Why complex signal has no imaginary spectrum I am learning about complex sampling.

I am confused why $~e^{ j 2\pi f~ n}~$ has only a real spectrum. I would have thought the $j ~\sin(2 \pi f n)$ would produce a single spike in imaginary spectrum just like there is a single spike in real axis from $\cos(2 \pi f n)$.

I understand that the spectrum is one sided because the negative complex exponentials cancel out, but why is there not a one sided real and imaginary spectrum?

Many thanks
The imaginary part of the spectrum corresponds to the odd part of the time-domain sequence. Since the given signal is even, i.e.,

$$x[n]=e^{jn\omega_0}=x^*[-n]=e^{-j(-n)\omega_0}=e^{jn\omega_0}$$

the imaginary part of the spectrum is zero, i.e., the spectrum is purely real-valued.

Note that for complex-valued signals, even in this context means $x[n]=x^*[-n]$, i.e., its real-part is even and its imaginary part is odd.

In sum, if a sequence $x[n]$ satisfies

$$x[n]=x^*[-n]$$

(i.e., the sequence is even and, consequently, its odd part is zero), then its DTFT is purely real-valued.