To solve this complex number problem, we are given that \(z\) is a complex number and the expression \(\frac{z-1}{z+1}\) is purely imaginary. We need to find the magnitude of \(z\), denoted as \(|z|\).

Understanding Purely Imaginary Complex Numbers

A complex number is typically written in the form \(a + ib\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit (\(i^2 = -1\)). A complex number is said to be purely imaginary if its real part \(a\) is equal to zero. In other words, a purely imaginary number looks like \(0 + ib\), or simply \(ib\), where \(b\) is a real number.

Setting up the Complex Number Expression

Let the complex number \(z\) be represented as \(z = x + iy\), where \(x\) and \(y\) are real numbers. The expression we are given is:

\(\frac{z-1}{z+1} = \frac{(x+iy)-1}{(x+iy)+1} = \frac{(x-1)+iy}{(x+1)+iy}\)

Simplifying the Expression Using Complex Conjugate

To determine the real and imaginary parts of this expression, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \((x+1)+iy\) is \((x+1)-iy\).

\(\frac{(x-1)+iy}{(x+1)+iy} \times \frac{(x+1)-iy}{(x+1)-iy}\)

Let's calculate the numerator and denominator separately.

Denominator:

\(((x+1)+iy)((x+1)-iy) = (x+1)^2 - (iy)^2 = (x+1)^2 - i^2y^2 = (x+1)^2 - (-1)y^2 = (x+1)^2 + y^2\)

Numerator:

\(((x-1)+iy)((x+1)-iy) = (x-1)(x+1) - (x-1)(iy) + (iy)(x+1) - (iy)(iy)\)

\(= (x^2 - 1) - i(x-1)y + i(x+1)y - i^2y^2\)

\(= (x^2 - 1) - ixy + iy + ixy + iy + y^2\)

\(= (x^2 - 1 + y^2) + i(-xy + y + xy + y)\)

\(= (x^2 + y^2 - 1) + i(2y)\)

So, the simplified expression is:

\(\frac{(x^2 + y^2 - 1) + i(2y)}{(x+1)^2 + y^2}\)

We can separate this into its real and imaginary parts:

\(\frac{x^2 + y^2 - 1}{(x+1)^2 + y^2} + i \frac{2y}{(x+1)^2 + y^2}\)

Using the Purely Imaginary Condition

We are given that the expression \(\frac{z-1}{z+1}\) is purely imaginary. This means its real part must be zero.

The real part is \(\frac{x^2 + y^2 - 1}{(x+1)^2 + y^2}\). Setting the real part to zero:

\(\frac{x^2 + y^2 - 1}{(x+1)^2 + y^2} = 0\)

For a fraction to be zero, the numerator must be zero, provided the denominator is non-zero. The denominator is \((x+1)^2 + y^2\), which is the squared magnitude of \(z+1\). This is zero only if \(x+1=0\) and \(y=0\), meaning \(x=-1\) and \(y=0\), or \(z=-1\). However, if \(z=-1\), the original expression \(\frac{z-1}{z+1}\) is undefined because the denominator is zero. Thus, \(z \neq -1\), and the denominator \((x+1)^2 + y^2\) is always non-zero.

Therefore, the numerator must be zero:

\(x^2 + y^2 - 1 = 0\)

\(x^2 + y^2 = 1\)

Calculating the Magnitude |z|

The magnitude of a complex number \(z = x + iy\) is defined as \(|z| = \sqrt{x^2 + y^2}\).

From our condition \(x^2 + y^2 = 1\), we can find \(|z|\):

\(|z| = \sqrt{x^2 + y^2} = \sqrt{1} = 1\)

Thus, if \(\frac{z-1}{z+1}\) is purely imaginary, the magnitude of \(z\) is 1.

Checking Options

Let's look at the given options:

• Option 1: \(\frac{1}{2}\)
• Option 2: \(\frac{2}{3}\)
• Option 3: 1
• Option 4: 2

Our calculated value for \(|z|\) is 1, which matches Option 3.

The final answer is 1.

Revision Table: Complex Number Concepts

Additional Information on Complex Number Properties

The condition that \(\frac{z-1}{z+1}\) is purely imaginary can also be interpreted geometrically. If \(\frac{z-1}{z+1} = ik\) for some real number \(k\), then \(\frac{z-1}{z+1} = -\overline{\left(\frac{z-1}{z+1}\right)}\) (unless the expression is 0). Let \(w = \frac{z-1}{z+1}\). If \(w\) is purely imaginary, then \(w = -\bar{w}\).

\(\frac{z-1}{z+1} = - \overline{\left(\frac{z-1}{z+1}\right)} = - \frac{\overline{z-1}}{\overline{z+1}} = - \frac{\bar{z}-1}{\bar{z}+1}\)

\((z-1)(\bar{z}+1) = -(z+1)(\bar{z}-1)\)

\(z\bar{z} + z - \bar{z} - 1 = -(z\bar{z} - z + \bar{z} - 1)\)

\(z\bar{z} + z - \bar{z} - 1 = -z\bar{z} + z - \bar{z} + 1\)

\(z\bar{z} - 1 = -z\bar{z} + 1\)

\(2z\bar{z} = 2\)

\(z\bar{z} = 1\)

Since \(z\bar{z} = |z|^2\), we have \(|z|^2 = 1\). Taking the square root, \(|z|=1\) (since magnitude is non-negative).

This confirms the result obtained by separating into real and imaginary parts. The locus of points \(z\) such that \(|z|=1\) is a circle of radius 1 centered at the origin in the complex plane. This is the unit circle.