Hello! I have a couple of doubts regarding a formula seen here :
$$\sum _{k=1}^{\infty } \frac {e^{kz}}{k}= -\log (1-e^{z}) /; Re(z)<0$$
What would happen if the real part of z Re(z) were equal to 0? Is there a formula for when Re(z)=0?
If the formula were
$$Im(\sum _{k=1}^{\infty } \frac {e^{kz}}{k})$$ being Im(z) the imaginary part of z, would it be equal to $$Im(-\log (1-e^{z}))$$ without mattering Re(z)?
Thanks!