MHB (e^ix)-1=(2ie^(ix/2))x(sin (x/2))

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The discussion focuses on proving the equation (e^ix)-1=(2ie^(ix/2))x(sin(x/2)) using properties of complex exponentials and trigonometric identities. It highlights the relationship between sine and exponential functions, specifically using sin(x) = (e^(ix) - e^(-ix))/(2i) to derive sin(x/2). Additionally, the participants calculate the sum Zn = 1 + e^(ix) + e^(2ix) + ... + e^((n-1)ix) as a geometric series, leading to expressions for Xn and Yn involving cosine and sine sums. The conversation clarifies the notation and confirms the multiplication sign in the equation. Overall, the thread emphasizes the connections between complex exponentials and trigonometric functions in mathematical proofs.
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Show that:
(e^ix)-1=(2ie^(ix/2))x(sin (x/2))

Calculate this sum
Zn=1+e^(ix)+e^(2ix)+...+e^((n-1)ix)
And deduce those values :
Xn=1+cos(x)+cos(2x)+...+cos((n-1)x)
Yn=sin(x)+sin(2x)+...+sin((n-1)x)
 
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minachi12 said:
Show that:
(e^ix)-1=(2ie^(ix/2))x(sin (x/2))
Is that "x" in the middle a multiplication sign? Don't do that!

You need to know that $sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$. So $sin(x/2)= \frac{e^{ix/2}- e^{-ix/2}}{2i}$, $sin(x/2)= \frac{e^{ix/2}- e^{-ix/2}}{2i}$, $2i e^{ix/2} sin(x/2)= e^{ix/2}(e^{ix/2}- e^{-ix/2})= e^{ix/2+ ix/2}- e^{ix/2- ix/2}= e^{ix}- 1$.

Calculate this sum
Zn=1+e^(ix)+e^(2ix)+...+e^((n-1)ix)
And deduce those values :
Xn=1+cos(x)+cos(2x)+...+cos((n-1)x)
Yn=sin(x)+sin(2x)+...+sin((n-1)x)
That's a geometric series, $\sum_{i= 0}^n r^i$ with $r= e^{ix}$. The sum of such a geometric series is $\frac{1- r^n}{1- r}$
 
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