prove that the equation

Z^3 + 2Z +4=0

has no roots in the circle uint.

thanks.

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- Apr 24th 2006, 01:37 PMunicorncomplex functions
prove that the equation

Z^3 + 2Z +4=0

has no roots in the circle uint.

thanks. - Apr 24th 2006, 02:10 PMThePerfectHackerQuote:

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. - Apr 24th 2006, 03:17 PMunicorn
thanks :)

- Apr 24th 2006, 11:54 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

on the unit circle :confused:

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 - Apr 25th 2006, 05:32 PMRebesques
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. :)