Let \( a \) and \( b \) be complex numbers such that \( a + b = 4 \) and \( a^2 + b^2 = 10 \). Find \( |a - b| \).

["Title:\nFind ( |a - b| ) Given ( a + b = 4 ) and ( a^2 + b^2 = 10 ) — Step-by-Step Solution", "---", "Meta Description:\nLearn how to compute ( |a - b| ) when complex numbers ( a ) and ( b ) satisfy ( a + b = 4 ) and ( a^2 + b^2 = 10 ) using algebraic identities and properties of complex modulus.", "---", "Introduction:\nComplex numbers often appear in equations involving sums and sums of squares. Given the conditions ( a + b = 4 ) and ( a^2 + b^2 = 10 ), we seek the modulus ( |a - b| ). This problem combines algebraic manipulation with complex number properties—ideal for students and enthusiasts exploring complex algebra.", "In this article, we will:", "- Use algebraic identities to find ( ab ),\n- Compute ( |a - b| ) using the modulus formula,\n- Interpret the meaning of finding a complex modulus from sum and sum-of-squares.", "---", "### Step 1: Use identity for ( a^2 + b^2 )", "Recall the identity:\n[\na^2 + b^2 = (a + b)^2 - 2ab\n]", "Substitute the given values:\n[\n10 = (4)^2 - 2ab = 16 - 2ab\n]", "Solve for ( ab ):\n[\n2ab = 16 - 10 = 6 \quad \Rightarrow \quad ab = 3\n]", "So, the product of ( a ) and ( b ) is ( 3 ).", "---", "### Step 2: Compute ( |a - b| )", "We want ( |a - b| ). For complex numbers, the modulus squared satisfies:\n[\n|a - b|^2 = (a - b)(\overline{a - b}) = (a - b)(\overline{a} - \overline{b})\n]", "However, for real-valued expressions involving sum and sum of squares, we can use:\n[\n|a - b|^2 = (a - b)^2 \quad \ ext{if } a, b \in \mathbb{R}, \ ext{ but not always true for complex } a,b\n]", "Instead, use the algebraic identity:\n[\n(a - b)^2 = a^2 - 2ab + b^2 = (a^2 + b^2) - 2ab = 10 - 2(3) = 10 - 6 = 4\n]", "Thus,\n[\n(a - b)^2 = 4 \quad \Rightarrow \quad a - b = \pm 2\n]", "But this gives ( (a - b)^2 = 4 ), so ( a - b ) is either ( 2 ) or ( -2 ), both real.", "However, since ( |z|^2 = z \cdot \overline{z} ), and we are dealing with complex numbers, the modulus ( |a - b| ) is the non-negative square root of ( |a - b|^2 ).", "But note:\n[\n|a - b|^2 = (a - b)\overline{(a - b)} = |a|^2 + |b|^2 - 2,\ ext{Re}(a\overline{b})\n]", "That approach is messy without magnitude info. Instead, use a key identity:", "[\n|a - b|^2 = |a + b|^2 - 4,\ ext{Re}(ab) \quad ? \quad \ ext{No — too flawed.}\n]", "Better: Use the discriminant method for complex numbers.", "From earlier:\n[\n(a - b)^2 = a^2 - 2ab + b^2 = (a^2 + b^2) - 2ab = 10 - 6 = 4\n]", "So,\n[\n(a - b)^2 = 4 \quad \Rightarrow \quad a - b = \pm 2\n]", "But since ( a ) and ( b ) are complex, squaring gives that ( a - b ) is a real number with square 4 — so ( a - b = 2 ) or ( -2 ), both real.", "Thus, ( a - b ) is real and equal to ( \pm 2 ).", "Therefore, its modulus is absolute value:\n[\n|a - b| = | \pm 2 | = 2\n]", "But wait — is this valid? Could ( a - b ) be complex?", "Yes! Our calculation shows ( (a - b)^2 = 4 ), but 4 is real, so ( a - b ) is a complex number whose square is real. This includes real numbers and purely imaginary numbers? No:\nIf ( z^2 = 4 ) (real), then ( z = \pm 2 ) are the only complex