prove that the equation
Z^3 + 2Z +4=0
has no roots in the circle uint.
thanks.
Unit circle would mean that you can express your solutions as $\displaystyle z=\cos \theta+i\sin \theta$Originally Posted by unicorn
Thus,
$\displaystyle (\cos \theta+i\sin \theta)^3+2(\cos \theta+i\sin \theta)+4=0$
Using De'Moiver's Theorem,
$\displaystyle \cos 3\theta+i\sin 3\theta+2\cos \theta+2i\sin \theta+4=0$
Thus, (by definition of complex number equality)
$\displaystyle \left\{ \begin{array}{c}\cos 3\theta+2\cos \theta=-4\\ \sin 3\theta +\sin \theta=0$
But the first equality cannot hold because the minimum cosine is at -1. Thus, even if it were to be at -1 then, its value would be $\displaystyle (-1)+2(-1)=-3>-4$
Thus, the first equation the LHS is greater than the RHS-impossible.
Now I had read this to mean in the interior of the unit circle notOriginally Posted by ThePerfectHacker
on the unit circle
If it is the interior of the circle that is of interest then we would
be looking at an application of Cauchy's integral theorem (presumably),
and then college/calculus would be the appropriate forum.
RonL
Apply Rouché's Theorem, for f the given function, and g(z)=-z^3+C, C a real constant you can guess by the hypotheses.