If `z=2 + t + i sqrt(3-t^2)`, where t is real and `t^2 lt 3`, show that the modulus of `(z+1)/(z-1)` is independent of t. Also show that the locus of the points z for different values of t is a circle and finds its centre and radius.

Answer

To solve the given problem, we will break it down into two parts:

1. Show that the modulus of \(\frac{z+1}{z-1}\) is independent of \(t\).
2. Show that the locus of points \(z\) for different values of \(t\) is a circle and find its center and radius.

### Step 1: Show that the modulus of \(\frac{z+1}{z-1}\) is independent of \(t\)

Given:
\[ z = 2 + t + i \sqrt{3 - t^2} \]

We can express \(z\) in terms of its real and imaginary parts:
- Real part: \(x = 2 + t\)
- Imaginary part: \(y = \sqrt{3 - t^2}\)

Now, we calculate \(z + 1\) and \(z - 1\):
\[ z + 1 = (2 + t + 1) + i \sqrt{3 - t^2} = (3 + t) + i \sqrt{3 - t^2} \]
\[ z - 1 = (2 + t - 1) + i \sqrt{3 - t^2} = (1 + t) + i \sqrt{3 - t^2} \]

Next, we find the modulus of \(\frac{z + 1}{z - 1}\):
\[ \left| \frac{z + 1}{z - 1} \right| = \frac{|z + 1|}{|z - 1|} \]

Calculating \(|z + 1|\):
\[ |z + 1| = \sqrt{(3 + t)^2 + (3 - t^2)} = \sqrt{(3 + t)^2 + 3 - t^2} \]
Expanding this:
\[ = \sqrt{(3 + t)^2 + 3 - t^2} = \sqrt{9 + 6t + t^2 + 3 - t^2} = \sqrt{12 + 6t} = \sqrt{6(2 + t)} \]

Calculating \(|z - 1|\):
\[ |z - 1| = \sqrt{(1 + t)^2 + (3 - t^2)} = \sqrt{(1 + t)^2 + 3 - t^2} \]
Expanding this:
\[ = \sqrt{(1 + t)^2 + 3 - t^2} = \sqrt{1 + 2t + t^2 + 3 - t^2} = \sqrt{4 + 2t} = \sqrt{2(2 + t)} \]

Now substituting back into the modulus:
\[ \left| \frac{z + 1}{z - 1} \right| = \frac{\sqrt{6(2 + t)}}{\sqrt{2(2 + t)}} = \sqrt{\frac{6}{2}} = \sqrt{3} \]

Thus, we have shown that:
\[ \left| \frac{z + 1}{z - 1} \right| = \sqrt{3} \]
which is independent of \(t\).

### Step 2: Show that the locus of points \(z\) for different values of \(t\) is a circle

From our expression for \(z\):
\[ z = 2 + t + i \sqrt{3 - t^2} \]

We can express \(z\) in terms of \(x\) and \(y\):
- \(x = 2 + t\)
- \(y = \sqrt{3 - t^2}\)

From \(x = 2 + t\), we can express \(t\):
\[ t = x - 2 \]

Substituting \(t\) into the expression for \(y\):
\[ y = \sqrt{3 - (x - 2)^2} \]
Squaring both sides:
\[ y^2 = 3 - (x - 2)^2 \]
Expanding:
\[ y^2 = 3 - (x^2 - 4x + 4) = 3 - x^2 + 4x - 4 = -x^2 + 4x - 1 \]

Rearranging gives:
\[ (x - 2)^2 + y^2 = 3 \]

This is the equation of a circle with:
- Center: \((2, 0)\)
- Radius: \(\sqrt{3}\)

### Final Result
1. The modulus \(\left| \frac{z + 1}{z - 1} \right|\) is independent of \(t\) and equals \(\sqrt{3}\).
2. The locus of points \(z\) is a circle with center \((2, 0)\) and radius \(\sqrt{3}\).