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

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- Apr 17th 2008, 04:21 AMNeloAngeloRoots
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 AMbobak
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

tell me how it goes

Bobak - Apr 21st 2008, 03:31 PMNeloAngelo

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 PMmr fantastic
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.