Niech u= -i,
Oblicz a) uv, u/v, ,
Bardzo proszę o pomoc. Potrzebuję tego na dziś wieczorem
Z góry dziękuję
u = - i
v = ( p(3) + i )/2
zatem
uv = - i *[ (1/2) p(3) + (1/2) i ] = - [p(3)/2] i + 1/2 = 1/2 - [ p(3)/2] i
============================================================
bo i^2 = - 1 oraz - i^2 = 1
u/v = - i / [ ( p(3) + i )/2 ] = - 2 i / [ p(3) + i ] =
= ( - 2i )*( p(3) - i)] / [ ( p(3) + i )*( p(3) - i ) ] =
= [ - 2 p(3) i - 2 ]/ [ 3 - i^2 ] = [ -2 - 2 p(3) i ]/ [ 3 + 1} =
= -1/2 - (1/2) p(3) i
=====================
v = p(3)/2 +(1/2) I
a = p(3)/2
b =1/2
I v I = p [ a^2 + b^2] = p [ ( p(3)/2)^2 = )1/2)^2 ] = p[ 3/4 + 1/4] =p [ 4/4] =
= p(1) = 1
I v I = 1
cos fi = a/ I v I = p(3)/2
sin fi = b / I v I = 1/2
zatem fi = 30 stopni
czyli
v = cos 30 st + i sin 30 st - postac trygonometryczna liczby zespolonej v
Z wzoru Moivre'a mamy
v^9 = [ cos 30 st + i sin 30 st ] ^9 =
= cos 9*30 st + i sin 9*30 st = cos 270 st + i sin 270 st = 0 - 1 i = - i
v^9 = - i
==============
u = -i = 0 - i
a = 0; b = - 1
I u I = p[ 0^2 + (-1)^2 ] = p(1) = 1
cos fi = a /I u I = 0 /1 = 0
sin fi = b/ I u I = -1/1 = - 1
zatem fi = 270 st
u = cos 270 st + i sin 270 st
p3st I u I = p3st ( 1) = 1
Pierwiastki 3 stopnia z liczby zespolonej u = - i to liczby zespolone
u1, u2, u3 :
u1 = cos [ 270 st/3] + i sin [ 270 st/3]
u1 = cos 90st + i sin 90 st = 0 + i *1 = i
u2 = cos [(270 st + 360 st)/3] + i sin [ ( 270 st + 360 st)/3]
u2 = cos 210 st + i sin 210 st = cos ( 180 + 30)st + i sin ( 180 + 30) st
u2 = - cos 30 st - i sin 30 st = - p(3)/2 - (1/2) i
u3 = cos[ (270 st + 720 st)/3] + i sin [ (270 st + 720 st)/3]
u3 = cos 330 st + i sin 330 st = cos ( 270 + 60) st + i sin(270 + 60)st
u3 = sin 60 st - i cos 60 st = p(3)/2 - (1/2) i
Odp. Pierwiastki 3 stopnia z liczby zespolonej u = - i
to u1 = i , u2 = - p(3)/2 - (1/2) I, u3 = p(3)/2 - (1/2) I
=====================================================
p(3) - pierwiastek drugiego stopnia z 3
p3st I u I - pierwiastej 3 stopnia z modułu liczby zespolonej u
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u = - i
v = ( p(3) + i )/2
zatem
uv = - i *[ (1/2) p(3) + (1/2) i ] = - [p(3)/2] i + 1/2 = 1/2 - [ p(3)/2] i
============================================================
bo i^2 = - 1 oraz - i^2 = 1
u/v = - i / [ ( p(3) + i )/2 ] = - 2 i / [ p(3) + i ] =
= ( - 2i )*( p(3) - i)] / [ ( p(3) + i )*( p(3) - i ) ] =
= [ - 2 p(3) i - 2 ]/ [ 3 - i^2 ] = [ -2 - 2 p(3) i ]/ [ 3 + 1} =
= -1/2 - (1/2) p(3) i
=====================
v = p(3)/2 +(1/2) I
a = p(3)/2
b =1/2
I v I = p [ a^2 + b^2] = p [ ( p(3)/2)^2 = )1/2)^2 ] = p[ 3/4 + 1/4] =p [ 4/4] =
= p(1) = 1
I v I = 1
cos fi = a/ I v I = p(3)/2
sin fi = b / I v I = 1/2
zatem fi = 30 stopni
czyli
v = cos 30 st + i sin 30 st - postac trygonometryczna liczby zespolonej v
Z wzoru Moivre'a mamy
v^9 = [ cos 30 st + i sin 30 st ] ^9 =
= cos 9*30 st + i sin 9*30 st = cos 270 st + i sin 270 st = 0 - 1 i = - i
v^9 = - i
==============
u = -i = 0 - i
a = 0; b = - 1
I u I = p[ 0^2 + (-1)^2 ] = p(1) = 1
cos fi = a /I u I = 0 /1 = 0
sin fi = b/ I u I = -1/1 = - 1
zatem fi = 270 st
u = cos 270 st + i sin 270 st
p3st I u I = p3st ( 1) = 1
Pierwiastki 3 stopnia z liczby zespolonej u = - i to liczby zespolone
u1, u2, u3 :
u1 = cos [ 270 st/3] + i sin [ 270 st/3]
u1 = cos 90st + i sin 90 st = 0 + i *1 = i
u2 = cos [(270 st + 360 st)/3] + i sin [ ( 270 st + 360 st)/3]
u2 = cos 210 st + i sin 210 st = cos ( 180 + 30)st + i sin ( 180 + 30) st
u2 = - cos 30 st - i sin 30 st = - p(3)/2 - (1/2) i
u3 = cos[ (270 st + 720 st)/3] + i sin [ (270 st + 720 st)/3]
u3 = cos 330 st + i sin 330 st = cos ( 270 + 60) st + i sin(270 + 60)st
u3 = sin 60 st - i cos 60 st = p(3)/2 - (1/2) i
Odp. Pierwiastki 3 stopnia z liczby zespolonej u = - i
to u1 = i , u2 = - p(3)/2 - (1/2) I, u3 = p(3)/2 - (1/2) I
=====================================================
p(3) - pierwiastek drugiego stopnia z 3
p3st I u I - pierwiastej 3 stopnia z modułu liczby zespolonej u