If `z=(3-sqrt(7)i)`, then find `|z^(-1)|`.

Answer

To find \(|z^{-1}|\) for \(z = 3 - \sqrt{7}i\), we can follow these steps:

### Step 1: Find \(z^{-1}\)
We start by calculating \(z^{-1}\), which is given by:
\[ z^{-1} = \frac{1}{z} = \frac{1}{3 - \sqrt{7}i} \]

### Step 2: Rationalize the denominator
To simplify \(\frac{1}{3 - \sqrt{7}i}\), we multiply the numerator and the denominator by the conjugate of the denominator, which is \(3 + \sqrt{7}i\):
\[ z^{-1} = \frac{1 \cdot (3 + \sqrt{7}i)}{(3 - \sqrt{7}i)(3 + \sqrt{7}i)} \]

### Step 3: Calculate the denominator
Now, we calculate the denominator using the difference of squares:
\[ (3 - \sqrt{7}i)(3 + \sqrt{7}i) = 3^2 - (\sqrt{7}i)^2 = 9 - (-7) = 9 + 7 = 16 \]

### Step 4: Write \(z^{-1}\)
Now we can write \(z^{-1}\):
\[ z^{-1} = \frac{3 + \sqrt{7}i}{16} \]

### Step 5: Find the modulus of \(z^{-1}\)
The modulus of a complex number \(a + bi\) is given by \(|z| = \sqrt{a^2 + b^2}\). In our case, we have:
\[ a = \frac{3}{16}, \quad b = \frac{\sqrt{7}}{16} \]
Thus,
\[ |z^{-1}| = \sqrt{\left(\frac{3}{16}\right)^2 + \left(\frac{\sqrt{7}}{16}\right)^2} \]

### Step 6: Calculate \(|z^{-1}|\)
Calculating the squares:
\[ \left(\frac{3}{16}\right)^2 = \frac{9}{256}, \quad \left(\frac{\sqrt{7}}{16}\right)^2 = \frac{7}{256} \]
Adding these:
\[ |z^{-1}| = \sqrt{\frac{9}{256} + \frac{7}{256}} = \sqrt{\frac{16}{256}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

### Final Answer
Thus, the modulus of \(z^{-1}\) is:
\[ |z^{-1}| = \frac{1}{4} \]