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| complex number https://forumdematematica.org/viewtopic.php?f=16&t=286 |
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| Autor: | kinu [ 04 abr 2012, 16:15 ] |
| Título da Pergunta: | complex number |
If \(a\;,b\;,c\) are different Complex no. and \(\mid a \mid = \mid b\mid =\mid c \mid>0\) If \(a+bc\;,b+ca\;,c+ab\in \mathbb{R}\). Then \(abc =\) |
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| Autor: | Fraol [ 01 jul 2013, 02:34 ] |
| Título da Pergunta: | Re: complex number |
After a while, we are here with this good old problem of complex numbers. A little challenging to say the truth. So, let's go: Let \(\left | a \right | = m > 0\), then \(m^2 = a \cdot \bar a\), by complex modulus properties. And by problem data \(m^2 = a \cdot \bar a = b \cdot \bar b = c \cdot \bar c\). If \(a + bc \in R\) then \(\bar a + \bar b \bar c \in R\) too, by complex numbers symetry. So, \(a + bc = \frac{m^2}{a \bar a} \cdot \bar a + \bar b \bar c \cdot \frac{bc}{bc}\) ( multiplying by 1 ). This implies that: \(a + bc = \frac{m^2}{a} + \frac{m^4}{bc} \Leftrightarrow a + bc = \frac{bcm^2 + am^4}{abc} = \frac{m^2(bc+am^2)}{abc}\) \(\Leftrightarrow \frac{a + bc}{bc+am^2} = \frac{m^2}{abc}\) and similarly: \(\frac{m^2}{abc} = \frac{b + ac}{ac+bm^2} = \frac{c + ab}{ab+cm^2}\). From the rules of proportionality, we have: \(\frac{m^2}{abc} = \frac{a + bc - (b+ac)}{bc+am^2 - (ac+bm^2}\). Which leads us to: \(\frac{m^2}{abc} = \frac{a + bc - b - ac)}{bc+am^2 - ac - bm^2}\). Computing a bit: \(\frac{m^2}{abc} = \frac{a(1-c)-b(1-c)}{a(m^2-c) - b(m^2 - c} = \frac{(a-b)(1-c)}{(a-b)(m^2-c)}\). So, we get \(\frac{m^2}{abc} = \frac{(1-c)}{(m^2-c)}\), again by proportionality: \(\frac{m^2}{abc} = \frac{1}{m^2}\) and ... Finally, we have \(abc = m^4\). |
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