How can I solve this equation $\in \mathbb{C}$? $1+3^{z}=0$

Question

How can i solve this equation?

$1+3^{(z)}=0 $, $z \in \mathbb{C}$

I try to do:

$ e^{z(ln|3|+i2\pi k)}= e^{1(ln|-1|+i(\pi+2\pi k))}$ $,k \in \mathbb{C}$

$ e^{z(ln|3|+i2\pi k)}= e^{i\pi(1+2 k))}$ $,k \in \mathbb{C}$

But i dont know how get z....

thanks

Answer 1

You are correct, but you don't need to add multiples of $2\pi i$ to both sides. If suffices to write $$3^z=e^{\pi i+2\pi i}\\e^{z\ln(3)}=e^{\pi i+2k\pi i}$$

from which it follows that $$z=\frac{(2k+1)\pi i}{\ln(3)}$$

---

You are right that it isnt neccesarily true that $e^z=e^w\Rightarrow z=w$. What is true though is that $e^z=e^w\Rightarrow z=w+2k\pi i$. To see this, note that we can do the following trick:

$$e^z=e^w\\e^{z-w}=1\\e^{z-w}=e^{2k\pi i}$$

Now that we have all the possible solutions to the last line, we may conclude that $z-w=2k\pi i$, that is, $z=w+2k\pi i$.

Answer 2

You have\begin{align}1+3^z=0&\iff3^z=-1\\&\iff e^{z\log 3}=-1\\&\iff z\log3=(2k+1)\pi i,\end{align}for some $k\in\Bbb Z$. So, the solutions of that equation are the numbers of the form $\dfrac{(2k+1)\pi i}{\log3}$, with $k\in\Bbb Z$.