To solve the problem, we start with the equation involving the complex number $z$:

$z - \frac{1}{z} = 3 + 5i$

We need to find the modulus of $z + \frac{1}{z}$.

Step 1: Express $z$ in Polar Form

Let $z = re^{i\theta}$, where $r = |z|$ is the modulus of $z$ and $\theta$ is the argument of $z$.

Then, the reciprocal of $z$ is:

$\frac{1}{z} = \frac{1}{re^{i\theta}} = \frac{1}{r}e^{-i\theta}$

Substitute these into the given equation:

$re^{i\theta} - \frac{1}{r}e^{-i\theta} = 3 + 5i$

Step 2: Separate Real and Imaginary Parts

Rewrite the equation using Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$:

$r(\cos\theta + i\sin\theta) - \frac{1}{r}(\cos(-\theta) + i\sin(-\theta)) = 3 + 5i$

Since $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$, this becomes:

$r(\cos\theta + i\sin\theta) - \frac{1}{r}(\cos\theta - i\sin\theta) = 3 + 5i$

Separate real and imaginary parts:

$\left(r - \frac{1}{r}\right)\cos\theta + i\left(r + \frac{1}{r}\right)\sin\theta = 3 + 5i$

From here, we can equate the real and imaginary parts:

$r\cos\theta - \frac{1}{r}\cos\theta = 3 \quad \text{(1)}$ $r\sin\theta + \frac{1}{r}\sin\theta = 5 \quad \text{(2)}$

Step 3: Solve for $r$ and $\theta$

Multiply equation (1) by $r$ and equation (2) by $r$ to get:

$r^2\cos\theta - \cos\theta = 3r \quad \text{(3)}$ $r^2\sin\theta + \sin\theta = 5r \quad \text{(4)}$

Divide equation (4) by equation (3) to solve for $\tan\theta$:

$\frac{r^2\sin\theta + \sin\theta}{r^2\cos\theta - \cos\theta} = \frac{5r}{3r}$

Simplify to get $\tan\theta$:

$\tan\theta = \frac{5}{3}$

Step 4: Express $z + \frac{1}{z}$ and Compute Its Modulus

Now, compute $z + \frac{1}{z}$:

$z + \frac{1}{z} = re^{i\theta} + \frac{1}{r}e^{-i\theta}$

Using Euler's formula again:

$z + \frac{1}{z} = r(\cos\theta + i\sin\theta) + \frac{1}{r}(\cos\theta - i\sin\theta)$

Combine like terms:

$z + \frac{1}{z} = \left(r + \frac{1}{r}\right)\cos\theta + i\left(r - \frac{1}{r}\right)\sin\theta$

The modulus of $z + \frac{1}{z}$ is:

$\left|z + \frac{1}{z}\right| = \sqrt{\left(\left(r + \frac{1}{r}\right)\cos\theta\right)^2 + \left(\left(r - \frac{1}{r}\right)\sin\theta\right)^2}$

Step 5: Use Previous Results to Simplify

Using our previous relations, recall $r^2 = \frac{3}{5}$ and $\tan\theta = \frac{5}{3}$.

Find $\cos\theta$ and $\sin\theta$ using $r$:

$\cos\theta = \frac{3}{r(r^2 + 1)} = \frac{3r}{3r}$ $\sin\theta = \frac{5}{r(r^2 + 1)} = \frac{5r}{5r}$

Substitute these into the modulus formula:

$\left|z + \frac{1}{z}\right| = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$

Therefore, the modulus of $z + \frac{1}{z}$ is:

$\boxed{5\sqrt{2}}$