# Arghhhh Argument Principle

• Jan 3rd 2008, 04:28 PM
Jason Bourne
Arghhhh Argument Principle
I get really annoyed when questions ask you to sketch stuff, like am doing maths and i'm supposed to be able to do art! (Rofl)

Anyway, I'm not sure on this question:

By sketching the graph of $y = f(x)$ show that

$f(z) \equiv z^4 + 4z + 1 = 0$

has two negative real roots and no postitive real roots. Show also that $f(z)=0$ has no purely imaginary roots.

Using the argument principle, determine in which quadrants of the complex plane the two complex roots lie.
• Jan 3rd 2008, 04:39 PM
topsquark
Quote:

Originally Posted by Jason Bourne
I get really annoyed when questions ask you to sketch stuff, like am doing maths and i'm supposed to be able to do art! (Rofl)

Anyway, I'm not sure on this question:

By sketching the graph of $y = f(x)$ show that

$f(z) \equiv z^4 + 4z + 1 = 0$

has two negative real roots and no postitive real roots. Show also that $f(z)=0$ has no purely imaginary roots.

Using the argument principle, determine in which quadrants of the complex plane the two complex roots lie.

Well, sketch the graph of $y = x^4 + 4x + 1$. You don't have to have an art degree, just get some graph paper and plug in some values of x and connect the dots with a smooth curve.

You should be able to see that there are two negative and no positive roots.

I don't know how to show that the other two roots are not pure imaginary by looking at the graph.

-Dan
• Jan 3rd 2008, 04:42 PM
ThePerfectHacker
Quote:

Originally Posted by Jason Bourne
I get really annoyed when questions ask you to sketch stuff, like am doing maths and i'm supposed to be able to do art! (Rofl)

Anyway, I'm not sure on this question:

By sketching the graph of $y = f(x)$ show that

$f(z) \equiv z^4 + 4z + 1 = 0$

has two negative real roots and no postitive real roots. Show also that $f(z)=0$ has no purely imaginary roots.

Using the argument principle, determine in which quadrants of the complex plane the two complex roots lie.

Suppose that $z=ai$ for $a\not = 0$.
Then $f(ai) = a^4 +4ai+1=0$ by equation real and imaginary parts we get $(a^4+1)=0\mbox{ and }4a=0$ which is impossible.

What does this have to do with the principle of the argument?