Prove that : $\frac{3 \omega-1}{\omega^{2}-2 i}=1+2 \omega i$

Step 1 ：Given the complex number equation: \(\frac{3 \omega-1}{\omega^{2}-2 i}=1+2 \omega i\)

Step 2 ：First, multiply both sides by the conjugate of the denominator on the left side to simplify the equation. The conjugate of a complex number a + bi is a - bi. So, the conjugate of \(\omega^{2}-2 i\) is \(\omega^{2}+2 i\).

Step 3 ：Then, we get the equation: \((3\omega - 1)(\omega^{2} + 2i) = (\omega^{2} - 2i)(2i\omega + 1)\)

Step 4 ：Solving this equation, we find that there are three possible values for \(\omega\) that satisfy the equation.

Step 5 ：Final Answer: The solutions to the equation are \(\boxed{0}\), \(\boxed{\frac{3}{13} + \frac{2i}{13} - \sqrt{1 - 26i}\frac{3 + 2i}{13}}\), and \(\boxed{\frac{3}{13} + \sqrt{1 - 26i}\frac{3 + 2i}{13} + \frac{2i}{13}}\).