Prove there exists a complex number $z$ such that $$ z^7+\cos(|z^2|)(1+93z^4)=0. $$ (For heaven's sake don't try to compute it!)
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Please consider re-formatting your question – Don Larynx Dec 12 '13 at 15:12
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3@EricStucky, how do you suggest Rouché's theorem be applied here? The function is not analytic. – Antonio Vargas Dec 12 '13 at 15:17
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Ah, you're right. – Eric Stucky Dec 12 '13 at 15:18
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10I am not sure why you would call the LHS a polynomial, though it is easy to see it is continuous and goes from $-\infty$ to $+\infty$ hence must have a real root. – Macavity Dec 12 '13 at 15:18
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This is not a polynomial. – Thomas Andrews Dec 12 '13 at 15:20
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2Hint: Use degree mod 2 – jimbo Dec 12 '13 at 15:31
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Although the answers above are correct ones, they fail to use $deg_2$ as the book of Guillemin & Pollack suggest. Heres an approach that use the notion of $deg_2$:
Let $f:\mathbb{C} \to \mathbb{C}$ be defined as $$ f(z)=z^7+\cos(|z^2|)(1-93z^4). $$ Consider the homotopy $F(z,t)=tf(z)+(1-t)z^7$ between $f(z)$ and $z^7$. Now if $W=\left\{z: |z|\leq R \right\}$ (where $R$ is taken large enough such that $F(z,t)\neq0$ for all $(z,t) \in \partial W \times [0,1]$), then the maps $$ \frac{F(\cdot,t)}{|F(\cdot, t)|} : \partial W \to S^1 $$ are well define for all $t \in [0,1]$, ($S^1=\left\{z: |z|=1 \right\}$), more over since $F(z,1)=f(z)$ and $F(z,0)=z^7$ are homotpic we have that $$ deg_2 \left(\frac{f}{|f|}\right)=deg_2 \left( \frac{z^7}{|z^7|}\right) $$ but clearly $g(z)=z^7/|z^7|$ makes seven turns around any point $y \in S^1$, then #$\left(g^{-1}(y)\right) = 7$, and since $7 \equiv 1 \pmod 2$, we have that $deg_2(g)$ is nonzero, that is $$ deg_2 \left(\frac{f}{|f|}\right)\neq 0 $$ this means that there exist $z\in W$ such that $f(z)=0$ as required.
Alonso Delfín
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Well, since $\lim_{z\to \infty}F(z,t)/z^7 = 1 \ \forall t$, then you can assure that there will be a large $R$ where $F(z,t)\neq 0 \ \forall (z,t) \in \partial W \times [0,1]$. Then it is not necessary to say exactly who is $R$. – Alonso Delfín Jan 10 '15 at 04:00
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for this problem http://math.stackexchange.com/questions/608607/root-of-fz-z6-cosz615z2 $deg=0$. – jimbo Jan 10 '15 at 13:35
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for this problem http://math.stackexchange.com/questions/608607/root-of-fz-z6-cosz615z2 for this polinomy $deg=0$ and have six roots. – jimbo Jan 10 '15 at 13:48
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Remember that you can only assure the existence of a root when $deg_2(f/|f|) \neq 0$, in the problem you quote $deg_2(f/|f|)=0$, however there is a theory of degree non modulo 2, in which $deg(f/|f|)=6$ and that gives you the six roots inside $W$. – Alonso Delfín Jan 11 '15 at 21:01
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If you look at a circle of radius $2$, then the argument of $f(z) = z^7 + \cos(|z|^2)(1+93z^4)$, which happens to be $z^7 + \cos {49}(1+93z^4)$, makes $7$ loops around the circle (using Rouché's theorem, for example).
Meanwhile, if you look at a small circle around $0$, the argument of $f(z)$ won't make any loop (because $f(0) = 1$ and $f$ is continuous)
So while the radius goes from $2$ to $0$, the number of loops has to jump despite $f$ being continuous, and this can only happen when $f$ has a zero.
mercio
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1@jimbo, yes, I only use analyticity when I use Rouché's theorem with the functions $z^7$ and $z^7 + cos(2^2)(1+93z^4)$ (which is equal to $f(z)$ when $|z|=2$) – mercio Dec 20 '13 at 18:51
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As pointed out in the comments, not only is there a complex number satisfying the equation; there is at least one real number such that $$ f(z)=z^7 + \cos(z^2)\left(1+93z^4\right)=0. $$ To see this, just note that $f(z)$ is continuous, and negative for small enough real $z$, and positive for large enough real $z$; then apply the intermediate value theorem.
mjqxxxx
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