Let's say you had a function $f \colon \mathbb{C} \longrightarrow \mathbb{C}$ defined by $$ f(x+ \iota y) = (3+6x)+ \iota (2−3y). $$
How would you go about finding its inverse? This provoked my interest and curiosity, as it seems you can't simply manipulate the equation to get $x+ \iota y$ on the right hand side, and then swap $f(x+ \iota y)$ with $x+ \iota y$ and solve for $f(x+\iota y)$, as that yields a different result to what my textbook gives. So, how do you tackle a problem like this?
Note that, for any complex numbers $x+\iota y$ and $u + \iota v$, we have $f ( x+ \iota y) = u + \iota v$ if and only if $x+ \iota y = f^{-1}(u+\iota v)$, or $f^{-1}(u+\iota v) = x+\iota y$.
Suppose that $f(x +\iota y) = u + \iota v$, then we have $$ (3+6x)+\iota (2−3y) = u + \iota v, $$ which implies $$ \begin{align} 3 + 6x &= u \\ 2 - 3y &= v, \end{align} $$ and these last two equations in turn imply $$ \begin{align} x &= \frac{ u-3 }{ 6 } \\ y &= \frac{ 2-v }{ 3 }. \end{align} $$ Therefore we have $$ f^{-1}(u+\iota v) = x+\iota y = \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 }, $$ that is, $$ f^{-1}(u+\iota v) = \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 }. $$ Hence the function $f^{-1} \colon \mathbb{C} \longrightarrow \mathbb{C}$ is defined by the formula $$ f^{-1} (x + \iota y) = \frac{ x-3 }{ 6 } + \iota \frac{ 2-y }{ 3 }. $$
PS:
There is one more piece of calculation / verification that is needed to be carried out.
We find that, for any complex $u + \iota v$, we have $$ \begin{align} \left( f \circ f^{-1} \right) ( u + \iota v) &= f \left( f^{-1} (u + \iota v) \right) \\ &= f \left( \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 } \right) \\ &= 3 + 6 \left( \frac{ u-3 }{ 6 } \right) + \iota \left[ 2 - 3 \left( \frac{ 2-v }{ 3 } \right) \right] \\ &= u + \iota v \\ &= i_{\mathbb{C}} ( u+\iota v), \end{align} $$ which implies that $$ f \circ f^{-1} = i_{\mathbb{C}}, $$ where $i_{\mathbb{C}}$ denotes the identity function on the set $\mathbb{C}$ of complex numbers.
Similarly, for any complex number $x + \iota y$, we have $$ \begin{align} \left( f^{-1} \circ f \right) ( x+ \iota y) &= f^{-1} \big( f( x+ \iota y) \big) \\ &= f^{-1} \big( (3+6x) + \iota (2-3y) \big) \\ &= \frac{ ( 3+6x ) - 3 }{ 6 } + \iota \frac{ 2- (2-3y) }{ 3 } \\ &= x + \iota y \\ &= i_{\mathbb{C}} (x + \iota y), \end{align} $$ which implies $$ f^{-1} \circ f = i_{\mathbb{C}}. $$
Use the fact that\begin{align}f(x+yi)=u+vi&\iff\left\{\begin{array}{l}3+6x=u\\2-3y=v\end{array}\right.\\&\iff\left\{\begin{array}{l}x=\frac u6-\frac12\\y=-\frac v3+\frac23\end{array}\right.\\&\iff x+yi=\frac u6-\frac12+\left(-\frac v3+\frac23\right)i.\end{align}
