Answer: 100

Small Hint:

$\frac{w-z}{z} = \frac{w}{z} - 1$ lies on the circle of radius $\frac{1}{10}$ centered at $-1$

Big Hint:

The argument is extremal when the ray from the origin is tangent to that circle, so the sine of the angle with the real axis is $\frac{1}{10}$

Solution:

Since $\frac{w-z}{z} = \frac{w}{z} - 1,$ and $\frac{w}{z}$ can be any complex number of modulus $\frac{1}{10},$ the point $\zeta = \frac{w-z}{z}$ ranges over the circle of radius $\frac{1}{10}$ centered at $-1.$

Because $\tan^2\theta$ is unchanged when $\theta$ shifts by $180^\circ,$ we want the largest angle $\alpha$ that a ray from the origin to this circle makes with the real axis. The extreme rays are tangent to the circle, where $\sin \alpha = \frac{\frac{1}{10}}{1} = \frac{1}{10}.$

Then $$\begin{aligned} \tan^2\theta &= \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{\frac{1}{100}}{\frac{99}{100}} \\ &= \frac{1}{99}, \end{aligned}$$ so $p + q = 1 + 99 = 100.$