Complex exponent problem Find all numbers in complex plane that solves equation


*

*$e^z=4i$


Since $e^u e^{iv}=re^{i\Theta}$ it must be that $e^u=r \to u=\ln4$ and $v=\Theta+n2\pi \to v=\pi/2+n2\pi$. So the equation holds for points $z=\log w=\ln4+i(\pi/2+n2\pi)$ when $n\in \mathbb{N}$


*$e^{1/z}=-1$


Let's set $w=1/z$ then $e^u e^{iv}=-1$ when $e^u=1 \to u=0$ and $v=\pi+n2\pi$. So $w=i(\pi+n2\pi)$ and that leads to $z=\frac{1}{i(\pi+n2\pi)}$


*$e^{2z}+e^z+1=0$. Isn't this $(e^{z})^2+e^z+(e^z)^0=0$. So if we set $e^z=a$, we have $a^2+a+a^0=0 \to a^2+a+1=0$


solution is $a=-\frac 12 \pm i\frac{\sqrt{3}}{2}=e^{\pm i 2 \pi/3+n2 \pi}$. That leads to solution $z= \ln 1+i(\pm \frac{2 \pi}{3}+n2\pi)=i(\pm \frac{2 \pi}{3}+n2\pi)$.

1) Solution:
$$e^z=4i<=>$$
$$z=\frac{\log(4i)}{\log(e)}<=>$$
$$z=\ln(4i)<=>$$
$$z=\ln\left(4e^{\left(\frac{1}{2}\pi\right)i}\right)<=>$$
$$z=\frac{1}{2}i\left(4\pi n+\pi -4i\ln(2)\right)$$
(n is the element of Z - the set of integers)
2) Solution:
$$e^{\frac{1}{z}}=-1<=>$$
$$\frac{1}{z}=\frac{\log(-1)}{\log(e)}<=>$$
$$\frac{1}{z}=\ln(-1)<=>$$
$$\frac{1}{z}=i\pi (2n+1)<=>$$
$$z=\frac{i}{\pi (2n+1)}, 2n+1\ne0$$
(n is the element of Z - the set of integers)
3) Solution:
Substitute x=e^z
$$1+e^z+e^{2z}=0<=>$$
$$x^2+x+1=0<=>$$
$$x^2+x=-1<=>$$
$$x^2+x+\frac{1}{4}=-\frac{3}{4}<=>$$
$$\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}<=>$$
$$--------$$
$$x+\frac{1}{2}=\frac{i\sqrt{3}}{2}$$
Or:
$$x+\frac{1}{2}=-\frac{i\sqrt{3}}{2}$$
$$--------$$
$$x=\frac{i\sqrt{3}}{2}-\frac{1}{2}$$
Or:
$$x+\frac{1}{2}=-\frac{i\sqrt{3}}{2}$$
Substitute x back, and we get the answers:
$$z_1=\frac{2}{3}i(3\pi n-\pi)$$
$$z_2=\frac{2}{3}i(3\pi n+\pi)$$
(n is the element of Z - the set of integers)
