Show $\left\lvert\frac{3z-i}{3+iz}\right\rvert=1$. I want to show that if $z$ is a complex number with $\lvert z \rvert=1$, we have:
$$\left\lvert\frac{3z-i}{3+iz}\right\rvert=1$$
Can someone help me?
 A: Hint:) Show that $|3z-i|=|3+iz|$ or
$$(3z-i)\overline{(3z-i)}-(3z-i)\overline{(3z-i)}=0$$
Note that $\overline{ab}=\overline{a}\overline{b}$ and $\overline{a+b}=\overline{a}+\overline{b}$.
A: Hint: $$|\overline{z}|\cdot|(3z-i)|=|\overline{z}(3z-i)|=|3|z|^2-i\overline{z}|=|3-i\overline{z}|=|3+iz|$$
Note that $|z|=|\overline{z}|=1$, and $\overline{a-ib}=\overline{a}+i\overline{b}$.
A: $|z|=1 \rightarrow |\bar{z}|=1$
$|\frac{3z-i}{3+iz}|=|\frac{3z-i}{3+iz}|*|\frac{1}{\bar{z}}|=|\frac{3z-i}{(3+iz)*\bar{z}}|=|\frac{3z-i}{3\bar{z}+i}|=|\frac{3z-i}{\bar{3z-i}}|=1$
A: Although many answers show some "slick" ways to prove the equality (which is fine!), but I am not that sharp so I try the stone age approach by letting $z=a+bi$, the fraction becomes: $$\frac{3a+3bi-i}{3+ai-b}$$
Now little rearranging gives $$\frac{3a+(3b-1)i}{3-b+ai}$$
Now taking absolute value of numerator:
$$9a^2+9b^2-6b+1$$and of denominator:
$$9-6b+b^2+a^2$$
The magic moment: $a^2+b^2=1$, right? So can you verify that numerator and denominator become identical?
