Assume its maximum value or minimum value in $\mathbb{C}$ or not? Consider the following statements:
(a) $|e^{\sin z}|$ does not assume its maximum value in $\mathbb{C}$.
(b)$|\sin (e^z)|$ does not assume ist minimum value in $\mathbb{C}$,
Then
A. only A is true
B. Only B is true
C. Both A and B is true
D. Both A and B is false.
I believe the question is related to maximum modulus principe and minimum modulus principle.
I think does not assume minimum value or maximum value in $\mathbb{C}$ means minimum is $-\infty$  and maximum is $+\infty$, this is the only case, but how can I get the idea of minimum and maximum of the given modulus.
 A: No, it doesn't min the minimum is -∞.
Think about simpler cases. And ask yourself those two simpler questions
(c) Does $g(x)=(x-1)²$ assumes its minimum value in ℝ?
(d) Does $f(x)=e^x$ assumes its minimum value in ℝ?
For question (c) answer is yes, and quite obvious. $g(x)$ assumes its minimum value at x=1: $g(1)=0$; and there is no $x\in ℝ$ for which $g(x)<g(1)=0$. So, since it assumes its minimum value at x=1, and $1\in ℝ$, you can say, yes, indeed, g(x) assumes its minimum value in ℝ.
Question (d) is slightly trickier. Where does f(x)=eˣ assumes its minimum value? Does it assume it in ℝ? Answer is no. f(x)→0 when $x→-∞$. There is no x∈ℝ for which you can say "f(x) is the minimum value of f". So $f(x)=e^x$ does not assume its minimum value in ℝ.
In more detail, you can say that minimum value of $f(x)$, if it exists, have to be 0. Because it cannot be, obviously any negative value. And it cannot be a positive one neither: any μ>0 you would supposed to be the minimum value can easily be disproved by considering x₀=log(μ)-log(2): $e^{x_0}=μ/2<μ$, proving that μ wasn't the minimum after all.
So, for proposition (d), IF f(x) assumes its minimum value in ℝ, then, it has to be f(x)=0, so there has to be a x∈R, such as $e^x=0$. And there isn't. So $f(x)=e^x$ doesn't assume its minimum value in ℝ.
So, now, back to (a) and (b):
Intuition says that minimum value, if it exists, would be 0 in both case. Because, 1st, they are modulus, so it could net be less than 0. And secondly, the range of reached values appears to be quite vast, so it is hard to believe that it exist a μ>0 such as $|e^{\sin(z)}|≥μ\, \forall z$ or $|\sin(e^z)|≥μ\, \forall z$.
But that has to be proven first. Can you prove that the only μ≥0 for which you can say $|e^{\sin(z)}|≥μ\, \forall z$ is μ=0?
In other words, that you can get any close to 0 you want.
And then, if this is true, that is if it is true that minimum, if it exists, is 0, the question becomes simply "is there a $z\in ℂ$ such as $|e^{\sin(z)}|=0$?" and "is there a $z\in ℂ$ such as $|\sin(e^z)|=0$?
