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For example, assume you have a function: $x(t) = 2e^{i\frac{7t\pi}{6}} + e^{i\frac{5t\pi}{6}}$

If real part is periodic, but the imaginary part is not, is this function still periodic? (I didn't check the imaginary part of this particular example.)

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    Nothing in this function is periodic. Do you really want the $t$ in the *denominator* of the exponent? – 2017-01-15
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    @ChristianBlatter No, it shouldn't be in the denominator actually, but I am just using it as an example. – 2017-01-15

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In order for $f$ to have period $p$, you absolutely need to have $f(x+p)=f(x)=f(x-p)$ for every $x$ in the domain of $f$. It is not enough that $\mathop{\frak Re}(f(x+p)) = \mathop{\frak Re}(f(x))$.

However, I suspect that when you speak about the "imaginary portion", you actually intend to ask whether you also need something like $f(x)=f(x+ip)$ in addition to $f(x)=f(x+p)$. And that is not required. Indeed, that would make the function doubly periodic with periods $p$ and $ip$.

For example, the sine function (defined on all of $\mathbb C$) is periodic with period $2\pi$, but is not doubly periodic.

(Well-behaved doubly periodic functions is much rarer than ordinary periodic ones; they are known as elliptic functions and are important in number theory).

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Periodic is periodic is periodic. It means there exists a $T>0$ such that $$x(t+T)=x(t)$$ for all $t$. If only the real part is periodic, then the function is not periodic.

We can say even further that even if both the real and imaginary parts are periodic, $x$ can still fail to be periodic. This will happen if the ratio of the two periods is irrational.

Here I'm assuming your function has the reals as its domain. It wasn't actually specified, but traditionally "$t$" tends to be a real number so this is what I assumed.