zeros and poles of $\operatorname{Im} f(z) / \operatorname{ Im z }>0$ using argument's principle

Hi everyone I find the following exercise and honestly I'm stuck and I can't see how to use the argument's principle in any way. I'd appreciate any help with the exercise. Thank you.

Let $f$ a mermorphic function defined on $\mathbb C$ and suppose that $\operatorname{Im} f(z) / \operatorname{ Im z }>0$ for all $z$ outside of the real axis.

Show that the poles and zeros of $f$ belongs to $\mathbb R$ and the zeros and poles are interlaced, that is, between to adjacent poles there is exactly a zero and between to adjacent zeros there is a pole (use the argument's principle).

(1) So $w= f(z)$ maps the upper half of the $z$ plane into the upper half of the $w$ plane, and maps the lower half of the $z$ plane into the lower half of the $w$ plane.
As $z(t)$ executes a small simple oriented loop $z=\gamma(t)$ in the $z$ plane the image curve $\Gamma (t) = w(t)= f(z(t))$ traces a closed curve in the $w$ plane . The Argument Principle tells us that this winding number of $\Gamma$ about $w=0$ equals $Z-P$, the difference between the number of zeros and poles of $f(z)$ inside $\gamma$.
The winding number of $\Gamma$ is evidently zero if $\Gamma$ lies strictly on one side of the real $w$ axis. Thus by (1) there are no zeros or poles of $f(z)$ inside $\gamma$ unless the small loop $\gamma$ straddles the real $z$ axis.
The remainder of the problem can be solved by (2) modeling the meromorphic function locally near a real zero or pole as $f(z)= c z^k$ and thinking about what restrictions that places on $c$ and $k$. Next (3) consider the image of a narrow rectangular curve $\gamma$ that encloses adjacent zeros of $f$.