Complex modulus of $\left|\frac{-3z+2i}{2iz+1}\right|$ given that $\left|z\right|=\frac{1}{\sqrt3}$

Solution 1:

If the problem troubled you for this long, you deserve a picture.

$\quad C = \{\frac 1{\sqrt3}e^{it} \mid t \in [0,2\pi]\}$, $\color{red}{f(C) = \{\sqrt3 e^{it} \mid t \in [0,2\pi]\}}$

Since $$ \mathbb C \ni z\mapsto f(z) = \frac{-3z + 2i}{2iz +1} = \sqrt3\frac{-\frac {\sqrt3}2\sqrt3z + i}{\frac {\sqrt3}2 + \sqrt3zi} \in \mathbb C$$ is a linear fractional transformation, the image of the circle $\{z \in \mathbb C : |z| = \frac 1 {\sqrt 3} \}$ is either a circle or a line. By plugging in $\pm \frac{i}{\sqrt3} $ and $\frac{-1}{\sqrt3}$ we learn that the image goes through the points \begin{align} f\left(\pm \frac{i}{\sqrt3}\right) &= \sqrt{3}\color{green}{i\frac{{\mp\frac{\sqrt3}2 + 1} }{\frac{\sqrt3}2 \mp 1}} = \color{green}{\pm i}\sqrt3, \\&\text{and } \\ f\left(\frac{-1}{\sqrt3}\right) &= \sqrt3\color{green}{\frac{\frac{\sqrt3}{2} +i}{\frac{\sqrt3}{2} - i}}. \end{align}

The $3$ $\color{green}{\text{green}}$ vectors are non-collinear unit vectors, so the image has to be a circle, which they uniquely determine: the circle around $0$ with a radius of $\sqrt3$.

Solution 2:

Hint:

$$ \begin{align} \left|\frac{-3z+2i}{2iz+1}\right|^2 &= \frac{(-3z+2i)(-3\bar z - 2i)}{(2iz+1)(-2i \bar z+1))} \\ &= \frac{9|z|^2+4+6i(z-\bar z)}{4|z|^2+1+2i(z-\bar z)} \\ &= \frac{7 + 6i(z-\bar z)}{\frac{7}{3}+2i(z-\bar z)} \\[5px] &= 3 \end{align} $$

Solution 3:

$$\left|\frac{-3z+2\mathrm i}{2\mathrm iz+1}\right| = \left|\frac{-3z\bar z+2\mathrm i\bar z}{(2\mathrm iz+1)\bar z}\right| = \left|\frac{1}{\bar z}\right|\left|\frac{-1+2\mathrm i\bar z}{2\mathrm iz+1}\right|=\sqrt{3}\underbrace{\left|-\frac{(\overline{1+2\mathrm iz})}{1+2\mathrm iz}\right|}_{=1}=\sqrt{3}$$