Show that $f(z)=z^4+iz+1$ has exactly one root in each quadrant.
Approach: If $z_0$ is a root, I deduced that $-\bar{z}_0$ is also a root. So if we have for example a root in the first quadrant, we also have one in the fourth quadrant. So it is enough to show that there are two roots in say two different quadrants. Then I used rouché's theorem to conclude that the roots are in between $1/2 < |z| < 2$. But I'm not sure how to proceed, i.e. how to show that there are in in two different quadrants is a solution. Thanks in advance.
You can ensure that no root path of $z^4+i\lambda z+1$ crosses the coordinate axes for $λ\in[0,1]$
And indeed for the real axis $z=x+i·0$ the real part $$x^4+1$$ will never be zero. For the imaginary axis $z=0+iy$ the real part is \begin{align} y^4-λy+1&=(y^2-\tfrac12)^2 +y^2-λy+\tfrac34 \\ &=(y^2-\tfrac12)^2+(y-\tfrac12λ)^2+\tfrac34-\tfrac14λ^2 \end{align} which is also always positive as $\tfrac34-\tfrac14λ^2>\tfrac12$.
Thus the single roots of $z^4+1=0$ stay single and in their quadrant for $z^4+iλz+1=0 $ for all $λ\in[0,1]$.
One can formalize these calculations as a winding number argument or Rouché argument with large enough quarter circles.