#### 78.5Question: Let $ a $ and $ b $ be complex numbers such that $ a^2 + b^2 = 1 $ and $ ab = \frac{1}{2} $. Find $ |a^4 + b^4| $.

["Understanding $ |a^4 + b^4| $ Given $ a^2 + b^2 = 1 $ and $ ab = \frac{1}{2} $", "When solving problems involving complex numbers $ a $ and $ b $ with symmetric conditions like $ a^2 + b^2 = 1 $ and $ ab = \frac{1}{2} $, it’s essential to simplify expressions using algebraic identities. This article explores how to compute $ |a^4 + b^4| $ under these constraints.", "### Step 1: Use Known Identities", "We begin with the identity for the sum of fourth powers:", "[\na^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2\n]", "Given:\n- $ a^2 + b^2 = 1 $\n- $ ab = \frac{1}{2} \Rightarrow (ab)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4} $, so $ a^2b^2 = \frac{1}{4} $", "Substitute into the identity:", "[\na^4 + b^4 = (1)^2 - 2 \cdot \frac{1}{4} = 1 - \frac{1}{2} = \frac{1}{2}\n]", "### Step 2: Consider Complex Modulus", "We now seek $ |a^4 + b^4| $. Since $ a^4 + b^4 = \frac{1}{2} $, a real and positive number, its modulus is simply the absolute value:", "[\n|a^4 + b^4| = \left| \frac{1}{2} \right| = \frac{1}{2}\n]", "This conclusion holds regardless of whether $ a $ and $ b $ are real or complex, because the sum $ a^4 + b^4 $ evaluates to a real positive number in this case.", "### Step 3: Verify Consistency of Given Conditions", "It’s important to confirm that such complex numbers $ a $ and $ b $ exist satisfying both $ a^2 + b^2 = 1 $ and $ ab = \frac{1}{2} $.", "Let $ s = a^2 + b^2 = 1 $, $ p = ab = \frac{1}{2} $. Then $ a $ and $ b $ are roots of the quadratic:", "[\nx^2 - sx + p = x^2 - x + \frac{1}{2}\n]", "Discriminant:", "[\n\Delta = (-1)^2 - 4 \cdot 1 \cdot \frac{1}{2} = 1 - 2 = -1\n]", "So $ a, b = \frac{1 \pm \sqrt{-1}}{2} = \frac{1 \pm i}{2} $", "Compute $ a^2 $ and $ b^2 $, but from earlier algebra, we already know $ a^2 + b^2 = 1 $ and $ a^2b^2 = \frac{1}{4} $, confirming consistency.", "Now compute $ a^4 + b^4 $ explicitly using $ a = \frac{1+i}{2} $, $ b = \frac{1-i}{2} $:", "[\na^2 = \left( \frac{1+i}{2} \right)^2 = \frac{1 + 2i - 1}{4} = \frac{2i}{4} = \frac{i}{2}\n]\n[\nb^2 = \left( \frac{1-i}{2} \right)^2 = \frac{1 - 2i - 1}{4} = \frac{-2i}{4} = -\frac{i}{2}\n]\n[\na^4 = \left( \frac{i}{2} \right)^2 = -\frac{1}{4}, \quad b^4 = \left( -\frac{i}{2} \right)^2 = -\frac{1}{4}\n]\n[\na^4 + b^4 = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2}\n]", "Wait — this contradicts the earlier result of $ \frac{1}{2} $. What's the issue?", "Ah, correction: $ a = \frac{1+i}{2} \Rightarrow a^2 = \frac{(1+i)^2}{4} = \frac{1 + 2i -1}{4} = \frac{2i}{4} = \frac{i}{2} $ — correct\nThen $ a^4 = (a^2)^2 = \left( \frac{i}{2} \right)^2 = -\frac{1}{4} $ — correct\nSimilarly $ b^2 = -\frac{i}{2},\ b^4 = -\frac{1}{4} $ — so sum is $ -1/2 $", "But earlier algebra gave $ a^4 + b^4 = \frac{1}{2} $ — contradiction?", "No — we made an error in the algebra if $ a^2 + b^2 = 1 $, but let's revisit:", "Wait — the identity is:", "[\na^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 = (1)^2 - 2 \cdot \left( \frac{1}{2} \right)^2 = 1 - 2 \cdot \frac{1}{4} = 1 - \frac{1}{2} = \frac{1}{2}\n]", "But direct computation gives $ -\frac{1}{2} $. Contradiction?", "No — in fact, $ ab = \frac{1}{2} $, but $ (ab)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4} $, so $ 2a^2b^2 = \frac{1}{2} $, and $ (a^2 + b^2)^2 = 1 $, so:", "[\na^4 + b^4 = 1 - \frac{1}{2} = \frac{1}{2}\n]", "But direct computation gave $ -\frac{1}{2} $ — so mistake must be in explicit computation.", "Recheck $ a = \frac{1+i}{2} $, $ a^2 = \frac{(1+i)^2}{4} = \frac{1 + 2i -1}{4} = \frac{2i}{4} = \frac{i}{2} $ — correct\nBut $ a^2 + b^2 = \frac{i}{2} + \left( \frac{1-i}{2} \right)^2 $", "Wait — $ b = \frac{1-i}{2} $, so $ b^2 = \left( \frac{1-i}{2} \right)^2 = \frac{1 - 2i + i^2}{4} = \frac{1 - 2i -1}{4} = \frac{-2i}{4} = -\frac{i}{2} $", "So $ a^2 + b^2 = \frac{i}{2} - \frac{i}{2} = 0 $, not 1 — error!", "Ah! The flaw: $ a $ and $ b $ are roots of $ x^2 - x + \frac{1}{2} = 0 $, but such roots satisfy $ a + b = 1 $, $ ab = \frac{1}{2} $, so $ a^2 + b^2 = (a+b)^2 - 2ab = 1 - 1 = 0 $, not 1.", "But the problem states $ a^2 + b^2 = 1 $ — contradiction.", "### Step 4: Reassess — Consistent Values?", "We have a contradiction unless both conditions are compatible.", "Given:\n- $ a + b = s $\n- $ ab = p = \frac{1}{2} $\n- $ a^2 + b^2 = s^2 - 2p = 1 $", "So:", "[\ns^2 - 2 \cdot \frac{1}{2} = 1 \Rightarrow s^2 - 1 = 1 \Rightarrow s^2 = 2 \Rightarrow s = \pm \sqrt{2}\n]", "Thus, $ a + b = \pm \sqrt{2} $, $ ab = \frac{1}{2} $", "Now compute $ a^2 + b^2 = s^2 - 2p = 2 - 1 = 1 $ — correct\nThen $ a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 = 1^2 - 2 \cdot \left( \frac{1}{2} \right)^2 = 1 - \frac{1}{2} = \frac{1}{2} $", "So $ a^4 + b^4 = \frac{1}{2} $ — real and positive", "Then $ |a^4 + b^4| = \left| \frac{1}{2} \right| = \frac{1}{2} $", "The earlier direct computation assumed $ a $ and $ b $ from $ x^2 - x + \frac{1}{2} = 0 $, but that gives $ a^2 + b^2 = 0 $, not 1 — so such roots do not satisfy $ a^2 + b^2 = 1 $. The correct $ a, b $ are roots of $ x^2 - \sqrt{2}x + \frac{1}{2} $? No — sum must be $ \pm\sqrt{2} $, product $ \frac{1}{2} $", "Let $ s = \sqrt{2} $, then roots:", "[\nx = \frac{\sqrt{2} \pm \sqrt{2 - 2}}{2} = \frac{\sqrt{2}}{2}\n]", "Wait — discriminant $ 2 - 2 = 0 $ — double root? No:", "$ s^2 = 2 $, $ 2p = 1 $, so discriminant $ s^2 - 4p = 2 - 2 = 0 $ — repeated roots? But $ ab = 1/2 $, $ a + b = \sqrt{2} $, so:", "[\nx = \frac{\sqrt{2}}{2}, \quad \frac{\sqrt{2}}{2"]