How can I show that modul of this complex number equal to 1?

Question

Let $z$ be a complex number such that $$(1+2 i) \left| z\right| -\frac{\sqrt{10}}{z}+2-i=0.$$ Prove that $\left| z\right|=1$.

I tried Put $z = a + bi$ $(a,b \in \mathbb{R})$, we have $$2-\frac{\sqrt{10}a}{a^2+b^2}+\sqrt{a^2+b^2}+i \left(-1+\frac{\sqrt{10}b}{a^2+b^2}+2\sqrt{a^2+b^2}\right) = 0.$$ Then $$\begin{cases} 2-\dfrac{\sqrt{10}a}{a^2+b^2}+\sqrt{a^2+b^2}=0,\\ -1+\dfrac{\sqrt{10}b}{a^2+b^2}+2\sqrt{a^2+b^2}=0 \end{cases} \Leftrightarrow \begin{cases} 2(a^2+b^2) +(a^2+b^2) \sqrt{a^2+b^2} - \sqrt{10}a=0,\\ -(a^2+b^2) +2(a^2+b^2) \sqrt{a^2+b^2} + \sqrt{10}b=0.\\ \end{cases}$$

$$ 5(a^2+b^2) \sqrt{a^2+b^2} = (2a+b)\sqrt{10}. $$ From here, we have $2 a + b >0$.

I think, I prove \begin{equation} (2a+b)\sqrt{10} \leqslant 5 \sqrt{a^2+b^2} \end{equation} Or $$5 (3 a-b) (a+3 b)\leqslant 0.$$ But, I can't above inequality. $$5(a^2+b^2) \sqrt{a^2+b^2} \leqslant 5\sqrt{a^2 + b^2}\Leftrightarrow a^2 + b^2 \leqslant 1.$$

Answer 1

Hint: note that $1+2i= i(2-i)$ then collect and rewrite:

$$ (2-i)(1+i|z|)= \frac{\sqrt{10}}{z} $$

Note that $|2-i|^2=5$ and take the square of the modulus on both sides:

$$ 5(1+|z|^2) = \frac{10}{|z|^2} $$

The latter is a simple quadratic in $|z|^2$ which gives $|z|=1$ in the end.