e^(-z)+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$1 + e^{- z} = 0$$
or
$$1 + e^{- z} = 0$$
or
$$e^{- z} = -1$$
or
$$e^{- z} = -1$$
- this is the simplest exponential equation
Do replacement
$$v = e^{- z}$$
we get
$$v + 1 = 0$$
or
$$v + 1 = 0$$
Move free summands (without v)
from left part to right part, we given:
$$v = -1$$
We get the answer: v = -1
do backward replacement
$$e^{- z} = v$$
or
$$z = - \log{\left(v \right)}$$
The final answer
$$z_{1} = \frac{\log{\left(-1 \right)}}{\log{\left(e^{-1} \right)}} = - i \pi$$
Sum and product of roots
[src]
$$i \pi$$
$$i \pi$$
$$i \pi$$
$$i \pi$$