Question

Sa se scrie sub forma trigonometrică redusă numerele complexe: a)-1 ; b) -1+i ; c) 1+iradical3

Answer 1

a)
z=-1
a=0
b=-1
r=√(a²+b²)=√1=1
t=3π/2
=> z=cos3π/2 + sin3π/2•i

b)
z=-1+i
a=-1
b=1
r=√(a²+b²)=√(1+1)=√2
t € (π/2; π)
sin t= b/r= 1/√2= √2/2
cos t= a/r= -1/√2= -√2/2
=> t= π-π/4= 3π/4
=> z=√2(cos 3π/4 + sin 3π/4•i)

c)
z=1+i√3
a=1
b=√3
r=√(a²+b²)=√(1+3)=√4
t € (0;π/2)
sin t= 1/2
cos t=√3/2
=> t= π/3
=> z=2(cosπ/3 + sinπ/3•i)