Question

Może mi to ktoś wytłumaczyć:

To są liczby zespolone. Proszę o obliczenie i wytłumaczenie jak to obliczyliście.

Answer 1

1)

[ p(2) i + p(2)]^5 = [ p(2)*( i + 1)]^5 = [ p(2)]^5 * [ i + 1)^5 = 4 p(2)*( - 4 i - 4) =

=  - 16 p(2) i - 16 p(2)

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bo

( i + 1)^5 = ( i +1)^2 * ( i + 1)^2 *( i + 1)  = ( i^2 + 2i + 1)*( i^2 +2i + 1)*(i + 1) =

= 2i *2i *( i + 1) = 4 i^2 *( i + 1) = - 4*( i + 1) = - 4 i - 4

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II sposób:

p(2)i + p(2) = p(2) *[ i + 1]

i + 1 = p(2)([ cos 45 st + i sin 45 st]

czyli

[ i + 1]^5 = [ p(2)]^5 * [ cos 5*45 st + i sin 5*45 st ] =

= 4 p(2) *[ cos (180 + 45) st + i sin ( 180 + 45)st ] =

= 4 p(2) *[ - cos 45 st - i sin 45 st ] =

= 4 p(2) * [ - p(2)/2 - i  * p(2)/2 ] =

= - 4 - 4 i

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czyli

[ p(2) i + p(2)]^5 = [ p(2)]^5 * [ i + 1]^5 = 4 p(2) *[ - 4 - 4 i ] =

= - 16 p(2) i   - 16 p(2)

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2)

[ 1 - i ]^8 = [ ( 1 - i)^2 ]^4 = [ 1 - 2i + i^2]^4 = [ 1 -2i - 1]^4 = [ (- 2i )^4 =

= [ ( -2 i)^2]^2 = [ 4 i^2 ]^2 = ( - 4]^2 = 16

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II sposób:

1 - i = p(2) *[ cos ( -45 st) + i sin ( - 45 st)] =

= p(2)*[ cos 45 st - i sin 45 st]

zatem

[ 1 - i]^8 = [ p(2)]^8 *[ cos 45 st - i sin 45 st]^8 =

= 16 *[ cos 8*45st - i sin 8*45 st] = 16*[ cos 360 st - i sin 360 st ] =

= 16*[ cos 0 st - i sin 0 st ] = 16*[ 1 - i * 0 ] = 16

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3)

( 2 + i)^4 = [ ( 2 + i)^2 ]^2 = [ 4 + 4 i + i^2 ]^2 =

= [ 4 + 4 i - 1]^2 = [ 3 + 4 i]^2 = 9 + 24 i + 16 i^2 =

= 9 +24 i - 16 = - 7 + 24 i

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4)

x^2 +2 x + 10 = 0

delta = 2^2 - 4*1*10 = 4 - 40 = - 36 = 36*i^2

p(delty) = 6 i

x1 = [ - 2 - 6 i ]/2 = - 1 - 3 i

x2 = [ - 2 + 6 i]/2 =  - 1 + 3 i

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5)

x^4 + x^3 + x^2 + x + 1 = 0

Jest to równanie zwrotne 4 stopnia.

0 nie jest pierwiastkiem tego równania, zatem można podzielić  obustronnie

przez  x^2

Mamy

x^2 + x + 1 + 1/ x + 1/x^2 = 0

[ x + 1/x]^2 - 2 + 1 + (x + 1/x) = 0

Stosujemy w tym przypadku podstawienie

z = x + 1/x

Mamy

z^2 + z - 1 = 0

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delta = 1^1 - 4*1*(-1) = 1 + 4 = 5

p(delty) = p(5)

z1 = [ -1 - p(5)]/2

z2 = [ - 1 + p(5)]/2

czyli  mamy

x + 1/x = [ - 1 - p(5)]/2            lub      x + 1/x = [ - 1 + p(5)]/2

Mnozymy obustronnie przez  2

2 x + 2/x + 1 + p(5) = 0             lub     2 x + 2/x + 1 - p(5) = 0

Mnożymy przez x

2 x^2 + [ 1 + p(5)] x + 2 = 0       lub     2 x^2 + ( 1 - p(5)] x +2 = 0

1)

2 x^2 + (1 + p(5)] x + 2 = 0

delta1 = [ 1 + p(5)]^2 - 4*2*2 = 1 + 2 p(5) + 5 - 16 = 2 p(5) - 10 < 0

delta1 = (-1)*( 10 - 2 p(5)] = ( 10 - 2 p(5)] *i ^2

p(delty1) = p [ 10 - 2 p(5)] * i

x1 = [ -1 - p(5) - p( 10 - 2 p(5))* i ]/4

x2 = [ -1 - p(5) + p( 10 - 2 p(5))* i ]/4

2)

2 x^2 = [ 1 - p(5)] x + 2 = 0

delta2 = [ 1 - p(5)]^2 - 4*2*2 = 1 - 2 p(5) + 5 - 16 =  - 2 p(5) - 10  < 0

delta2 =  - 1*[ 10 + 2 p(5)] = [ 10 + 2 p(5)] *i^2

p ( delta2) = p [ 10 + 2 p(5)] *i

x3 = [ -1 + p(5) - p ( 10 + 2 p(5))* i ] / 4

x4 = [ -1 + p(5) + p( 10 + 2 p(5))* i ]/ 4

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