1. ## complex functions

prove that the equation
Z^3 + 2Z +4=0
has no roots in the circle uint.
thanks.

2. 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 $\displaystyle z=\cos \theta+i\sin \theta$
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.

3. thanks

4. Originally Posted by ThePerfectHacker
Unit circle would mean that you can express your solutions as $\displaystyle z=\cos \theta+i\sin \theta$
Now I had read this to mean in the interior of the unit circle not
on the unit circle

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

5. 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.