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The following showed up in a solution to a problem I'm working on. $$(1+in)e^{inx}.$$ To finish the solution I have to show that the displayed equation above is real and I don't know how. I came up with the following $$ e^{inx}+e^{-inx}=2\cos(nx)$$ $$in\left(e^{inx}-e^{-inx}\right)=-2n\sin(nx)$$ but then got stuck.

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$(1+in)e^{inx}$ is certainly not real... –  Lepidopterist Mar 5 '13 at 3:27
    
Consider for example $n=1, x=0$. –  Alex Becker Mar 5 '13 at 3:29
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up vote 4 down vote accepted

You have $$ (1+in)e^{inx}=\cos(nx)-n\sin(nx)+i(n\cos(nx)+\sin(nx)). $$ This is real if and only if $$ n\cos(nx)+\sin(nx)=0. $$ Since $\cos(nx)=0$ implies $\sin(nx)=\pm 1$, this case will never yield solutions so we can divide by $\cos(nx)$.

Equivalently: $$ \tan(nx)=-n. $$ Drawing the plot of $\tan (nx)$, you'll se that there are infinitely many solutions to this equation. But only countably many. Exactly one in each $$ (k\pi/n-\pi/2n,k\pi/n,)\qquad\forall k\in\mathbb{Z}. $$

To be explicit, your expression is real if and only if $x$ belongs to the set: $$ -\frac{\arctan n}{n}+\frac{\pi}{n} \mathbb{Z} $$

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+1 I admire it when people understand the "real question" being asked. I didn't in this case. –  Lepidopterist Mar 5 '13 at 3:33
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@Lepidopterist And "real" is the right word. Thanks. –  1015 Mar 5 '13 at 3:38
    
(+1) for the part "Since $\cos(nx) = 0 \implies \sin(nx) = \pm 1 \ldots$". It is seldom to see people cover potential holes of their logic so cleanly. –  achille hui Mar 5 '13 at 4:31
    
@achillehui Thanks. I've had the kind of teachers who take a lot of points away when such holes are not covered. –  1015 Mar 5 '13 at 4:41
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Since $$(1+in)e^{i n x} = (1+in)(\cos (nx) + i \sin (nx)) = \cos(nx)-n \sin(nx) + i (n \cos(nx)+\sin(nx)))$$ For this to be real, you need $n \cos(nx) = - \sin(nx)$, solutions ($n,x$) can be found by looking at the intersection of the line $y = -nx$ with the unit circle. (It follows that for any $n$, there are two 'x' solutions modulo $\frac{2 \pi}{n}$.)

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