Inverse functions using complex numbers Recently I came across a method to find the inverse function, in which if we assume $z=x+iy$ we can say that the inverse of $y=f(x)$ is $iz^*$
However, when I try to find the functions when a function is equal to its inverse, $z=iz^*$ and expand it, I get only the case where the real and imaginary part of z is same , i.e, $y=x$ Why am I missing out on the other clear solution $y=1/x?$
Also are there any other solutions to this equation?
PS:By $z^*$,  I mean the complex conjugate of z.
 A: To a function $f\colon\mathbb R\to\mathbb R$ you are associating the complex graph $\Gamma_f=\left\{\,x+\mathrm i y\,|\,f(x)=y\,\right\}\subset\mathbb C$. Since $\mathrm i(x+\mathrm i y)^* = y+\mathrm i x$, this allows us to obtain the graph of the inverse of an invertible $f$ as
$$
\Gamma_{f^{-1}} = \left\{\, \mathrm iz^*\,|\, z\in\Gamma_f\,\right\}.
$$
Now a function is its own inverse $f=f^{-1}$ if and only if $\Gamma_f = \Gamma_{f^{-1}}$.
The equation you wrote down is a lot stronger: asking that $z=\mathrm i z^*$ for $z\in\Gamma_f$ is equivalent to $x+\mathrm i y=y+\mathrm i x$, so that $f$ can only be given by $f(x)=x$.
The correct condition for $f$ being its own inverse would be that for all $z\in \mathbb C$ we have
$$
z\in \Gamma_f \,\Longleftrightarrow\, \mathrm i z^* \in \Gamma_f.
$$
So $z$ and $\mathrm i z^*$ don't need to be equal, they just need to be either both on the graph or both not on the graph for any $z\in\mathbb C$.
Here's a simple example: Consider $f(x)=-x$ so that $\Gamma_f$ consists of all $z=x-\mathrm i x$. These do not satisfy $z=\mathrm i z^*$, but they do satisfy
$$
\mathrm i z^* = -x+\mathrm i x = (-x) - \mathrm i (-x) \in \Gamma_f.
$$
