Define $h: \mathbb{C}\to\mathbb{R}$ by:$$h(z)=\operatorname{Re}(z)+\operatorname{Im}(z)$$ d define $f$ by the formula $f(z)=\log (h(z))$
(a) Prove that $h$ is surjective. Easy!
(b) Find the implied domain of $f$ and sketch it in a complex plane diagram.
(c) Find the implied range of $f$
Help me with this;(
I have no idea how to prove it. When I put complex equation in $Im(z)$ part, they make output of some number with $i$. So I was thinking the codomain is not $\Bbb R$.
So, you have your function $h:\mathbb{C} \to \mathbb{R}$ given by:
$$h(z) = \Re(z) + \Im(z)$$
To prove surjectivity, all you need to do is prove that for every real number, there exists a complex number $z$ such that the sum of its real & imaginary parts yields that real number.
So, let $r \in \mathbb{R}$. Then, define $z = r + i0$. So, clearly:
$$h(z) = h(r+i0) = \Re(z)+\Im(z) = r + 0 = r$$
Since that gives us the correct result and $r$ was arbitrarily chosen, it follows that $h(\mathbb{C}) = \mathbb{R}$. That proves surjectivity.
Let me also say something about the injectivity of this function. So, clearly, this function is not injective. Why?
Let $r \in \mathbb{R}$. Then, notice that $z_1 = \frac{r}{2} +i\cdot \frac{r}{2}$ and $z_2 = r + i0$ yield exactly the same result when used as inputs for the function? So, clearly, the preimages are not unique and, therefore, this function is not injective.
This is just a bit of extra work on my part but, hopefully, it should teach you something about how to approach such problems in the future.