What is the locus of $Z = \frac{4-z}{4+z}$ if $|z|=4$? The question is: Given $$Z=\dfrac{4-z}{4+z},$$ if $|z|=4$, then find the locus of $Z$.
The solution in the textbook is:
$$Z=\dfrac{4-z}{4+z}\iff z=\dfrac{4(1-Z)}{1+Z},$$ so,
$|z|=\left|\frac{4(1-Z)}{1+Z}\right|$ by taking the modulus on both sides. Then, $1=\left|\frac{1-Z}{1+Z}\right|$ since $|z|=4$. Now, $|Z+1|=|Z-1|$ implies that the locus of $Z$ is the perpendicular bisector of the join of $1$ and $-1$, which is simply the $y$-axis.
My question:
Isn't taking the modulus on both sides a forwards implication only? I would assume that we would need a double implication to show that $|Z+1|=|Z-1|$ is equivalent to our original locus to ensure that taking the modulus hasn't lost/added information. Suppose we have $z=1+i$, then wouldn't $|z|=|1+i|$ be completely different?
Note: I understand that there are other ways of doing it, such as letting $z=x+iy$ and $x^2+y^2=16$. I just wanted to know why the textbook's method is valid or seems to get the same answer.
 A: In general, I completely agree with your analysis.  $z_1 = z_2 \implies |z_1| = |z_2|$, while the implication does not go in the reverse direction.
It is true that, with sophisticated intuition, one might recognize situations where this criticism is not critical, but for the beginning student, I personally would not have offered such a solution.
Instead, my approach would have been that by the analysis given in your posting, the $y$-axis represents all candidate solutions.  That is, the analysis proves that no other solution is possible other than the $y$-axis.
Then, I would let $Z_0 = 0 + iy$.
Then, I would compute $z = \frac{4(1 - Z_0)}{1 + Z_0} = \frac{4(1-iy)}{1 + iy}.$
Edit
In general, for such problems, there is a potential trap here.  Any value of $Z_0$ that causes a denominator to equal $0 + i(0)$ must be rejected.  In the given problem, that possibility does not exist, since for any value of $y$, $(1 + iy) \neq (0 + i[0]).$
Then, I would compute $|z| = 4 \times \frac{|1 - iy|}{|1 + iy|} = 4.$
This would verify that every candidate solution, which will have form $0 + iy$, is in fact a solution to the problem.
A: You are right in saying that, as in your example, that $|z|=|1+i|$ does not imply that $z=1+i$. But in this question, it is a condition on $Z$ that $|z|=4$ so only the "forwards implication" is required.
So $$|z|=4\implies\left|\frac{4(1-Z)}{1+Z}\right|=4$$
A: 
ensure that taking the modulus hasn't lost/added information

All the modulus does is strip the information about complex argument of the number. All the division does is divide the moduli and subtract the arguments. Likewise for multiplication: it multiplies the moduli and adds the arguments.
Consider two numbers $a=Ae^{i\phi}$ and $b=Be^{i\psi}$ where $A,B\ge0,$ and $\phi,\psi\in[0,2\pi).$ For them we have:
$$|a|=A,\tag1$$
$$|b|=B,\tag2$$
$$\left|\frac ab\right|=\left|\frac ABe^{i(\phi-\psi)}\right|=\frac AB=\frac{|a|}{|b|},\tag3$$
$$\left|ab\right|=\left|ABe^{i(\phi+\psi)}\right|=AB=|a||b|.\tag4$$
Thus, if we are given that
$$\left|\frac ab\right|=1,\tag5$$
we can tell that
$$|a|=|b|,\tag6$$
which is exactly the last step in your textbook's solution.
A: Let $$Z=\frac{4-z}{4+z} \implies z=4\frac{1-Z}{1+Z}$$
Next $$z\bar z=16\frac{1-Z}{1+Z} \frac{1-\bar Z}{1+\bar Z}=16 \implies Z+\bar Z=0$$
$Z=X+iY$, gives the locus of $Z$-points as $X=0$, that is $Y$ axis.
