I noticed someone do this from one of the questions is asked on here i had:
$$e^z = -0.5$$
$$e^z = 0.5e^{i\pi}$$
which magically became:
$$z = \ln\left(\frac12\right) + iπ + 2ikπ$$
does this mean that if i have:
$$e^z = -r = re^{i\pi} = \ln(r) + iπ + 2ikπ$$
Thanks for any help you can offer me, i have really been strugling with this,
edit:
Oh my bad i meant to write:
$$e^z = -r = re^{i\pi} = \ln|r| + iπ + 2ikπ$$
Let $r \in R$ and let $z \in C$, where $e^{z}=-r$. Since $e^{i\pi}=-1$ and $e^{2i\pi}=1$, it follows that $e^{z}=e^{z}\cdot (1)^k=e^z\cdot (e^{2i\pi})^k=e^{z}\cdot e^{2ki\pi}=r\cdot(-1)=r\cdot e^{i\pi}$ where $k\in Z$. Now taking the logarithm of both sides of the equation $e^{z} \cdot e^{2ki\pi}=r\cdot e^{i\pi}$, we have that $ln(e^{z} \cdot e^{2ki\pi})=ln(r\cdot e^{i\pi})$. So $z+2ki\pi =ln(r)+i\pi$ and thus $z=ln(r)+i\pi-2ki\pi$. Since k is an arbitrary integer, we can rewrite this expression as $z=ln(r)+i\pi+2ki\pi$. Hope this helps.