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Is complex valued function like $y(t) = t^2 + i\cdot t^2$ a periodic function?

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You can simply factorize your function as $y(t)=(1+\mathrm i)t^2$. So the question boils down to whether $t^2$ is a periodic function. That would mean that there's an $a\ne 0$ so that for all $t$, $(t+a)^2 = t^2$. Now applying the binomial formula and subtracting $t^2$ on both sides gives the equation $2ta+a^2=0$ or, if $a\ne 0$, $t=-a/2$, which clearly cannot be true for arbitrary $t$. Therefore the function is not periodic.

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  • $\begingroup$ Thanks a lot for help. That's all I wanted to know. $\endgroup$ – mvr950 Sep 17 '12 at 21:40
  • $\begingroup$ Sorry I've one more question. if $a \neq 0, t = \frac{-a}{2}, \text{ which clearly cannot be true for arbitrary } t$ why's that? I mean why can't $t \neq \frac{-a}{2}$ for arbitrary $t$? $\endgroup$ – mvr950 Sep 17 '12 at 21:59
  • $\begingroup$ @mvr950: I'm not sure what you mean. The equation which would have to be true for arbitrary $t$ is $t=-a/2$. This equation is clearly not fulfilled for $t\ne-a/2$ because that's exactly what $t\ne-a/2$ means. But it would have to be fulfilled for all values of $t$ if $t^2$ were periodic. $\endgroup$ – celtschk Sep 17 '12 at 22:09
  • $\begingroup$ Sorry I made a spelling mistake in my last post. I meant why can't $t = \frac{-a}{2}$ for arbitrary $t$? Why can't I use $t = \frac{-a}{2}$? $\endgroup$ – mvr950 Sep 17 '12 at 22:16
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    $\begingroup$ Because for a given $a$, there's exactly one $t$ which fulfils the equation $t=-a/2$, namely the one which you get by dividing $a$ by $2$ and changing the sign. For example, for $a=1$, the only $t$ which fulfils $t=-a/2$ is $t=-1/2$, but e.g. $t=1$ doesn't fulfill that equation. Thus $t^2$ is not $1$-periodic. The same is true for any $a$, therefore $t^2$ is not periodic at all. The critical point is that for fixed $a$, the equation would have to be fulfilled for all $t$. There's no $a$ such that $t=-a/2$ is fulfilled for all $t$. $\endgroup$ – celtschk Sep 17 '12 at 22:20
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No non-constant polynomial $p(x)$ can ever be periodic. If it were, there would be infinitely many solutions to $p(x)-p(0) = 0$.

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This is nothing but $(1+i)t^{2}$. Why you would think this is periodical?

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  • $\begingroup$ I just wanted to make sure. Thanks for your answer. $\endgroup$ – mvr950 Sep 17 '12 at 21:28

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