Задача 35503 ...

sova

2019-04-08 09:00:28

z_(1)=-4i; z_(2)=3-9i;

1)
Rez_(1)=x_(1)=0; Imz_(1)=y_(1)=-4
Rez_(2)=x_(2)=3; Imz_(2)=y_(2)=-9

2)
Cм.рис.

3)
|z_(1)|=4
argz_(1)=phi

sin(phi)=y_(1)/|z_(1)|=-4/4=-1
cos(phi)=x_(1)/|z_(1))=0/4=0
phi=-π/2

z_(1)=4*(cos(-π/2)+i*sin(-π/2)) - в тригоном. форме

z_(1)=4e^(-iπ/2)- в показ форме


|z_(2)|=sqrt(3^2+(-9)^2)=sqrt(90)=3sqrt(10)

argz_(2)=ψ

sinψ=y/|z_(2)|=-9/3sqrt(10)=-3sqrt(10)/10
cosψ=x/|z_(2))=3/3sqrt(10)=sqrt(10)/10
tgψ=-3
ψ=arctg (-3)

z_(2)=3sqrt(10)*(arctg (-3)+i*arctg (-3))- в тригоном. форме

z_(2)=3sqrt(10)*e^(-iarctg3)в показ форме
4)

z=2*z_(1)-10z_(2)=2*(-4i)-10*(3-9i)=-8i-30+90i= [b]-30+81i[/b]


5)
z=z_(1)*z_(2)=(-4i)*(3-9i)=-12i+36i^2=[b]-36-12i[/b]

6)

z=z_(1)/z_(2)=(-4i)/(3-9i)= (4i)*(3+9i)/(3-9i)*(3+9i)=(12i+36i^2)/(3^2-(9i)^2)=(12i-36)/90=(12/90)i-(36/90)=[b](-4/10)+(4/30)i[/b]

7)
Применяем формулу Муавра

z^(7)_(1)=4^(7)*(cos7*(-π/2)+i*sin7*(-π/2))=

=4^(7)*(cos(-7π/2)+i*sin(-7π/2))=4^(7)*(cos((-3π/2)+i*sin(-3π/2))=

=[b]4^(7)*i[/b]

8)

z^(1/2)_(1)=(4)^(1/2)*cos((-π/2)/2)+(2πk/2))+i*sin(((-π/2)/2)+(2πk/2))

k=0,1

при k=0
(z^(1/2)_(1))_(0)=2(cos(-π/4)+i*sin(-π/4))=2*[b]([/b](sqrt(2)/2) + i*(-sqrt(2)/2) [b])[/b]= [b]sqrt(2)-i*sqrt(2)[/b]

при k=1
(z^(1/4)_(1))_(1)=2*(cos((-π/4)+π)+i*sin((-π/4)+π))=

=2* [b]([/b](-sqrt(2)/2) + i*(sqrt(2)/2) [b])[/b]= [b]-sqrt(2)+i*sqrt(2)[/b]