Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail From: Moebius Newsgroups: sci.math Subject: Re: Division of two complex numbers Date: Tue, 21 Jan 2025 11:37:20 +0100 Organization: A noiseless patient Spider Lines: 28 Message-ID: References: <3OCPm1fExu73aCqhbosWeWpssJM@jntp> <1f331uj8cjsge$.rox7zzvx5o63$.dlg@40tude.net> <206lh7vmixbg$.65ym8wmigyp0$.dlg@40tude.net> <9bds61cjwkps.4limb46c5wk9$.dlg@40tude.net> Reply-To: invalid@example.invalid MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Tue, 21 Jan 2025 11:37:21 +0100 (CET) Injection-Info: dont-email.me; posting-host="167941bcdca3a92f25c5377ee1395966"; logging-data="13122"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX18TVAHwTBKdrNWsdMAS3ptv" User-Agent: Mozilla Thunderbird Cancel-Lock: sha1:eVfmjjkxdqwyfzg2tp4o1MUYWd0= In-Reply-To: Content-Language: de-DE Bytes: 2189 Am 21.01.2025 um 11:27 schrieb Moebius: > Außerdem macht "(a, b)^2 = a^2 + 2ab + b^2" nach wie vor keinen Sinn. Let's write the complex number (a, b) in the usual form a + ib. Then (a + ib)^2 = a^2 + i2ab + (ib)^2 by the binomial formula (!). Hence (a + ib)^2 = a^2 + i2ab + (-1)b^2 = (a^2 - b^2) + i2ab . And hence (a, b)^2 = (a^2 - b^2) + i2ab (!) or rather (a, b)^2 = (a^2 - b^2, 2ab) would be correct. ( Knapp daneben ist auch vorbei. :-P ) .. .. ..