<s>[INST] Complex modulus of $\left|\frac{-3z+2i}{2iz+1}\right|$ given that $\left|z\right|=\frac{1}{\sqrt3}$ The following question was on my first year algebra exam way back in 1989.

If $\left|z\right|=\frac{1}{\sqrt3}$, then find $\left|\frac{-3z+2i}{2iz+1}\right|$. 

I couldn't figure it out then, and 28 years on, I still can't. It was only worth 4 marks, so the solution must be simpler than all the things I have tried over the years to no avail.
 [/INST] Since $ \mid Z \mid = \dfrac{1}{\sqrt{3}}$, considering $Z > 0$ then it follows that $Z = \dfrac{1}{\sqrt{3}}$. \ \
Now by the absolute value properties $ | \dfrac{-3z + 2i}{2iz + 1} | = \dfrac{| -3z + 2i |}{| 2iz + 1 |} $ \ \
Since $| -3z +2i | = \sqrt{(-3z+2i) (-3z - 2i)} = \sqrt{9z^2 +4} = \sqrt{9( \dfrac{1}{ \sqrt{3}})^2 + 4} = \sqrt{7}$
and $|2iz + 1 | = \sqrt{(1 +2iz)(1 - 2iz)} = \sqrt{1 + 4z^2} = \sqrt{\dfrac{7}{3}} $ \ \
Hence $ | \dfrac{ -3z + 2i}{2iz + 1} | = \dfrac{| -3z + 2i|}{|2iz + 1|} = \dfrac{\sqrt{7}}{\sqrt{\dfrac{7}{3}}} = \sqrt{3} $
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