How can I find such a $z$? (Injectivity of $T(z) = \lambda z + \mu\bar{z}$) We consider $T:\mathbb{C}\to\mathbb{C}$ defined by $T(z)=\lambda z+\mu\overline{z}$, where $\lambda ,\mu\in\mathbb{C}$. I want to prove that if $T$ is injective, then $\lambda\cdot\overline{\lambda}\neq\mu\cdot\overline{\mu}$.
Clearly, if $\mu =0$, then $\lambda\neq 0$ because $T$ is injective.
Now, if $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ and $\mu\neq 0$, then I wanted to find a $z\in\mathbb{C}\setminus\{0\}$ such that $z=-\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}$, because we would have that:
$T(z)=-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}+\mu\overline{z}=\overline{z}(-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}+\mu)=0$
('Cause this would prove that $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ implies $T$ is not injective.)
So, the question is, can we always find such a $z\in\mathbb{C}\setminus\{0\}$?
 A: View $T$ as an $\Bbb R$-linear map.
Since $T(1)=\lambda+\mu$ and $T(i)=\lambda-\mu$ the determinant of $T$ is 
$$
\det(T)=(\lambda_1+\mu_1)(\lambda_1-\mu_2)-(\lambda_2+\mu_2)(-\lambda_2+\mu_2)=
\lambda_1^2+\lambda_2^2-\mu_1^2-\mu_2^2=|\lambda|-|\mu|,
$$
where the subscript 1 denotes real part and the subscript 2 denotes imaginary part.
Therefore $T$ is injective if and only if
$$
|\lambda|\neq|\mu|.
$$

Note : the first version of this answer contained an error of computation. This justifies the comment of Jonas Meyer.
A: Here's an alternative proof. 
Suppose that $T$ is injective. Note that $\mu \neq \bar{\lambda}$, otherwise $T(z) = \lambda z + \overline{\lambda z}$ in which case $T(z) = 0$ for any $z \in \mathbb{C}$ such that $\Re(\lambda z) = 0$. Now note that $T(\bar{\lambda}) = |\lambda|^2 + \mu\lambda$ and $T(\mu) = \lambda\mu + |\mu|^2$. Therefore $T(\bar{\lambda}) - T(\mu) = |\lambda|^2 - |\mu|^2$; as $\mu \neq \bar{\lambda}$ and $T$ is injective, $|\lambda|^2 \neq |\mu|^2$.
A: To "generalize" your reformulation to an equivalent statement: Given $a\in\mathbb C$ with $|a|=1$, can we find $z\in\mathbb C\setminus\{0\}$ such that $z=a\overline z$?
The answer is yes.  E.g., restricting to $|z|=1$, we can reformulate this as finding a solution $x\in\mathbb R$ to the equation $e^{ix}=e^{i\theta}e^{-ix}$, which is easy.
