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I would like to know the clarification of the following.

If the follwoing one is correct, we can find cosix and sin ix series very quickly.

We know that, Cosx + i sinx = e^ix

If we replace x by ix on both sides, we get; cos ix + i sin i x = 1/e^x

we know that e^-x series. In this by taking real part and imaginary part and then equating to LHS, can we get cos ix and sinix series. If it is wrong? why it is wrong...justify. Thanks in advance

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It's wrong because $\cos ix$ and $\sin ix$ are not strictly real, so they won't correspond to the real/imaginary parts of the right side. (The RHS doesn't even have an imaginary part for real $x$.) –  anon Oct 20 '11 at 9:04
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It's not clear to me what the point of inserting $\mathrm i$ is here. The Euler formula itself is a good basis for relating the series for the exponential and sine and cosine; why this transformation attempt? –  joriki Oct 20 '11 at 9:10
    
(Well, technically $\cos ix$ is always real, but $\sin ix$ is purely imaginary.) –  anon Oct 20 '11 at 9:25
    
Thank you for all –  jiha Oct 20 '11 at 10:37

2 Answers 2

I assume you're looking for series expansions of $\cos(ix)$ and $\sin(ix)$. Given Euler's formula, $$e^{ix}=\cos(x)+i\sin x,$$ and the fact that $\cos,\sin$ are real for real $x$, we may equate the real and imaginary parts of both sides of the above equation in order to find series for $\cos x$ and $\sin x$. Using the series expansion of the exponential function, and the fact that powers of $i$ cycle, we arrive at ($\color{Red}{\text{can you tell why?}}$): $$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots;\qquad \sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.$$ (One issue here is circularity: the above formulas are typically used in deriving Euler's formula in the first place! But I will assume this issue is of no concern for the purposes of this question.)

From here, we may replace $x$ with $ix$ in the expansions of cosine and sine to obtain the series expansions of $\cos(ix)$ and $\sin(ix)$, in which case we get ($\color{Red}{\text{can you check this for yourself?}}$): $$\cos(ix)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdots;\qquad \sin(ix)=i\left(x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots\right).$$

We see from the above that $\cos(ix)+i\sin(ix)$ is purely real; indeed it is exactly the series expansion of $e^{-x}$ we would expect, and we know that is also purely real for real $x$. This means that taking the real and imaginary parts of $e^{-x}$ will not give us $\cos(ix)$ and $\sin(ix)$: the only way taking the real and imaginary parts of an expression of the form $a+bi$ will gives us back $a$ and $b$ respectively is if both $a$ and $b$ are real numbers, which is not the case here.

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Very good explanation. Thank you so much. –  jiha Oct 20 '11 at 10:35

Since $\cos(x)=\frac12(\mathrm e^{\mathrm ix}+\mathrm e^{-\mathrm ix})$ and $\sin(x)=\frac1{2\mathrm i}(\mathrm e^{\mathrm ix}-\mathrm e^{-\mathrm ix})$, $\cos(\mathrm ix)=\frac12(e^{-x}+e^{x})=\cosh(x)$ and $\sin(\mathrm ix)=\frac12\mathrm i(e^{x}-e^{-x})=\mathrm i\sinh(x)$. Is there anything more to say?

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Good and interesting. Thank you. –  jiha Oct 20 '11 at 10:35

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