Question

найти модуль и аргумент z=5+i/2+3i

Answer 1

Ответ:

(5+i)/(2+3i) = (5+i)*(2 — 3i)/((2 + 3i)(2 — 3i)) = (10 + 2i — 15i — 3i^2)/(4-9i^2) = (13 — 13i)/13 = 1 — i

|z| = sqrt(1^2 + (-1)^2) = sqrt(1 + 1) = sqrt(2)

arg z = arctg((-1)/1)= -п/4