Question

Dział Liczby Zespolone.

Zadanie . Podane liczby zespolone zapisz w postaci trygonometrycznej.

1) Z=

2) Z=sinα - icosα

3) Z=(1+cosα)+isinα

4) Z=1-ictgα

5) Z=(

6) Z=(

Uwaga. Przy zadaniu 5 i 6 należy skorzystać ze wzorów de Moivre'a !

Answer 1

1)

z = p(3) + i

I Z I =p [ [ p(3)]^2 + 1^2] = p [ 3 + 1] = p(4) = 2

cos fi = a/ I z I = p(3)/2

sin fi = b/z = 1/2

fi = 30 st = pi/6

zatem

z = 2*( cos (pi/6) + i sin (pi/6))

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5) z = ( p(3) - i )^32

z1 = p(3) - i

I z1 I = p [ ( p(3)^2 + 1^2] = p [3 + 1] = 2

cos fi1 =p(3)/2

sin fi 1 = -1/2

fi 1 = 2 pi - pi/6 = (11/6) pi

z = (z1)^32 = 2^32 *( cos 32*(11/6) pi + i sin 32*(11/6) pi ) =

= 2^32 *( cos ( 58 + 2/3) pi + i sin( 58 + 2/3)pi ) =

= 2^32 *( cos(2/3) pi + i sin (2/3) pi ) =

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= 2^32* ( - 1/2 + i *p(3)/2 )

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6)

z = ( - 2 + 2 i )^8

z1 = -2 + 2 i

I z1 I = p [ (-2)^2 +2^2] = p(8) = p(4*2) = 2 p(2)

cos fi  = -2/ 2p(2) = - 1/p(2) = -p(2)/2

sin fi = 2/2 p(2) = 1/p(2) = p(2)/2

fi = pi/2 + pi/4 = (3/4) pi

z1 = 2 p(2) *( cos (3/4) pi + i sin (3/4) pi )

z = ( z1) ^8 = [ 2 p(2)]^8 *( cos 8*(3/4) pi + i sin 8*(3/4) pi) =

= 2^12 *( cos 6 pi + i sin 6 pi ) = 2^12 *( cos 0 + i sin 0 ) =

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= 2^12 *( 1 + 9*0 ) = 2^12 = 2^10 * 2^2 = 1024 *4 = 4096

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