Fixed points on complex function $T(z) = (1+i)z + 3-4i$

Question

If $T(z) = (1+i)z + 3-4i$ is a complex function, and I got this fixed point: $z=4+3i$

Is this correct? Are there more fixed points?

$(1+i)z+3-4i=z$

$3-4i=z-(1+i)z$

$3-4i=-iz$

$z=-\frac{3-4i}{i}=4+3i$

Answer 1

Your proof is correct and there are no other fixed points. You can see this abstractly by saying that the fixed point equation is a degree one polynomial and therefore has at must one solution. More concretely, each deduction you made was "reversible" so each implication was actually an equivalence.

You had

$$ \begin{align*} (1+i)z+3-4i=z &\implies 3-4i=z-(1+i)z\\ &\implies 3-4i=-iz\\ &\implies z=-\frac{3-4i}{i}=4+3i \end{align*} $$

but at each step the reverse implication holds as well. For instance, in the last step you divided by $-i$, so to go the other way you multiply by $-i$. Hence, you really have

$$ \begin{align*} (1+i)z+3-4i=z &\Longleftrightarrow 3-4i=z-(1+i)z\\ &\Longleftrightarrow 3-4i=-iz\\ &\Longleftrightarrow z=-\frac{3-4i}{i}=4+3i. \end{align*} $$

In other words, $z$ is a fixed point of $T$ if and only if $z = 4+3i$. Be warned that not all manipulations you will do are reversible, however. $(-1)^2 = 1^2$ but you cannot reverse this to get $-1=1$!