Помогите. Комплексные числа.. Created by Awinerx. matematika-ru

Помогите. Комплексные числа...

Awinerx @Awinerx

August 2022 1 2 Report

Помогите. Комплексные числа.

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moboqe

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1)sqrt(7+2i)=*
r=sqrt(49+4)=sqrt(53)
tg(phi)=(2/7)⇒phi=arctg(2/7)
*=sqrt(53)*( cos(arctg(2/7))+i*sin(arctg(2/7)) )
z_k=sqrt(r)*( cos((phi+2pi*k)/2)+i*sin((phi+2pi*k)/2) )
k=0
z_0=(53)^(1/4)*( cos(arctg(2/7)/2)+i*sin(arctg(2/7)/2) )
k=1
z_1=(53)^(1/4)*( cos((arctg(2/7)2+2pi)/2) +i*sin((arctg(2/7)2+2pi)/2) )

2)sqrt(4-3i)=*
r=sqrt(16+9)=5
tg(phi)=(-3/4)⇒phi=2pi-arctg(3/4)
*=5( cos(2pi-arctg(3/4))+i*sin(2pi-arctg(3/4)) )
z_k=sqrt(r)*( cos((phi+2pi*k)/2)+i*sin((phi+2pi*k)/2) )
k=0
z_0=sqrt(5)*( cos((2pi-arctg(3/4))/2)+i*sin((2pi-arctg(3/4))/2) )
k=1
z_1=sqrt(5)*( cos((4pi-arctg(3/4))/2)+i*sin((4pi-arctg(3/4))/2)

3)sqrt(-7+24i)=*
r=sqrt(49+576)=25
tg(phi)=(-24/7)⇒phi=pi-arctg(24/7)
*=25*( cos(pi-arctg(24/7))+i*sin(pi-arctg(24/7)) )
z_k=sqrt(r)*( cos((phi+2pi*k)/2)+i*sin((phi+2pi*k)/2) )
k=0
z_0=5*( cos( (pi-arctg(24/7))/2 )+i*sin( (pi-arctg(24/7))/2 ) )
k=1
z_1=5*( cos( (3pi-arctg(24/7))/2 )+i*sin( (3pi-arctg(24/7))/2 ) )

4)z^2-6z-34=0
D=36+136=172
z1=(6+sqrt(172))/2=3+sqrt(43)
z2=3-sqrt(43)

5)z^2-4z+5=0
D=16-20= -4
sqrt(D)=sqrt(-4)=sqrt(4*i^2)=2i
z1=(4+2i)/2=2+i
z2=(4-2i)/2=2-i

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oharakterizujte ekonomicheskoe razvitie v rossii v konce 90 h gg

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Awinerx July 2022 | 0 Ответы

ocenite itogi vyborov 1993 g

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Answers & Comments

Verified answer

oharakterizujte ekonomicheskoe razvitie v rossii v konce 90 h gg

ocenite itogi vyborov 1993 g

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