Suppose $z$ and $\omega$ are two complex numbers such that $|z|≤1$ and $|\omega|≤1$ and $|z+i\omega|=|z-i\omega|=2$. Find $|z|$ and $|\omega|$.
My attempt:
I squared the two given equations $$|z+iω|^2=4 $$ Therefore $|z+iω||\overline z-i\overline ω|=4$. On simplifying, $$z\overline z+iω\overline z-i\overline\omega z+ω\overline ω=4\tag{1}$$
Proceeding similarly for $|z-iω|=2$ we get another equation,call it $(2)$.
On adding $(1)$ and $(2)$ we get $$z\overline z+ω\overline ω=4\;\text{or}\; |z|²+|ω|²=4 $$
which cannot simultaneously satisfy $|z|≤1$ and $|ω|≤1$ but the answer given is $|z|=|ω|=1$. What is the mistake in my method?
Let's try this the old school way. Suppose $z = r_1 e^{i\theta}$ and $w = r_2 e^{i\psi}$ with $0 \le r_1, r_2 \le 1$ and $0 \le \psi \le \theta \le 2\pi$ without loss of generality. Then $$\begin{align} |z+iw|^2 &= |(r_1 \cos \theta - r_2 \sin \psi) + i(r_1 \sin \theta + r_2 \cos \psi)|^2 \\ &= (r_1 \cos \theta - r_2 \sin \psi)^2 + (r_1 \sin \theta + r_2 \cos \psi)^2 \\ &= r_1^2 \cos^2 \theta + r_1 \sin^2 \theta + r_2^2 \sin^2 \psi + r_2^2 \cos^2 \psi - 2r_1 r_2 \cos \theta \sin \psi + 2r_1 r_2 \sin \theta \cos \psi \\ &= r_1^2 + r_2^2 + 2r_1 r_2 (\sin \theta \cos \psi - \cos \theta \sin \psi) \\ &= r_1^2 + r_2^2 + 2r_1 r_2 \sin (\theta - \psi). \end{align}$$ Similarly, $$|z-iw|^2 = r_1^2 + r_2^2 - 2r_1 r_2 \sin (\theta - \psi).$$ Thus $$4 = r_1^2 + r_2^2$$ which is impossible. Essentially, this is the same as your solution, just written explicitly with real-valued variables throughout.
The difference of the two equations yields $$0 = 4 r_1 r_2 \sin (\theta - \psi),$$ which of course is only possible if $|z| = 0$ or $|w| = 0$ or $\theta - \psi = k \pi$ for some integer $k$. It is trivial to dismiss the first two cases as these lead to immediate contradictions. The last case would imply that $z = -w$ or $z = w$; in either case, this implies $$|z \pm iw| = |z(1 \pm i)| = \sqrt{2}|z| \le \sqrt{2} \cdot 1 = \sqrt{2},$$ again showing impossibility.