Вроде как то так............
1) z1 + z2 = 5 - 12i + 1 + i = 6 - 11i
2) z1 - z2 = 5 - 12i - 1 - i = 4 - 13i
3) z1*z2 = (5 - 12i)(1 + i) = 5 - 12i + 5i - 12i^2 = 5 - 7i + 12 = 17 - 7i
4)
5) Напишем z2 в тригонометрической форме:
z2 = 1 + i = √2*(1/√2 + i*1/√2) = √2(cos(pi/4) + i*sin(pi/4))
z2^3 = (√2)^3*(cos(3pi/4) + i*sin(3pi/4)) = √8*(-1/√2 + i*1/√2) = -2 + 2i
Другой способ, просто разложением скобок.
<span>z2 = (1+i)^3 = 1^3 + 3*1^2*i + 3*1*i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i
</span>
6) z1 = 5 - 12i <span>= 5 - 2*2*3*i = 5 - 2(2i*3) = (2i)^2 + 3^2 - 2(2i*3) =
= -4 + 9 - 2(2i*3) = (-3 + 2i)^2 = (3 - 2i)^2</span>
√z1 = -3 + 2i
√z1 = 3 - 2i
M^2 -22/13mn+121/169n^2=(m-11/13n)^2
400/441t^2 +40/21nt+n^2=(20/21t+n)^2