I'm preparing for an exam, and I have come over the following problem: I have the function f, where C is a complex number $$f: C \to C,$$ $$z=a+bi \mapsto f(z)=a+i(a+b)$$ Is the function of the form $f(z) = z·w$, where $w$ does not depend on $z$?
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 $c=d=1$. Seeing as this does not hold for any z, I concluded with the answer 'no', to the question asked in this task. My teacher, however, concludes with the opposite, using the following logic:
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 $$ \begin{matrix}a&-b\\b&a\end{matrix}$$ 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.
Please let me know if something was unclear, or if I should explain more clearly what I have done. Thank you in advance!