Bijectivity of T(z) = λ z + μ z*

Question

Let $T = \lambda z + \mu \bar z$ where $\lambda, \mu \in \mathbb{C}$. Show that $T$ is bijective exactly when $\lambda \bar\lambda \ne \mu \bar\mu$.

Answer 1

Think of $ z = a + i b $ as a vector $ (a,b) $ and write $ T(z) $ as a matrix in terms of the real and complex parts of $ \lambda $ and $ \mu $ . T is bijective when the determinant of T is nonzero.

Answer 2

Let $t$ in $\mathbb C$, then $t=T(z)$ if and only if $$ \lambda z+\mu\bar z=t,\qquad \bar\lambda\bar z+\bar\mu z=\bar t, $$ that is, $$ \begin{pmatrix}t\\ \bar t\end{pmatrix}=A\cdot\begin{pmatrix}z\\\bar z\end{pmatrix},\qquad A=\begin{pmatrix}\lambda&\mu\\\bar\mu&\bar\lambda\end{pmatrix}. $$ The matrix $A$ is invertible if and only if its determinant is not zero, that is, $|\lambda|\ne|\mu|$. Can you carry on from here?