Suppose z is a complex number satisfying z−1/z=3+5i . Find the length (or modulus) of z+1/z . Express your solution in closed form.

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:

\[
\boxed{5\sqrt{2}}
\]

Explore how to find the modulus of complex number z + 1/z using the given equation z - 1/z = 3 + 5i. This problem involves complex numbers, polar form, and Euler's formula. Suitable for advanced high school students and beyond.

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