Why does the Taylor series expansion of eiθ equal the expansions of cosθ+isinθ when their 3 graphs look nothing alike?

Answer 1

To begin with look at the series expansion for e^ix , cos x ,sin x.

e^ix= 1+ ix + (ix)^2/2! +(ix)^3/3! +(ix)^4/4! + (ix)^5/5! + (ix)^6/6! + (ix)^7/7! + ....

cos x= 1- x^2/2! + x^4/4! - x^6/6! +....

sin x = x - x^3/3! + x^5/5! - x^7/7! +....

So going back to the expansion for e^ix =1 + ix + (i)^2 x^2/2! + i(i)^2 x^3/3! +(i)^4 x^4/4! + i(i)^4 x^5/5! + (i)^6 x^6/6! + i(i)^6 x^7/7! +...

Therefore e^ix = 1+ix - x^2/2! - ix^3/3! + x^4/4! +ix^5/5! - x^6/6! - ix^7/7! +....

e^ix = 1-x^2/2! +x^4/4! -x^6/6! + i( x -x^3/3! +x^5/5! -x^7/7!)

Therefore Re(e^ix) = cos x and Im(e^ix)= sin x So e^ix = cos x +i sin x.

Answer 2

Because the graph of cos(θ) is the graph of (e^iθ + e^-iθ)/2; and the graph of sin(θ) is the graph of (e^iθ - e^-iθ)/(2i) as shown below: