I'm preparing for an exam, and I have come over the following problem: I have the function $f$, where C is a set of complex numbers $$f: C \to C,$$ $$z=a+bi \mapsto f(z)=a+i(a+b)$$ Is the function of the form $f(z) = z + w$, where $w$ does not depend on $z$? Is it of the form $z\cdot w$ (where $w$ does not depend on $z$)?
For the first part, my answer is 'no', seeing as $w=ai$ for the identity to hold, and it is then clear that $w$ depends on $z$. My teacher's solution says the same thing, but that the answer to the question is 'yes'. Am i wrong, or is my teacher wrong?
For the second part of the exercise: I have written $w=c+di$, which gives $z·w=(ac-bd)+ (bc+ad)i$. This gives us two equations, namely: $$ac-bd=a$$ and $$bc+ad=a+b$$
I tried solving this system of linear equations for $c$ and $d$, and ended up with $c=d=1$. However this only holds if $b=0$. I am a bit stuck, and can't really see clearly how to proceed, or how to come to a general conclusion.
Please let me know if something was unclear, or if I should explain more clearly what I have done. Thank you in advance!
EDIT (included my teacher's answer, which I do not understand):
In his solution, my teacher ends up with the same system of linear equations as I did in my post. From there he says that the system has a unique solution if and only if the determinant of the matrix $$ \begin{matrix}a&-b\\a&b\end{matrix}$$ is non-zero. That is $a^2+b^2 \neq 0$.Then he says: But if $a^2+b^2=0 \Rightarrow a=b=0 \Rightarrow f(z) = 0 = 0 · w$ Hence the answer to the question must be 'yes'. If anyone could help me understand what is meant, I would greatly appreciate it. NB: There is a possibility that my teacher has read the task wrong, and therefore answers 'yes' when the answer to both questions is 'no'.
Concerning the first question: you are right. The question is equivalent to this one: is the function $z\mapsto f(z)-z$ constant? No, it is not: $f(z)-z=i\operatorname{Re}z$.
Concerning the other question, just note that $f(1)=1+i=(1+i)\times1$. So, if there was such a $w$, $w=1+i$. But $f(i)=i\neq(1+i)\times i$. Therefore, there is no such $w$ (again).