Let z=x+it,then |z|=√x^2+y^2=1; compare real terms i.e;x+√x^2+y^2=1; and compare imaginary part with imaginary part i.e;y=7 put in 1 and get value of x as |z| is required;find √x^2+y^2
aravind m t
8 Years ago
real partpart of is 24..we will get by solving =625....|z|^2 √a^2+7^2 +a=1;hence
aravind m t
8 Years ago
real partpart of is 24..we will get by solving |z|^2 √a^2+7^2 +a=1....;hence|z|^2 =625