Suppose $z$ and $omega$ are two complex numbers such that $|z|≤1$ and $|omega|≤1$ and $|z+iomega|=|z-iomega|=2$. Find $|z|$ and $|omega|$.

Let $v=iw$. The parallelogram law gives:

$$|z+v|^2+|z-v|^2= 2(|z|^2+|v|^2).$$

Hence

$$8=4+4=2(|z|^2+|v|^2) le 2(1+1)=4,$$

a contradiction.

Conclusion: $z$ and $w$ with the above properties do not exist.

Let's try this the old school way. Suppose $z = r_1 e^{itheta}$ and $w = r_2 e^{ipsi}$ with $0 le r_1, r_2 le 1$ and $0 le psi le theta le 2pi$ 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.

You are correct in that there are no solutions, but there are easier ways to see it. $$2=|z+iw|<=|z|+|w|<=1+1implies |z|=|w|=1$$

The only way that $|z+iw|$ can equal $2$ is if $z$ and $iw$ have the same argument. By the same token, $z$ and $-iw$ have the same argument. Then $iw$ and $-iw$ have the same argument, which is absurd.