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Home » Stop Setting Off the Stove! Mexican Rice in a Rice Cooker Tastes Epic Every Time - Carbonext

Stop Setting Off the Stove! Mexican Rice in a Rice Cooker Tastes Epic Every Time - Carbonext

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Divide by 18:

📌 $ rac{(x+3)^2}{4.5} - rac{(y-1)^2}{2} = 1 $.

📌 Center at $ (-3, 1) $. Final answer: oxed{(-3,\ 1)}

📌 Question: Let $ z $ and $ w $ be complex numbers such that $ z + w = 2 + 4i $ and $ z \cdot w = 13 - 2i $. Find $ |z|^2 + |w|^2 $.

📌 Solution: Use $ |z|^2 + |w|^2 = |z + w|^2 - 2 ext{Re}(z \overline{w}) $. Compute $ |z + w|^2 = |2 + 4i|^2 = 4 + 16 = 20 $. Let $ z \overline{w} = a + bi $, then $ ext{Re}(z \overline{w}) = a $. From $ z + w = 2 + 4i $ and $ zw = 13 - 2i $, note $ |z|^2 + |w|^2 = (z + w)(\overline{z} + \overline{w}) - 2 ext{Re}(z \overline{w}) = |2 + 4i|^2 - 2a = 20 - 2a $. Also, $ zw + \overline{zw} = 2 ext{Re}(zw) = 26 $, but this path is complex. Alternatively, solve for $ |z|^2 + |w|^2 = |z + w|^2 - 2 ext{Re}(z \overline{w}) $. However, using $ |z|^2 + |w|^2 = (z + w)(\overline{z} + \overline{w}) - 2 ext{Re}(z \overline{w}) = |z + w|^2 - 2 ext{Re}(z \overline{w}) $. Since $ z \overline{w} + \overline{z} w = 2 ext{Re}(z \overline{w}) $, and $ (z + w)(\overline{z} + \overline{w}) = |z|^2 + |w|^2 + z \overline{w} + \overline{z} w = |z|^2 + |w|^2 + 2 ext{Re}(z \overline{w}) $, let $ S = |z|^2 + |w|^2 $, then $ 20 = S + 2 ext{Re}(z \overline{w}) $. From $ zw = 13 - 2i $, take modulus squared: $ |zw|^2 = 169 + 4 = 173 = |z|^2 |w|^2 $. Let $ |z|^2 = A $, $ |w|^2 = B $, then $ A + B = S $, $ AB = 173 $. Also, $ S = 20 - 2 ext{Re}(z \overline{w}) $. This system is complex; instead, assume $ z $ and $ w $ are roots of $ x^2 - (2 + 4i)x + (13 - 2i) = 0 $. Compute discriminant $ D = (2 + 4i)^2 - 4(13 - 2i) = 4 + 16i - 16 - 52 + 8i = -64 + 24i $. This is messy. Alternatively, use $ |z|^2 + |w|^2 = |z + w|^2 + |z - w|^2 - 2|z \overline{w}| $, but no. Correct approach: $ |z|^2 + |w|^2 = (z + w)(\overline{z} + \overline{w}) - 2 ext{Re}(z \overline{w}) = 20 - 2 ext{Re}(z \overline{w}) $. From $ z + w = 2 + 4i $, $ zw = 13 - 2i $, compute $ z \overline{w} + \overline{z} w = 2 ext{Re}(z \overline{w}) $. But $ (z + w)(\overline{z} + \overline{w}) = 20 = |z|^2 + |w|^2 + z \overline{w} + \overline{z} w = S + 2 ext{Re}(z \overline{w}) $. Let $ S = |z|^2 + |w|^2 $, $ T = ext{Re}(z \overline{w}) $. Then $ S + 2T = 20 $. Also, $ |z \overline{w}| = |z||w| $. From $ |z||w| = \sqrt{173} $, but $ T = ext{Re}(z \overline{w}) $. However, without more info, this is incomplete. Re-evaluate: Use $ |z|^2 + |w|^2 = |z + w|^2 - 2 ext{Re}(z \overline{w}) $, and $ ext{Re}(z \overline{w}) = ext{Re}(rac{zw}{w \overline{w}} \cdot \overline{w}^2) $, too complex. Instead, assume $ z $ and $ w $ are conjugates, but $ z + w = 2 + 4i $ implies $ z = a + bi $, $ w = a - bi $, then $ 2a = 2 \Rightarrow a = 1 $, $ 2b = 4i \Rightarrow b = 2 $, but $ zw = a^2 + b^2 = 1 + 4 = 5

📌 eq 13 - 2i $. So not conjugates. Correct method: Let $ z = x + yi $, $ w = u + vi $. Then:

📌 $ x + u = 2 $, $ y + v = 4 $,

📌 $ (x + yi)(u + vi) = xu - yv + (xv + yu)i = 13 - 2i $.

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\begin{bmatrix}
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