{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-complex-modulus-conjugate"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-complex-modulus-conjugate-01","title":{"zh-Hant":"複數模與共軛：證明 z·z̄=|z|² 與共軛保持運算","en":"Complex modulus and conjugate: proving z·z̄=|z|² and that conjugation preserves the operations"},"summary":{"zh-Hant":"對 z=a+bi、w=c+di（a、b、c、d 為實數）證明：（一）z·z̄=(a²+b²)+0i，即模平方的實數嵌入；（二）共軛保持加法：(z+w)‾=z̄+w̄；（三）共軛保持乘法：(zw)‾=z̄·w̄；（四）由此得 |zw|²=|z|²|w|²。再以 z=3−4i、w=1+2i 精確核對，並區分模平方與模。","en":"For z=a+bi and w=c+di (a, b, c, d real) prove: (1) z·z̄=(a²+b²)+0i, the real embedding of the modulus squared; (2) conjugation preserves addition: (z+w)‾=z̄+w̄; (3) conjugation preserves multiplication: (zw)‾=z̄·w̄; (4) hence |zw|²=|z|²|w|². Then check everything exactly for z=3−4i and w=1+2i, distinguishing the modulus squared from the modulus."},"mko_ids":["mko-complex-modulus-conjugate"]}],"title":"Concept derivation guides"}