Question

Find the modulus and argument of the complex number in the picture

Answer 1

Since
`e^(ix)=\cos x+i\sin x`, you have


`((\sin x-i\cos x)^3)/((\cos x+i\sin x)^5)=((-ie^(ix))^3)/((e^(ix))^5)=(ie^(3ix))/(e^(5ix))=ie^(-2ix)`

So,
`|ie^(-2ix)|=|e^(-2ix)|=1` and the argument is
`-2x`.