Complex number moduli inequalities

Question

I've been asked to prove that if $\begin{equation}\left|z\right| \le 1 , z \in \mathbb{C} \end{equation}$ then $$\begin{equation}\left|\frac{3z-i}{3+iz}\right|\leq1.\end{equation}$$

I've tried letting z = a +bi and using the fact that |z| = $\sqrt{a^2+b^2}$ and expanding the expression within the modulus brackets but I'm having trouble showing that the remaining fraction is less than or equal to one.

Any help would be much appreciated, I know this is a basic question but I can't seem to make the link between properties of complex moduli and how they solve this inequality.

Answer 1

Let $z = a+bi$. Then we will proove that: $$|{3z-i}|\leq|{3+iz}|$$ $$\sqrt{9a^2+(3b-1)^2}\leq \sqrt{a^2+(3-b)^2}$$ $$9a^2+9b^2-6b+1\leq a^2+b^2+9-6b$$ $$8(a^2+b^2)\leq 8$$

Which is true because $|z|\leq 1$

Answer 2

Hint

For $z=x+iy,$ and real $a$

$$|az-i|^2-|a+iz)|^2=(ax)^2+(ay-1)^2-[(a-y)^2+(x)^2]$$

$$=(a^2-1)(x^2+y^2-1)$$

Now for real $p,q>0$ $$p^2<=> q^2\iff \dfrac pq<=>1$$

Answer 3

To prove the inequality, since $|3z-i|,|3+iz|>0,$ it is enough to prove $|3z-i|^2\le|3+iz|^2.$ Since $z$ is complex, it can be written as $a+bi$ for some $a,b\in \mathbb R.$ Then, the equation becomes $|3a+(3b -1)i|^2\le|(3-b)+ai|^2.$

By the definition of $|\cdot|,$ this is equivalent to $(3a)^2+(3b-1)^2\le(3-b)^2+a^2.$ Further transforming, this becomes $a^2+b^2\le 1,$ or $|a+bi|\le 1.$