# complex functions

• April 24th 2006, 01:37 PM
unicorn
complex functions
prove that the equation
Z^3 + 2Z +4=0
has no roots in the circle uint.
thanks.
• April 24th 2006, 02:10 PM
ThePerfectHacker
Quote:

Originally Posted by unicorn
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 $z=\cos \theta+i\sin \theta$
Thus,
$(\cos \theta+i\sin \theta)^3+2(\cos \theta+i\sin \theta)+4=0$
Using De'Moiver's Theorem,
$\cos 3\theta+i\sin 3\theta+2\cos \theta+2i\sin \theta+4=0$
Thus, (by definition of complex number equality)
$\left\{ \begin{array}{c}\cos 3\theta+2\cos \theta=-4\\ \sin 3\theta +\sin \theta=0$
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 $(-1)+2(-1)=-3>-4$
Thus, the first equation the LHS is greater than the RHS-impossible.
• April 24th 2006, 03:17 PM
unicorn
thanks :)
• April 24th 2006, 11:54 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Unit circle would mean that you can express your solutions as $z=\cos \theta+i\sin \theta$

Now I had read this to mean in the interior of the unit circle not
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
• April 25th 2006, 05:32 PM
Rebesques
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. :)