solutions — because any complex number ( z = x + iy ) squares to:\n[\nz^2 = x^2 - y^2 + 2ixy\n]\nSetting this equal to real 4 requires ( xy = 0 ). So either ( x = 0 ) or ( y = 0 ).", "- If ( y = 0 ): ( z = \pm 2 ) (real)\n- If ( x = 0 ): ( z = \pm 2i ), then ( z^2 = -4 <br/>\ne 4 ), not valid", "So only solutions are ( \pm 2 ), both real.", "Thus, ( a - b ) is real, ( |a - b| = 2 )", "But wait — can we get a direct formula?", "---", "### Step 3: Use identity for ( |a - b|^2 )", "We know:\n[\n|a - b|^2 = (a - b)(\overline{a} - \overline{b}) = |a|^2 + |b|^2 - a\overline{b} - \overline{a}b\n]", "Too complicated without magnitudes.", "But here’s a key insight: Use the identity:\n[\n|a - b|^2 = |a + b|^2 - 4,\ ext{Re}(ab) \quad ? \quad \ ext{No — not generally true.}\n]", "Wait — instead, recall:\n[\n(a + b)^2 = a^2 + 2ab + b^2 \Rightarrow (a + b)^2 = (a^2 + b^2) + 2ab\n]\nAlready used.", "Alternatively, use:\n[\n|a - b|^2 = |a|^2 + |b|^2 - 2,\ ext{Re}(a\overline{b})\n]", "Still messy.", "But here is a smarter algebraic path:", "Let ( s = a + b = 4 ), ( p = ab = 3 )", "Then ( a ) and ( b ) are roots of the quadratic:\n[\nx^2 - sx + p = x^2 - 4x + 3 = 0\n]", "Solve:\n[\nx = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2} = 3, 1\n]", "So ( a = 3, b = 1 ) or vice versa — both real.", "Thus, ( a - b = 2 ) or ( -2 ), so ( |a - b| = 2 )", "Since both are real, modulus is 2.", "Even without solving, since ( a ) and ( b ) are roots of a quadratic with real coefficients and sum/product real, and discriminant ( D = 4 \geq 0 ), roots are real — so ( |a - b| = \sqrt{(a + b)^2 - 4ab} )", "Wait — in real numbers,\n[\n(a - b)^2 = (a + b)^2 - 4ab = 16 - 12 = 4 \Rightarrow |a - b| = 2\n]", "And since the discriminant is non-negative, no complex part — so this extends to complex numbers only if roots are real or differ by 2.", "But in general, for any complex numbers, if ( (a - b)^2 = 4 ), then ( a - b = \pm 2 ), so modulus is 2.", "Thus, regardless of whether ( a ) and ( b ) are real or complex, as long as ( (a - b)^2 = 4 ), we have ( |a - b| = 2 )", "---", "### Final Answer:", "[\n\boxed{|a - b| = 2}\n]", "---", "### Bonus Interpretation:", "In real analysis, such problems often reduce to computing ( (a - b)^2 = (a + b)^2 - 4ab ), a well-known identity. This formula holds even when ( a ) and ( b ) are complex — the square of their difference is still ( a^2 - 2ab + b^2 ), and if that equals 4, then ( a - b = \pm 2 ), so modulus is 2.", "The key insight is recognizing that the sum and sum of squares allow computation of ( ab ), and then using the algebraic identity for the square of the difference.", "---", "Related Keywords:\ncomplex numbers, ( a + b = 4 ), ( a^2 + b^2 = 10 ), find ( |a - b| ), algebraic identities, complex modulus, discriminant identity, quadratic roots, sum of squares formula", "For more:\nSee "quadratic formula for complex numbers" and "sum vs symmetric functions in complex algebra"", "---", "This solution combines algebraic manipulation, properties of modulus, and roots of quadratics to deliver a rigorous and clear answer for learners at intermediate to advanced high school or early college level."]