http://kalvado.livejournal.com/ ([identity profile] kalvado.livejournal.com) wrote in [personal profile] green_fr 2012-06-04 02:37 pm (UTC)

Ну а теперь представь тоже самое в другом базисе, х'=sqrt(1/2)(x+iy), y'=sqrt(1/2)(x-iy)
Этособственно переход откруга к линии
(37 comments)

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