How can I show that modul of this complex number equal to 1? 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. 
$$5(a^2+b^2) \sqrt{a^2+b^2} \leqslant 5\sqrt{a^2 + b^2}\Leftrightarrow a^2 + b^2 \leqslant  1.$$

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.
