<s>[INST] How many zeros does the polynomial have in the right half plane? The polynomial is $f(z) = z^4+\sqrt{2}z^3+2z^2-5z+2$
If you check the image of the imaginary axis, you see that there are no zeros, so we can use the right semicircle from $iR$ to $-iR$,and make $R$ arbitrarily big.  The problem now is picking the right function to use Rouche's theorem with. I'm confused because if I were to pick something like $z^4+2$, then it would have 1 zero on the right-half plane, and two on the imaginary axis.  Do we count those two as well?  
But then, how do you show that $|z^4+2|>|\sqrt{2}z^3+2z^2-5z|$ on the imaginary-axis part of the semi-circle. It doesn't seem like a clean argument.
 [/INST] We can use the following way.
For all real $k$ we have:
$$z^4+\sqrt2z^3+2z^2-5z+2=\left(z^2+\frac{1}{\sqrt2}z+k\right)^2-\left(\left(2k-\frac{3}{2}\right)z^2+(\sqrt2k+5)z+k^2-2\right).$$
Now, let $2k-\frac{3}{2}>0$ and $$(\sqrt2k+5)^2-4\left(2k-\frac{3}{2}\right)(k^2-2)=0.$$ 
We see that $k=2.65...$ is valid.
Id est, for this value of $k$ we obtain
$$z^4+\sqrt2z^3+2z^2-5z+2=$$
$$=\left(z^2+\frac{1}{\sqrt2}z+k\right)^2-\left(\left(2k-\frac{3}{2}\right)z^2+(\sqrt2k+5)z+k^2-2\right)=$$
$$=\left(z^2+\frac{1}{\sqrt2}z+k\right)^2-\left(\sqrt{2k-\frac{3}{2}}z+\sqrt{k^2-2}\right)^2=$$
$$=\left(z^2+\left(\tfrac{1}{\sqrt2}-\sqrt{2k-\tfrac{3}{2}}\right)z+k-\sqrt{k^2-2}\right)\left(z^2+\left(\tfrac{1}{\sqrt2}+\sqrt{2k-\tfrac{3}{2}}\right)z+k+\sqrt{k^2-2}\right).$$
Now, easy to see for this value of $k$ we have
$$\left(\tfrac{1}{\sqrt2}-\sqrt{2k-\tfrac{3}{2}}\right)^2-4\left(k-\sqrt{k^2-2}\right)<0$$ and
$$\left(\tfrac{1}{\sqrt2}+\sqrt{2k-\tfrac{3}{2}}\right)^2-4\left(k+\sqrt{k^2-2}\right)<0,$$ which says that our polynomial has no real roots and since
$$\tfrac{1}{\sqrt2}-\sqrt{2k-\tfrac{3}{2}}<0$$ and
$$\tfrac{1}{\sqrt2}+\sqrt{2k-\tfrac{3}{2}}>0,$$
we see that two roots exactly placed in the right half of the plane. 
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