DMCA
Contact
Privacy Policy

Home » robert frost - RTA

robert frost - RTA

📅 May 9, 2026 👤 scraface

May 09, 2026

Content is being prepared. Please check back later.

📚 You May Also Like These Articles

📌 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 $.

🔥 Popular Posts

hilton garden inn lake buena vista
grand pacific palisades hotel & resort carlsbad
hyatt regency schaumburg chicago schaumburg il
peppermill
hilton wilmington christiana
hilton los angeles north glendale
juno beach fl
grand hotel du palais royal
the westin san jose
hilton garden inn washington dc georgetown area

📝 Recent Posts

Restore Point Creator
Lexmark Cs310n Driver
Asus Rt Ax86u Pro Firmware
Dxo Pure Raw Download
Hdd Raw Copy Tool Torrent
Pinokio Ai Download
3utools Download for Windows 10
Portable Chrome
Vmware Vcenter Converter Standalone Download
Mozilla Xp Download

© 2026 RTA · Powered by WordPress

Disclaimer: This website operates exclusively as an informational archive. Content consists of publicly available international death records, condolence statements, and memorial information. The site does not publish real-time news coverage, live reporting, or entertainment content.