Task/25407502
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см приложение
* * * i =√(-1) ⇒ i² = -1 * * *
Дано :
z₁=1 -√3 i ;
z₂ = -3 -4 i .
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1)
Представить в тригонометрической форме
z =a +bi ; z =ρ(cosφ +isinφ) , где ρ =√(a²+b²) ; φ =arctq(b/a)
z₁=1 -√3 i =2(сos(-π/3)+isin( -π/3) ) или 2(cos(5π/3) +i sin(5π/3) ) .
* * *ρ =√(1² +(-√3)² ) =√(1 +3 ) =√4 =2 ; tqφ = -√3/1 = - √3 ⇒ φ = -π/3 * * *
z₂ = -3 -4 i =5(cos(π +arctq4/3) +sin(π +arctq4/3) i)
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2)
z₁*z₂ =(1 -√3 i )*(-3 -4 i) = -3 - 4i +3√3 i +4√3 * (i²) = - (3+4√3) +(3√3 - 4) i .
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3)
z₁/ z₂ = (1 -√3 i )/ (-3 -4 i) = (1 -√3 i)(-3 +4 i) / (-3 -4 i) (-3 +4 i) =
((4√3 -3) +(4+3√3) i ) / (9 -16 i²) = ((4√3 -3) +(4+3√3) i ) / 25 =
(4√3 -3) /25+(4+3√3)/ 25 i .
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4)
* * *z^k =( ρ(cosφ +isinφ) )^k =ρ^k (coskφ +isinkφ)_ формула Муавра * * *
(z₁)⁴ = ( 2(cos(5π/3) +sin(5π/3) i) )⁴ =2⁴(cos(20π/3) + sin(20π/3) i ) =
2⁴(cos(2π/3) + sin(2π/3) i ) = 2⁴(-cos(π/3) + sin(π/3) i ) =
16( -1/2 +(√3)/2 i ) = - 8 +8√3 i.
* * *(z₁)⁴ =(1 -√3 i)⁴ =1⁴ - 4*√3 i +6(√3 i )² - 4*(√3 i)³ + (√3 i)⁴ =
1 +6(√3 i )² + (√3 i)⁴ - 4√3 i - 4(√3 i)³ =1-18 +9 - 4√3 i +12√3 i = -8+8 √3 * * *
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5)
√z₂ = √( 5(cos(π +arctq4/3) + i sin(π +arctq4/3) ) ) =
√(5 ( cos(π +arctq4/3 +2πk) + i sin(π +arctq4/3 +2πk) ) ) =
(√5)*(cos(π +arctq4/3 +2πk) /2 + i sin(π +arctq4/3 +2πk) /2 )
если k =0, то
√5 *( cos(π +arctq4/3) /2 + i sin (π +arctq4/3) /2) ) =
√5 *( cos(π/2 +(1/2)arctq4/3) + sin(π/2 +(1/2)arctq4/3) i ) =
если k =1, то
√5 *( cos(3π +arctq4/3) /2) +i sin ( (3π +arctq4/3) /2) ) ) =
√5 *( cos(3π/2 +(1/2)arctq4/3) + sin ( 3π/2 +(1/2)arctq4/3) i )