Show that $f(z)=2z+z^2$ with $|z|<1$ is a one-to-one function.
By using $z=x+iy$, I get $$f(z)=x^2+2x-y^2+i(2y+2xy)$$
So to prove that this function is one-to-one, suppose $f(z_1)=f(z_2)$ where $z_1=x_1+iy_1,z_2=x_2+iy_2$, by comparing the real part and imaginary part, $$x_1^2+2x_1-y_1^2=x_2^2+2x_2-y_2^2$$ $$2y_1+2x_1y_1=2y_2+2x_2y_2$$.
So how can I reach the result $x_1=x_2,y_1=y_2$? And how can I relate the information $|z|<1$ in proving?
$\endgroup$ 12 Answers
$\begingroup$Hint:
Step 1) Notice $z_1^2 + 2z_1 - z_2^2 - 2z_2 = (z_1 - z_2)(z_1+z_2+2).$
Step 2) If $z_1 + z_2 = -2$, then $2 > |z_1|+|z_2| > |z_1 + z_2| = 2$ which is a contradiction.
Step 3) Conclude $z_1 = z_2.$
$\endgroup$ $\begingroup$Hint: The composition of two functions that are one-to-one is one-to-one. Consider $z+1$ on the unit disk, and $z^2-1$ on the right half plane $\{\textrm{Re}(z)>0\}$. Show that they are one-to-one on their corresponding domains, and take their composition by checking first that it is defined on these domains.
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