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

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

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.

• Thanks a lot for help. That's all I wanted to know. – mvr950 Sep 17 '12 at 21:40
• 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 $t$? – mvr950 Sep 17 '12 at 21:59
• @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. – celtschk Sep 17 '12 at 22:09
• 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 $t = \frac{-a}{2}$? – mvr950 Sep 17 '12 at 22:16
• 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 $t$. – celtschk Sep 17 '12 at 22:20

No non-constant polynomial $p(x)$ can ever be periodic. If it were, there would be infinitely many solutions to $p(x)-p(0) = 0$.

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

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