express in the form of Ae^(iθ) Im(50 cos(2π100t) for t ∈ [0.2π

Question

i hve no idea about this problem if you could help me taht would be great,,appretiate it thanks

Answer 1

Ae^niθ=A(cos(nθ)+isin(nθ))=A(cosθ+isinθ)ⁿ. I assume that "Im" refers to the imaginary part.

Im(Ae^niθ)=Asin(nθ)=A×Im[(cosθ+isinθ)ⁿ].

Im(cosθ+isinθ)ⁿ=Im(cosⁿθ+nicos^n-1θsinθ+...+ⁿC_rcos^n-rθ(isinθ)^r+...+(isinθ)ⁿ).

Unfortunately the rest of your question cannot be interpreted.

I would guess that some substitutions could be made replacing θ by t, A by 50 (?). What does 2π100t mean? What is the complete interval for t starting [0.2π?

If 2π100t is meant to be 2π+100t it would make sense. cos(2π+100t)=cos(100t) because of the periodic nature of cosine.

50e^100it=50(cos(100t)+isin(100t)); 50ie^100it=50(icos(100t)-sin(100t)).

Imaginary part of 50ie^100it is 50cos(100t). When t=0.2π, 50cos(100t)=50cos(20π). cos(2πn)=cos(0)=1, so 50cos(20π)=50.

When t=0.3π, 50cos(100t)=50cos(30π)=50. Im(50ie^100it)=Im(50icos(100t)).