If α and β are different complex numbers with |α | = 1, then what is \(\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right|\) equal to?
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Understanding the Complex Number Problem
The question asks us to find the value of the modulus of a specific complex expression: \(\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right|\). We are given two crucial pieces of information:
- \(\alpha\) and \(\beta\) are different complex numbers (\(\alpha \neq \beta\)).
- The modulus of \(\alpha\) is equal to 1, i.e., \(|\alpha| = 1\).
We need to use the properties of complex numbers and their moduli to simplify the given expression.
Key Properties of Complex Numbers and Modulus
To solve this problem, we will use the following properties:
- The modulus of a ratio: \(\left| {\frac{z_1}{z_2}} \right| = \frac{|z_1|}{|z_2|}\) for \(z_2 \neq 0\).
- The relationship between modulus squared and conjugate: \(|z|^2 = z \bar z\).
- The property of complex conjugates: \(\overline{z_1 + z_2} = \bar{z_1} + \bar{z_2}\) and \(\overline{z_1 z_2} = \bar{z_1} \bar{z_2}\).
- Given \(|\alpha| = 1\), we have \(|\alpha|^2 = 1^2 = 1\). Using \(|z|^2 = z \bar z\), this means \(\alpha \bar \alpha = 1\). This also implies \(\bar \alpha = \frac{1}{\alpha}\) (since \(|\alpha|=1 > 0\), \(\alpha \neq 0\)).
Step-by-Step Solution for Modulus Calculation
We want to find \(\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right|\). Using the property \(\left| {\frac{z_1}{z_2}} \right| = \frac{|z_1|}{|z_2|}\), we have:
\( \left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right| = \frac{{|\alpha - \beta|}}{{|1 - \alpha \bar \beta|}} \)
To compare the numerator and the denominator, let's calculate the square of their moduli. We know that \(|z|^2 = z \bar z\).
Calculating the Square of the Numerator Modulus
\( |\alpha - \beta|^2 = (\alpha - \beta) \overline{(\alpha - \beta)} \)
Using the conjugate property \(\overline{z_1 - z_2} = \bar{z_1} - \bar{z_2}\):
\( |\alpha - \beta|^2 = (\alpha - \beta)(\bar \alpha - \bar \beta) \)
Expand the product:
\( |\alpha - \beta|^2 = \alpha \bar \alpha - \alpha \bar \beta - \beta \bar \alpha + \beta \bar \beta \)
Using \(|\alpha|^2 = \alpha \bar \alpha\) and \(|\beta|^2 = \beta \bar \beta\):
\( |\alpha - \beta|^2 = |\alpha|^2 - \alpha \bar \beta - \bar \alpha \beta + |\beta|^2 \)
Now, substitute the given condition \(|\alpha| = 1\), so \(|\alpha|^2 = 1\):
\( |\alpha - \beta|^2 = 1 - \alpha \bar \beta - \bar \alpha \beta + |\beta|^2 \quad (*)\label{eq:num_sq} \)
Calculating the Square of the Denominator Modulus
\( |1 - \alpha \bar \beta|^2 = (1 - \alpha \bar \beta) \overline{(1 - \alpha \bar \beta)} \)
Using the conjugate properties \(\overline{z_1 - z_2} = \bar{z_1} - \bar{z_2}\) and \(\overline{z_1 z_2} = \bar{z_1} \bar{z_2}\):
\( \overline{(1 - \alpha \bar \beta)} = \bar 1 - \overline{(\alpha \bar \beta)} = 1 - \bar \alpha \overline{\bar \beta} = 1 - \bar \alpha \beta \)
So, the square of the denominator modulus is:
\( |1 - \alpha \bar \beta|^2 = (1 - \alpha \bar \beta)(1 - \bar \alpha \beta) \)
Expand the product:
\( |1 - \alpha \bar \beta|^2 = 1 \cdot 1 - 1 \cdot (\bar \alpha \beta) - (\alpha \bar \beta) \cdot 1 + (\alpha \bar \beta)(\bar \alpha \beta) \)
\( |1 - \alpha \bar \beta|^2 = 1 - \bar \alpha \beta - \alpha \bar \beta + \alpha \bar \beta \bar \alpha \beta \)
Rearrange the terms in the last part: \(\alpha \bar \beta \bar \alpha \beta = (\alpha \bar \alpha)(\bar \beta \beta)\). Using \(|z|^2 = z \bar z\), we have \(\alpha \bar \alpha = |\alpha|^2\) and \(\bar \beta \beta = |\beta|^2\).
\( |1 - \alpha \bar \beta|^2 = 1 - \bar \alpha \beta - \alpha \bar \beta + |\alpha|^2 |\beta|^2 \)
Now, substitute the given condition \(|\alpha| = 1\), so \(|\alpha|^2 = 1\):
\( |1 - \alpha \bar \beta|^2 = 1 - \bar \alpha \beta - \alpha \bar \beta + 1 \cdot |\beta|^2 \)
\( |1 - \alpha \bar \beta|^2 = 1 - \bar \alpha \beta - \alpha \bar \beta + |\beta|^2 \quad (**) \)
Comparing Numerator and Denominator Moduli Squared
Comparing equation \((*)\) and equation \((**)\), we see that:
\( |\alpha - \beta|^2 = 1 - \alpha \bar \beta - \bar \alpha \beta + |\beta|^2 \)
\( |1 - \alpha \bar \beta|^2 = 1 - \bar \alpha \beta - \alpha \bar \beta + |\beta|^2 \)
Thus, \(|\alpha - \beta|^2 = |1 - \alpha \bar \beta|^2\).
