MAT334-2018F > Quiz-7

Q7 TUT 0102

**Victor Ivrii**:

Muyao's idea is correct but you need to be more specific where is $f(x)$ for different values of $x$. For that you need also find where $\Im(f(x))$ changes sign,

**Victor Ivrii**:

Following Yangbo from the place I wrote "WRONG". $2x^4-x^2-1=0$ is biquadratic equation and we hget $x^=\frac{1}{2},-1$ and real roots are $\pm\frac{1}{\sqrt{2}}$. So all roots of $\Re f(x)$, $\Im f(x)$ differ and are simple, $f(x)$ does not vanish and $\Re(f(x))$ or $\Im (f(x))$ change sign when $x$ passes the corresponding value. Thus the following scheme holds (see picture) and increment of $\arg (f(x))$ is $\approx -2\pi$ (the error tends to $0$ as $R\to \infty$) and thus the increment of $\arg (f(z))$ along this pass is $4\pi-2\pi=2\pi$ (it must be a multiple of $2\pi$) and there is indeed one root in the upper half-plane. And three in the lower half-plane.

Addition: observe that $\overline{f(z)}=f(-\bar{z})$ and therefore $z$ is a root implies that $-\bar{z}$ (its symmetric about imaginary axis) is a root as well. Therefore one root in the upper half-plane must be purely imaginary and at least one root in the lower half-plane must be purely imaginary. Indeed, consider $g(y):=f(yi)=2y^4-2y^3-y^2-2y-1$ which is real-valued. As $y\to \pm \infty$ it tends to $+\infty$ and $f(0)=-1$ so it must have real roots as $y>0$ and as $y<0$. As $y>0$ it has just one root since $f(z)$ has just one root there (we are talking about roots of real polynomial $g(y)$).

As $y<0$ there is one real root of $g(y)$. Indeed: we see that $g'(y)= 8y^3 -6y^2 -2y -2<0$ for $y<0$.

Let us ask Wolfram-Alpha to find (approximately) roots of $f(z)$ then Wolfram-Alpha query

Because it is approximate, it thinks that two roots have tiny real parts but we know that they are just imaginary.

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