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assuming the power series for $f(x)=e^x$ holds for complex numbers, show that $e^{ix}=\cos x+i\sin x$


How can I solve this problem?

What is the complex number and what does it mean in this problem?

Please help me. Thanks.

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In order to be able to answer your question we have to know what is your definition of $\cos$ and $\sin$. There is a long way from rectangular triangles to the exponential function. – Christian Blatter Jun 10 '13 at 14:46
you mean the domain of x? – Guido Di Pietro Jun 10 '13 at 14:48
If you also know the power series for the sine and cosine, then just take the power series for $e^x$, plug in $ix$, and separate the real and imaginary terms. – Andreas Blass Jun 10 '13 at 14:48
oh thanks. nice advice. – Guido Di Pietro Jun 10 '13 at 14:56

By assumption, the formula

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots$$

is valid for all $x\in\mathbb{C}$. Thus we are free to replace $x$ by $ix$ and obtain

$$\begin{align*} e^{ix} &= 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} - \cdots =\\&= \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right) + i\left(1 - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right) =\\&= \cos{x} + i\sin{x}, \end{align*}$$

which is what we wanted to show.

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@Guido I think Tharis brought a really elegant Answer to your question, would you mind to fairly click ok! – al-Hwarizmi Jun 10 '13 at 20:53
already did :). – Guido Di Pietro Jun 10 '13 at 23:06

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