Some complex logarithms: please could somebody check my work? I am doing some exercises from my book, this one asks me to find suitable $z \in \mathbb C$. 

Please could someone check my work?

1) $z$ such that $e^{z}=-2$: This means that $-2 = iArg(z) + \log |z|$ and it is therefore clear that $z \in e^{-2}\{e^{2\pi k} | k \in \mathbb Z\}$
2) $z$ such that $e^z = i$: The same thing again. This time, $z \in \{e^{{\pi \over 2} + 2\pi k} | k \in \mathbb Z\}$
3) $z$ such that $e^z = -i$: The same thing again. This time, $z \in \{e^{{-\pi \over 2} + 2\pi k} | k \in \mathbb Z\}$
4) $z$ such that $\sin z = 100$: Here  I let $x = e^{iz}$ and use that $\sin z = {x + {1\over x}\over 2i}$. This gives the quadratic equation $x^2-200i x - 1 = 0$. I use the formula for the quadratic equation to get
$x_{1/2} = {200 i \pm \sqrt{(200i)^2 +4}\over 2}$. It is purely imaginary so the argument is $2\pi k$. Without using a calculator it's not clear to me whether I really have to determine the absolute value?
Then there were a few more similar ones but it's not so clear to me what I should learn from these. Aren't these exercise purely computational and not insighful?

What is it that I should learn from these exercises?

 A: By taking the logarithm of both sides, $$e^z=a$$ becomes $$z=\log(a)+i2k\pi=\log|a|+i\arg(a)+i2k\pi,$$ as adding periods to the argument doesn't change the value of the exponential.
$$\log(-2)=\log2+i\pi,$$
$$\log i=\log1+i\frac\pi2=i\frac\pi2,$$
$$\log(-i)=\log1-i\frac\pi2=-i\frac\pi2.$$
Then
$$\frac1{2i}\left({x\color{red}-\frac1x}\right)=100$$
indeed gives
$$x^2-200ix-1=0$$and
$$x=100i\pm\sqrt{9999}i$$
leading to
$$iz=\log((100+\sqrt{9999})i)+i2k\pi=\log(100+\sqrt{9999})+\log i+i2k\pi$$or
$$iz=\log((100-\sqrt{9999})i)+i2k\pi=\log(100-\sqrt{9999})+\log i+i2k\pi.$$
Divide by $i$.
A: In your 1-3, you didn't really use the exponential or logarithm. For example, problem 1,
$e^z=-2 \implies z=\ln{|-2|+i\text{arg}(-2)}$
So $z=\ln{2}+i(2k\pi+\pi)$
For problem 4, You should use $\sin{z}=\sin{(x+yi)}=\sin{x}\cosh{y} + i\cos{x}\sinh{y} $.
So $\sin{x}\cosh{y}=100$ and $\cos{x}\sinh{y}=0$.
If $\sinh{y}=0$, $\frac{e^y-e^{-y}}{2}=0$. This means $y=0$ and $\cosh{y}=1$ which is impossible since $\sin{x}\ne 100$.
Hence we must have $\cos{x}=0$, which gives $x=2k\pi\pm \frac{\pi}{2}$. Plug this into $\sin{x}\cosh{y}=100$, we get $\cosh{y}=\pm 100$. But $\cosh{y}=\frac{e^x+e^{-x}}{2}$ is positive, so it has to be $y=2k\pi+\frac{\pi}{2}$ and $\cosh{y}=\frac{e^x+e^{-x}}{2}=100$. Solving this equation gives you $e^x=100\pm \sqrt{9999}$. So $x=\ln{(100\pm \sqrt{9999})}$.
Combining above:
$$z=\ln{(100\pm \sqrt{9999})}+i(2k\pi+\frac{\pi}{2})$$
A: to give one example, you will learn, or, as appears, have already learned, to find the full solution set (in $\mathbb{C}^2$) of 
$$
z^w = 1
$$
where $z\ne 0$ is a general complex number. (you will also have learned to list the special cases that can occur). this may then lead you to wonder how the various branchings and degeneracies of such a "many-valued functional relation" fit together.
two exercises that might appeal, if they are not in your list, are to evaluate (in the general case) the functions: $\log \log z$ and $e^{e^z}$
then you can describe what is left out of a simplistic, or optimistic, statement like: 
$$
e^{e^{\log \log z}} = z
$$
