Image of square under Mobius Transformation
Am i right that there is no Mobius transformation $$ h(z) $$ that sends a square to rectangle "A" with vertices in $$ 0, 2, i, i+2 $$because if we inscribe a cricle into square then it has 4 points $$ z_1, z_2, z_3, z_4 $$on a boundry on square therefore
$$ h(z_1), h(z_2), h(z_3), h(z_4)$$ also belongs to boundry of image of square under h and image of square is generlised circle whih is contradiction because we can't construct a cricle through theses points?
complex-analysis complex-numbers mobius-transformation
asked Nov 28 '18 at 9:45
Pablo
996
add a comment |
Am i right that there is no Mobius transformation $$ h(z) $$ that sends a square to rectangle "A" with vertices in $$ 0, 2, i, i+2 $$because if we inscribe a cricle into square then it has 4 points $$ z_1, z_2, z_3, z_4 $$on a boundry on square therefore
$$ h(z_1), h(z_2), h(z_3), h(z_4)$$ also belongs to boundry of image of square under h and image of square is generlised circle whih is contradiction because we can't construct a cricle through theses points?
complex-analysis complex-numbers mobius-transformation
asked Nov 28 '18 at 9:45
Pablo
996
Right. Could perhaps be better explained - if you inscribe a circle in that square the image of the circle would be a circle inscribed in A, which is impossible.
– David C. Ullrich
Nov 28 '18 at 15:16
add a comment |
Am i right that there is no Mobius transformation $$ h(z) $$ that sends a square to rectangle "A" with vertices in $$ 0, 2, i, i+2 $$because if we inscribe a cricle into square then it has 4 points $$ z_1, z_2, z_3, z_4 $$on a boundry on square therefore
$$ h(z_1), h(z_2), h(z_3), h(z_4)$$ also belongs to boundry of image of square under h and image of square is generlised circle whih is contradiction because we can't construct a cricle through theses points?
complex-analysis complex-numbers mobius-transformation
asked Nov 28 '18 at 9:45
Pablo
996
Am i right that there is no Mobius transformation $$ h(z) $$ that sends a square to rectangle "A" with vertices in $$ 0, 2, i, i+2 $$because if we inscribe a cricle into square then it has 4 points $$ z_1, z_2, z_3, z_4 $$on a boundry on square therefore
$$ h(z_1), h(z_2), h(z_3), h(z_4)$$ also belongs to boundry of image of square under h and image of square is generlised circle whih is contradiction because we can't construct a cricle through theses points?
complex-analysis complex-numbers mobius-transformation
complex-analysis complex-numbers mobius-transformation
asked Nov 28 '18 at 9:45
Pablo
996
asked Nov 28 '18 at 9:45
Pablo
996
asked Nov 28 '18 at 9:45
Pablo
996
asked Nov 28 '18 at 9:45
Pablo
996
asked Nov 28 '18 at 9:45
Pablo
996
996
Right. Could perhaps be better explained - if you inscribe a circle in that square the image of the circle would be a circle inscribed in A, which is impossible.
– David C. Ullrich
Nov 28 '18 at 15:16
add a comment |
Right. Could perhaps be better explained - if you inscribe a circle in that square the image of the circle would be a circle inscribed in A, which is impossible.
– David C. Ullrich
Nov 28 '18 at 15:16
Right. Could perhaps be better explained - if you inscribe a circle in that square the image of the circle would be a circle inscribed in A, which is impossible.
– David C. Ullrich
Nov 28 '18 at 15:16
Right. Could perhaps be better explained - if you inscribe a circle in that square the image of the circle would be a circle inscribed in A, which is impossible.
– David C. Ullrich
Nov 28 '18 at 15:16
add a comment |
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Right. Could perhaps be better explained - if you inscribe a circle in that square the image of the circle would be a circle inscribed in A, which is impossible.
– David C. Ullrich
Nov 28 '18 at 15:16