I'm tasked with finding the zeros of $f(z)=z^3+1$ that lie inside the first quadrant using the Argument Principle, which I have simplified below:
$$N=\frac{1}{2\pi}[arg(f(z))]_C$$
where N represents the number of zeros and $[arg(f(z))]_C$ represents the change in the argument of $f(z)$ over C. The solution given by the book seems to be quite trivial, however, I'm having trouble piecing the process together. My analysis of the problem is below:
$$arg(f(z))=\phi \quad \tan(\phi)=\frac{Im(f(x,y))}{Re(f(x,y))} \quad arg(z)=\theta$$
We begin at the origin, $z=0 \rightarrow f(z) =1$, and move far down the real axis to a point $R$. At $R$ $z=x=R \rightarrow f(z)=R^3+1$ where our argument of $z$ has not changed, hence, neither has the argument of $f(z)$. However, we now move along a radial length to the point $iR$ on the imaginary axis. Along this radial curve, $\theta$ changes by $\pi/2$ since we have gone from having only a real component to now only having an imaginary component, and as a result, $\phi$ changes by $3\pi/2$. Lastly, we move from $iR$ back to $z=0$ along the imaginary axis. This is where I lose it and cannot deduce how $\theta$ changes. On this last leg we go from having a purely imaginary number to one that is zero, which I can't seem to make sense of, in terms of the argument of z. Care to provide any insight?