Since the modulus of a complex number is always non-negative, we can take the square root of both sides:
\( \sqrt{|\alpha - \beta|^2} = \sqrt{|1 - \alpha \bar \beta|^2} \)
\( |\alpha - \beta| = |1 - \alpha \bar \beta| \)
Final Calculation
Now, substitute this back into the original expression:
\( \left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right| = \frac{{|\alpha - \beta|}}{{|1 - \alpha \bar \beta|}} \)
Since \(|\alpha - \beta| = |1 - \alpha \bar \beta|\), and the denominator \(1 - \alpha \bar \beta \neq 0\) (because if it were 0, then \(\alpha \bar{\beta} = 1\). Since \(|\alpha|=1\), \(\alpha\bar{\alpha}=1\). This would imply \(\alpha\bar{\beta} = \alpha\bar{\alpha}\). As \(\alpha \neq 0\), we can divide by \(\alpha\) to get \(\bar{\beta} = \bar{\alpha}\), which means \(\beta = \alpha\). But the problem states \(\alpha \neq \beta\). Therefore, \(1 - \alpha \bar \beta \neq 0\)), the ratio is:
\( \left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right| = \frac{{|\alpha - \beta|}}{{|\alpha - \beta|}} = 1 \)
Thus, the value of the given expression is 1.
Revision Table: Complex Number Modulus
| Concept | Formula/Property | Example |
|---|---|---|
| Modulus of a complex number \(z = x + iy\) | \(|z| = \sqrt{x^2 + y^2}\) | If \(z = 3 + 4i\), \(|z| = \sqrt{3^2 + 4^2} = 5\) |
| Modulus squared | \(|z|^2 = z \bar z\) | If \(z = 3 + 4i\), \(\bar z = 3 - 4i\), \(z \bar z = (3+4i)(3-4i) = 3^2 - (4i)^2 = 9 - (-16) = 25\). \(|z|^2 = 5^2 = 25\). |
| Modulus of a product | \(|z_1 z_2| = |z_1| |z_2|\) | \(|(1+i)(2+i)| = |1+i||2+i| = \sqrt{2} \sqrt{5} = \sqrt{10}\). \((1+i)(2+i) = 2+i+2i-1 = 1+3i\), \(|1+3i| = \sqrt{1^2+3^2} = \sqrt{10}\). |
| Modulus of a quotient | \(\left| {\frac{z_1}{z_2}} \right| = \frac{|z_1|}{|z_2|}\) (for \(z_2 \neq 0\)) | \(\left| {\frac{1+i}{2+i}} \right| = \frac{|1+i|}{|2+i|} = \frac{\sqrt{2}}{\sqrt{5}} = \sqrt{\frac{2}{5}}\) |
| Complex Conjugate of a sum/difference | \(\overline{z_1 \pm z_2} = \bar{z_1} \pm \bar{z_2}\) | \(\overline{(3+4i) + (1-i)} = \overline{4+3i} = 4-3i\). \(\overline{(3+4i)} + \overline{(1-i)} = (3-4i) + (1+i) = 4-3i\). |
| Complex Conjugate of a product/quotient | \(\overline{z_1 z_2} = \bar{z_1} \bar{z_2}\), \(\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z_1}}{\bar{z_2}}\) | \(\overline{(1+i)(2+i)} = \overline{1+3i} = 1-3i\). \(\overline{(1+i)} \overline{(2+i)} = (1-i)(2-i) = 2-i-2i-1 = 1-3i\). |
| Property for \(|z|=1\) | \(z \bar z = 1 \implies \bar z = 1/z\) | If \(z = i\), \(|z|=1\). \(\bar z = -i\). \(1/z = 1/i = -i\). |
Additional Information on Complex Numbers
Complex numbers are an extension of real numbers, expressed in the form \(z = x + iy\), where \(x\) and \(y\) are real numbers, and \(i\) is the imaginary unit satisfying \(i^2 = -1\). The real part is \(\text{Re}(z) = x\) and the imaginary part is \(\text{Im}(z) = y\).
- Geometric Interpretation: A complex number \(z = x + iy\) can be represented as a point \((x, y)\) in the complex plane (also called the Argand plane). The modulus \(|z| = \sqrt{x^2 + y^2}\) represents the distance of the point \((x, y)\) from the origin \((0,0)\).
- Unit Circle: Complex numbers with modulus 1 lie on the unit circle centered at the origin in the complex plane. The condition \(|\alpha|=1\) means \(\alpha\) lies on this unit circle.
- Complex Conjugate: The conjugate of \(z = x + iy\) is \(\bar z = x - iy\). Geometrically, \(\bar z\) is the reflection of \(z\) across the real axis.
- Properties Used: The solution heavily relied on the property \(|z|^2 = z \bar z\) and the property that if \(|z|=1\), then \(\bar z = 1/z\). These are fundamental in simplifying expressions involving moduli of complex numbers on the unit circle. The expression \(\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}\) is related to Mobius transformations, and this specific form has a special property related to the unit circle. If \(|\alpha|=1\), this transformation maps the unit circle to itself.
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