Solving Complex Number Conditions: \({\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = 0\)

The question asks for a condition on the complex number \(z = x + iy\) given that the real part of the expression \({\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}}\) is equal to zero.

A complex number \(z\) is generally expressed in the form \(x + iy\), where \(x\) is the real part and \(y\) is the imaginary part. The modulus or magnitude of \(z\) is denoted by \(|z|\) and is calculated as \(|z| = \sqrt{x^2 + y^2}\).

We are given the condition: \({\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = 0\)

First, let's substitute \(z = x + iy\) into the expression \({\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}}\):

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

To find the real part of this complex fraction, 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 the denominator separately.

Numerator:

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

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

Since \(i^2 = -1\), this simplifies to:

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

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

Denominator:

\({((x + 1) + iy)((x + 1) - iy) = (x + 1)^2 - (iy)^2}\)

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

Since \(i^2 = -1\), this simplifies to:

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

So, the expression becomes:

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

For the expression \({\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}}\) to be defined, the denominator cannot be zero. The denominator is \({x^2 + y^2 + 2x + 1 = (x + 1)^2 + y^2}\). This is zero only if \({x + 1 = 0}\) and \({y = 0}\), which means \({x = -1}\) and \({y = 0}\), corresponding to the complex number \(z = -1\). Thus, the given condition implies that \(z \neq -1\).

The real part of \(\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}\) is \({\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = \frac{{x^2 + y^2 - 1}}{{x^2 + y^2 + 2x + 1}}\).

We are given that this real part is 0:

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

For a fraction to be zero, the numerator must be zero, provided the denominator is not zero (which we've already established requires \(z \neq -1\)).

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

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

Recall that the modulus of \(z = x + iy\) is \(|z| = \sqrt{x^2 + y^2}\). Squaring both sides gives \(|z|^2 = x^2 + y^2\).

So, the condition \(x^2 + y^2 = 1\) is equivalent to \(|z|^2 = 1\).

Since the modulus \(|z|\) is always non-negative, taking the square root gives \(|z| = 1\).

Thus, if \({\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = 0\), then it must be true that \(|z| = 1\) (and \(z \neq -1\)). The statement \(|z|=1\) is one of the given options.

Let's check the options based on this derivation:

Analysis of Options:

• Option 1: \(z = 1 + i\). The modulus is \(|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}\). This does not satisfy \(|z|=1\).
• Option 2: \(|z| = 2\). This does not match our derived condition \(|z|=1\).
• Option 3: \(z = 1 - i\). The modulus is \(|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\). This does not satisfy \(|z|=1\).
• Option 4: \(|z| = 1\). This is exactly the condition we derived from the given premise \({\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = 0\).

Therefore, if \({\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = 0\), the condition \(|z| = 1\) must be correct.

Revision Table: Complex Number Equation

Additional Information: Geometric Interpretation and Purely Imaginary Numbers

The condition \({\rm{Re}}(w) = 0\) means that the complex number \(w = \frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}\) lies on the imaginary axis in the complex plane (or is the origin if \(w=0\)). A complex number with a real part of zero is called a purely imaginary number.

The expression \(\frac{{{\rm{z}} - a}}{{{\rm{z}} - b}}\) (where a and b are complex numbers) has a geometric interpretation. \({\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = 0\) means that the argument of \(\frac{{{\rm{z}} - 1}}{{{\rm{z}} + (-1)}}\) is \(\pm \frac{\pi}{2}\) or \(\pm \frac{3\pi}{2}\) radians. This corresponds to the locus of points \(z\) such that the angle formed by the vectors from \(-1\) to \(z\) and from \(1\) to \(z\) is \(90^\circ\) or \(270^\circ\). This locus is a circle with the segment connecting \(1\) and \(-1\) as a diameter. This circle is centered at the origin \((0,0)\) and has a radius of 1. The equation of this circle is \(x^2 + y^2 = 1\), which is \(|z|=1\). However, the point \(z = -1\) makes the expression undefined, so it is excluded from the locus. The point \(z = 1\) makes the real part \({\rm{Re}}(0/2) = 0\), so \(z=1\) is included.

The set of points \(z\) satisfying \({\rm{Re}}\left( {\frac{{{\rm{z}} - 1}}{{{\rm{z}} + 1}}} \right) = 0\) is the unit circle excluding the point \(-1\). However, the options ask for a property that is correct for any \(z\) satisfying the premise. The property \(|z|=1\) is true for all points on the unit circle, including those excluding \(-1\). Therefore, \(|z|=1\) is the correct statement.