= Solution
For $z\ne0$, division by $4z$ gives $p(z)=1/(2z)$ and $q(z)=1/(4z)$. Thus zero is singular, but $zp=1/2$ and $z^2q=z/4$ are analytic. \b[Zero is a regular singular point.]
Apply the <Frobenius method> with $y=z^r\sum_{n=0}^\infty a_nz^n$, $a_0\ne0$. The lowest power $z^{r-1}$ has coefficient $[4r(r-1)+2r]a_0$, so the <indicial equation> is $2r(2r-1)=0$. Its <characteristic exponents at a regular singular point> are $0$ and $1/2$. At the next powers,
$$
2(n+r)(2(n+r)-1)a_n+a_{n-1}=0,\qquad n\geq1.
$$
With $a_0=1$, the two <Frobenius solutions> are therefore
$$
y_1(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^n}{(2n)!},
\qquad
y_2(z)=z^{1/2}\sum_{n=0}^{\infty}\frac{(-1)^nz^n}{(2n+1)!}.
$$
Both analytic power-series factors converge everywhere by the <ratio test>. Recognizing the trigonometric series gives
$$
\boxed{y(z)=A\cos\sqrt z+B\sin\sqrt z.}
$$
On a punctured neighbourhood choose a consistent square-root branch for the second solution; the first is entire in $z$. These solutions are independent, since their leading orders are $1$ and $z^{1/2}$. A direct <square-root reduction of a regular-singular differential equation> also verifies the closed forms: for $s=\sqrt z$ and $Y(s)=y(s^2)$, $4zy''=Y_{ss}-Y_s/s$ and $2y'=Y_s/s$, so the equation reduces to $Y_{ss}+Y=0$.
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