Expressing the sine function in terms of exponential

Question

Prove $e^{iz} - e^{-iz} = \sin z$.

I used $$\begin{align*} \sin z & = z - z^3/3! + z^5/5! - z^7/7! + \dots & (i) \\ e^{iz} & = 1 - z^2/ 2! - iz^3/3! + \dots & (ii) \\ e^{-iz} & = 1 - z^2/ 2! + iz^3/3! +z^4/4! + \dots & (iii) \end{align*}$$

However, I couldn't get the subtraction of $(iii)$ from $(ii)$ to yield $\sin z$.

Answer 1

I like to write series with a summation sign rather than individual terms.

In this case, $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $ so

$\begin{array}\\ e^{ix} &= \sum_{n=0}^{\infty} \frac{(ix)^n}{n!}\\ &= \sum_{n=0}^{\infty} \frac{(ix)^{2n}}{(2n)!} +\sum_{n=0}^{\infty} \frac{(ix)^{2n+1}}{(2n+1)!}\\ &= \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!} +i\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\\ \end{array} $

From this, putting $-x$ for $x$ and noting that $(-x)^{2n} = x^{2n}$ and $(-x)^{2n+1} = -x^{2n+1}$,

$e^{-ix} =\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!} -i\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} $

If we add these, $e^{ix}+e^{-ix} = 2\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!} =2 Cos(x) $ and $e^{ix}-e^{-ix} = 2i\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = 2i Sin(x) $.

Answer 2

$e^{iz}-e^{-iz}=\sin(z)$ is false.

The correct formula is $$\frac{e^{iz}-e^{-iz}}{2i}=\sin{z}$$

Also, your formulas (ii) and (iii) are missing the first-order terms. The correct equations are:

$$e^{iz}=1+iz-\frac{z^2}{2}-i\frac{z^3}{3!}+\cdots$$ $$e^{-iz}=1-iz+\frac{z^2}{2}+i\frac{z^3}{3!}-\cdots$$

See if you can go from there.