Solution
= Solution
Since $z+dz=\varphi(z,\theta)$,
$$
dz^r=
\left.\frac{\partial\varphi^r(z,\theta)}{\partial\theta^a}
\right|_{\theta=0}\theta^a+O(\theta^2).
$$
Thus
$$
\boxed{dz^r=\mu_a{}^r(z)\theta^a}.
$$
= Solution
Since $z+dz=\varphi(z,\theta)$,
$$
dz^r=
\left.\frac{\partial\varphi^r(z,\theta)}{\partial\theta^a}
\right|_{\theta=0}\theta^a+O(\theta^2).
$$
Thus
$$
\boxed{dz^r=\mu_a{}^r(z)\theta^a}.
$$