MAT334-2018F > Quiz-7

Q7 TUT 5101

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Jeffery Mcbride:
You're right, $\displaystyle f( iy)$ is totally irrelevant here. My mistake. Do you mean the fact that f([-R, R]) lies completely in the upper half plane. Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$

When $\displaystyle z\ \rightarrow \ \pm \infty$, Re(f(z)) $\displaystyle \rightarrow \ \infty$ and Im(f(z)) $\displaystyle \rightarrow \ \infty \$and the endpoints of $\displaystyle f([ -R,\ R]) \$lie in the first quadrant for R really big. So the change in argument is going to be 0.

This can be coupled with what I said about the curve.

$\displaystyle f\left( Re^{it}\right)$ =  $\displaystyle e^{4it} \$from [0, $\displaystyle \pi$] because the 4 term dominates. So the change in argument is 4$\displaystyle \pi$.

So in total we get 2 zeros as the total change in argument is 4$\displaystyle \pi$.

Victor Ivrii:
Yes, $f(x)$ belongs to upper half-plane for $x\in (-\infty,\infty)$. But why?

Jeffery Mcbride:
Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$