复数

Viola Maria Grazia
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引用次数: 0

摘要

在本文中,我们从另一个角度来研究这些复合物。定义复杂的字段定义红外/ (x + 1)我们认为,像所有的知道,我≔倍根号{−1},我们看来复合物,如红外(i)现在,我们知道,飞机上的复数是rapresented但只有图的矢量空间复合体换句话说他们可以通过以下方式rapresented我们取一个点在现实与极坐标平面pcosO i + psinO j, p是正实数和O(0, 2π(我们知道如果moltiplicate复杂' a '和我,A将旋转一个角度π/ 2逆时针所以我们点在真实飞机变得复杂点(pcosO-ipsinO)我我们注意到复杂的直线上的配合物都是y = 0(- 0)我们也看到,(我,0)是解决系统y = 0 & & y = x + 1(2, 0),(2, 0)是系统的解决方案y = 0 & & y = x + 4 (- i + 1,0)、(i + 1, 0)是系统的解决方案y = 0 & & y = (x - 1) + 1等等这样一些复合物rapresented这样的…-1+i -1 0 -i -2i -3i…3i 2i i 0 1-i 1-i 1-i 1+ i1 1 2-i Pi-i…i和i等的位置取决于旋转和实平面的表示,在这种情况下,作者保持角度的相同方向,即逆时针方向。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Complex Numbers
In this article we study the complexes in an other point of view. Definition The Complex field is defined like IR/(x+1) we assume, like all know, i ≔ sqrt{−1} And we view complexes like IR(i) Now, we know that the complex numbers are rapresented on the plane but it is only the graph of the vectorial space of the complexes in other word they can be rapresented in the following way We take a point in the real plane with polar coordinate pcosO i + psinO j , where p is positive real number and O is in [0,2pi[ And know that if we moltiplicate a complex ‘A’ with i, ‘A’ will rotate by an angle of pi/2 anticlockwise So our point in real plane becomes the complex point (pcosO-ipsinO)i We note that the complexes are all on the complex straight line y=0 We saw also that (-i,0),(i,0) is the solution of the system y=0 && y=x+1 (-2i,0),(2i,0) is the solution of the system y=0 && y = x+4 (-i+1,0),(i+1,0)is the solution of the system y=0 && y=(x-1)+1 etc etc So the some complexes are rapresented on the line like this ... -1+i -1 0 -i -2i -3i ... 3i 2i i 0 1 1-i 1-2i 1+i 1 2 2-i Pi Pi-i ... The position of -i ad i etc etc depends by the rotation and the rapresentation of real plane in this the author keep the same direction of the angles i.e. anticlockwise.
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