Задача 37754 Дана функция w=z^2+1/2z найти значения...

neznaika01

2019-05-29 14:49:39

Дана функция w=z^2+1/2z найти значения функции при 1)z=–4–i 2) z=–3+4i

sova

2019-05-29 18:03:59

w=(z^2+1)/2z
w=(z^2)/2z + (1/2z)
w=(1/2)z + (1/2)*(1/z)

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

=-2 -(1/2)i - (1/2) ( 4-i)/17= -2 - (1/2)i - 2 +(1/34)i= = [b]-4 -(8/17)i[/b]

2)
w(-3+4i)=(1/2)*(-3+4i) +(1/2)*(1/(-3+4i))=

=(-3/2)+2i-(1/2)*(3+4i)/(9-16i^2)=

=(-3/2)+2i-(3/25)-(2i/25)=

[b](-81/50)+(48/50)i[/b]