# Roshe theorem questiom.

1. Jul 14, 2010

### nhrock3

find the number of roots on the right side of the field
for
$$f(z)=z^6+7z^4-3z^2+z+1$$

??

usually i have stuff like 1<|z|<4

but here instead they say "on the right side of the field"
meaning on the right frim the imaginary axes

i cant interpret it into a circles like i am use to

2. Jul 14, 2010

### jackmell

First, just plot the roots using Mathematica:

Code (Text):
myeqn = z^6 + 7 z^4 - 3 z^2 + z - 1;
myzeros = N[z /. Solve[myeqn == 0, z]]
mypoints = Point @@ {{Re[#], Im[#]} & /@ myzeros}
Show[{Graphics[{{Red, PointSize[0.01], mypoints},
Circle[{0, 0}, 1]}]}, PlotRange -> {{-5, 5}, {-5, 5}},
Axes -> True]

Not doing that to act smart. I think it's crucial to learn Mathematica and use it in your math education. So we know where they are. Now prove it using the Argument Principle:

First, what would happen if I just plug in the value $ai$ for real a into that expression. Will it ever be zero? Next consider the change in argument of

$$f(z)=z^6+7z^4-3z^2+z-1=z^6\left(1+7/z^2-3/z^4+1/z^5-1/z^6\right)$$

along the closed half-circle contour in the right half-plane and allow R to go to infinity. Where do you start with such a thing? Well, $\arg(zw)=\arg(z)+\arg(w)$ so that as R goes to infinity, how will the argument change for $z^6$ along the curved part of the half-circle from $-\pi/2$, around to $\pi/2$? You can do that. And as R goes to infinity, how will the argument change for that other expression as you go around the half-circle? Finally, how will the argument change from the point say $f(a i)$ down the imaginary axis, to the point $f(-ai)$ as a goes to infinity? Now, does your work with the Argument Principle agree with the plot above?