Behaviour of the curve $(y^4 − 4y^2 + 3)+i(-y^3+2y)$ in the complex plane

Question

I was reading an example of the application of the Argument Principle and I encountered the following reasoning, which I do not understand.

At a certain point in the example, we consider $f(iy)=(y^4 − 4y^2 + 3)+i(-y^3+2y)$ and the following asseveration is made: since $\lim_{y\to \infty}\frac{-y^3+2y}{y^4 − 4y^2 + 3}=0$, then $f(iy)$ reaches the positive real axis as $y\to\infty$. (You can see the context of this reasoning on page 6 of this PDF http://legacy-www.math.harvard.edu/archive/115_fall_06/argument_principle.pdf).

I do not understand why this is true. I fact, I have ploted the curve (in this context $y$ takes values from $0$ to $\infty$), and it does not seem like it reaches the positive real axis as $y\to\infty$.

If anyone could clarify why this reasoning is valid it would be very helpful.

Thanks.

Answer 1

Not an answer

$(p=\text{Re }z=t^4-4 t^2+3,q=\text{Im }z=2 t-t^3)$

eliminating the parameter $t$ I got the cartesian equation of the curve $$-p^3+p^2-4 p q^2+5 p+q^4-4 q^2+3=0$$ that I used to get the plot in the image below.

If you manage to plot the curve for huge values of $q$ you will see that $\arg z\to 0$ as the limit says $$\lim_{y\to \infty}\frac{-y^3+2y}{y^4 − 4y^2 + 3}=0\to \arg z\to 0$$

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Answer 2

$$(y^4 − 4y^2 + 3)+i(-y^3+2y)=y^3[(y-4/y+3/y^2)-i(1-2/y^2)] $$

$$\arg f(iy)= \tan^{-1} \frac{2/y^2-1}{y-4/y+3/y^2} \approx -\tan^{-1} \frac{1}{y}\to 0.$$