Question

Compute the modulus and argument of:

2(cos(2π/3)+isin(2π/3))

Answer 1

for a number a+bi it's complex modulus is sqrt(a^2+b^2)
so we get
sqrt(4*cos(2pi/3)^2+4*sin(2pi/3)^2)
we know that sin^2(x)+cos^2(x)=1
so we get
sqrt(4)=2
basically you already have the modulus in your description,
cause when you convert to polar coordinates the first thing r(cosA+isinA), r - is the so called modulus. So don't repeat that demonstration. just say it's 2
and for the argument is just 2pi/3. So basically you already have those things if your number is like r(cosx+i*sinx). So you don't have to do anything at all.