Consider the polynomial $2z^3-3z^2+15$.
I have proved that this polynomial has one unique zero in the first quadrant. Now I am asked to see how many zeroes this polynomial has in the first half of the first quadrant (the region where $\pi/4>\arg(z)>0$).
I have tried to use contour integration and then the principal argument without succeeding.
I don't know how do you try the principal argument, but you can try this maybe.
The principal argument say that $$\frac{1}{2 \pi i} \int_{\gamma} \frac{f'}{f} dz = N $$ where $N$ is the amount of zeros that is within the region delimited by $\gamma$, if $f$ is meromorphic (what you function is) and if it doesn´t have polos o zeros in $\Omega$.
I encourage you to try with the following curve.
We can make this curve in three steps.
$\gamma_1(t)=t(1+i) $ where $t \in [R,0]$ (in that order so it goes down), then
$\gamma_2(t)= t $ where $t \in [0,R]$ and finally
$\gamma_3(t)=Re^{it}$ where $t\ \in [0, \pi/4]$
Then if you make the limits when $R$ tends to $\infty$ you obtain the domain you want.