Is $e^{(1+j2t)}$ a periodic function?

Question

Is $e^{(1+j2t)}$ a periodic function? I think yes,because $e^{(1+j2t)}=e^1 \times e^{j2t}$,

$e^{j2t}=cos(2t)+jsin(2t)$,and it is periodic,and $e^1$ is a constant, so i think a constant value multiply a periodic function is still a periodic function,is my thinking right?

Answer 1

Yes ! , $f(t)=e^{1+2jt}$ is periodic with period $\pi$, $$f(t+\pi)=e^{1+2j(t+\pi)}= e^{1+2jt} e^{2j \pi}=f(t), \forall t\in R$$ As $j=\sqrt{-1},e^{2j\pi}=cos(2\pi)+jsin(2\pi)=1. $

Answer 2

Yes, $f(t)=e^{1+j2t}$ is a periodic function because

$$f(t)=e^{1+j2t}=e^1e^{j2t}$$ which is a constant times a periodic function. The period is $\pi$ since $f(t+\pi)=f(t)$. We also know that for a constant $c$

1. $cf(t)$ is periodic with period $\pi$.
2. Every constant function is periodic and has no fundamental period.

therefore you are correct to deduce that a constant times a periodic function remains a periodic function.