 = -32$. first rewrite -32 in the form of $\displaystyle Re^{i(\theta +2\pi n)}$

tell me how it goes

Bobak
• Apr 21st 2008, 03:31 PM
NeloAngelo
Quote:

Originally Posted by bobak
all the coefficients of the quartic are real so the complex conjugate is a root.
so we have two factors $\displaystyle (x-(2-i))$ and $\displaystyle (x-(2+i))$.
so $\displaystyle (x-(2-i))(x-(2+i))$ is a factor of your quartic, divide your quartic by this factor and then solve the quadratic to get the two remaining roots.

you want the solve the equation $\displaystyle z^5 = -32$. first rewrite -32 in the form of $\displaystyle Re^{i(\theta +2\pi n)}$

tell me how it goes

Bobak

Thx for the feedback
How did you get those answers on the 1st part? Did you use combination of de Moivre equation, quadratic formulae and long division
o how would it look if I applied -32 to $\displaystyle Re^{i(\theta +2\pi n)}$ ?? Just lookin at tht formula confused the hell outta me :)
• Apr 21st 2008, 03:45 PM
mr fantastic
Quote:

Originally Posted by NeloAngelo
Thx for the feedback
How did you get those answers on the 1st part? [snip]

Two things you should know:

1. If $\displaystyle \alpha$ is a root then $\displaystyle z - \alpha$ is a linear factor.

2. If a polynomial has real coefficients and $\displaystyle \alpha$ is a root, then $\displaystyle \bar{\alpha}$, the conjugate of $\displaystyle \alpha$, is also a root.

Some of the questions you're asking suggest it would be a good thing for you to go back and thouroughly review this whole topic.