Suppose that $\phi _{i},$ $\left( i=1,2,3\right) $ triple of holomorphic complex functions satisfy $\left\{ \begin{array}{c} \phi _{1}^{2}+\phi _{2}^{2}+\phi _{3}^{2}=0 \\ \left\vert \phi _{1}\right\vert ^{2}+\left\vert \phi _{2}\right\vert ^{2}+\left\vert \phi _{3}\right\vert ^{2}\neq 0% \end{array}% \right. $.
Define $\left\{ \begin{array}{c} f=\left( \phi _{1}-i\phi _{2}\right) \\ g=\frac{\phi _{3}}{\phi _{1}-i\phi _{2}} \end{array} \right. $
and show that $g$ is meromorphic and $f$ is holomorphic and that the poles of order $m$ of $g$ are zeros of order $2m$ of $f$.
I already tried using the equivalences for poles and zero of a function but I could not.
We compute $$g^2f=\frac{\phi_3^2}{\phi_1-i\phi_2}=-\frac{\phi_1^2+\phi_2^2}{\phi_1-i\phi_2}=-(\phi_1+i\phi_2) $$ and see that this is holomorphic. To show that every pole of order $m$ of $g$ is a root of order $2m$ of $f$, it suffices to show that the right hand side has no zero that is also a pole of $g$. Indeed, if $\phi_1(z)+i\phi_2(z)=0$ and $\phi_1(z)-i\phi_2(z)=0$, then $\phi_1(z)=\phi_2(z)=0$, hence $\phi_3(z)=0$ by the first condition, contradicting the second condition.