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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.

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10 Answers 10

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

$\quad C = \{\frac 1{\sqrt3}e^{it} \mid t \in [0,2\pi]\}$, $\color{red}{f(C) = \{\sqrt3 e^{it} \mid t \in [0,2\pi]\}}$ Animation of the image of the given circle.

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 $\sqrt3$.

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Hint:

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

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    $\begingroup$ Not really a hint ;-), but nicely done. $\endgroup$
    – StackTD
    Apr 5, 2017 at 15:42
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    $\begingroup$ @StackTD Thanks. I got a bit carried away, indeed ;-) It's still a couple of fill-in-the-blanks short of a truly complete answer. $\endgroup$
    – dxiv
    Apr 5, 2017 at 15:46
  • $\begingroup$ Indeed. It's fine! $\endgroup$
    – StackTD
    Apr 5, 2017 at 15:46
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$$\left|\frac{-3z+2\mathrm i}{2\mathrm iz+1}\right| = \left|\frac{-3z\bar z+2\mathrm i\bar z}{(2\mathrm iz+1)\bar z}\right| = \left|\frac{1}{\bar z}\right|\left|\frac{-1+2\mathrm i\bar z}{2\mathrm iz+1}\right|=\sqrt{3}\underbrace{\left|-\frac{(\overline{1+2\mathrm iz})}{1+2\mathrm iz}\right|}_{=1}=\sqrt{3}$$

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The modulus can be divided over division of complex numbers. Essentially:

$$\left|\frac{-3z+2i}{2iz+1}\right| = \frac{\left|-3z+2i\right|}{\left|2iz+1\right|}$$ Let $z=x+iy$, then: $$=\dfrac{|-3x+(2-3y)i|}{|1-2y+2xi|}$$

Now use $x^2 + y^2 = \dfrac{1}{3}$, we have

$$=\sqrt{\dfrac{9x^2 + 4-12y+9y^2}{1-4y+4y^2+4x^2}}\\ =\sqrt{\dfrac{7-12y}{1+\frac{4}{3}-4y}}\\ = \sqrt{3}$$

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    $\begingroup$ Trying $z = 1/\sqrt{3}$ and $z = i/\sqrt{3}$, the end result turns out to be $\sqrt{3}$. That's not proof that it will always be $\sqrt{3}$, but I would imagine we actually can say more, in this case. $\endgroup$
    – pjs36
    Apr 5, 2017 at 15:22
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We have $$ \biggl| \dfrac{2i-3z}{2iz+1} \biggr|^2 = \dfrac{2i-3z}{2iz+1} \cdot \dfrac{-2i-3\overline{z}}{-2i\,\overline{z}+1} = \dfrac{4+6i(z-\overline{z}\,)+9|z|^2}{4|z|^2+2i(z-\overline{z}\,)+1} = \dfrac{7+6i \cdot 2i(z-\overline{z}\,)}{\tfrac{7}{3} + 2i \cdot 2i (z-\overline{z}\,)} = 3, $$ so $|(2i-3z)/(2iz+1)| = \sqrt{3}$.

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    $\begingroup$ dxiv beat me to it :) $\endgroup$
    – Mark Twain
    Apr 5, 2017 at 15:51
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This could be done thinking about the geometric effect of multiplying by complex numbers and using pythagoras.

Since all we know to start with is $|z|=1/\sqrt3$, you can assume $z=1/\sqrt3$ without losing any generality because rotating the frame of reference will not change a length.

For the numerator, $-3z$ has length $\sqrt3$ and lies along the negative real axis, $-3z+2i$ adds 2 imaginary units to this. By pythagoras for this right-angled triangle $|-3z+2i|=\sqrt7$

For the denominator, $2iz$ rotates our chosen $z$ to the positive imaginary axis and doubles its length to $2/\sqrt3$ and $2iz+1$ adds one real unit so again by pythagoras $|2iz+1|=\sqrt{4/3+1}=\sqrt{7/3}$

So $$\left| \frac{-3z+2i}{2iz+1} \right|=\frac{|-3z+2i|}{|2iz+1|}=\frac{\sqrt7}{\sqrt{7/3}}=\sqrt3$$

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Note that $z \overline{z} = \frac{1}{3}$

Then Numerator $= \left|-3z+2i \right| =\left|-\frac{z}{z\overline{z}} +2i \right| =\left|-\frac{1}{\overline{z}} +2i \right| = \frac{|-1+2\overline{z}i|}{|\overline{z}|}$

Now we know that $ |u| = |\overline{u}|$

So $|-1+2\overline{z}i| = |-1-2zi| = |1+2zi|$

So Numerator $= \frac{|1+2zi|}{|z|}$

Its clear now that the given expression $ = \boxed{\sqrt 3}$

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    $\begingroup$ You mean $z\bar z = \frac 13$. $\endgroup$ Apr 5, 2017 at 16:30
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Tricky answer: if there is a unique solution to this problem, then the answer must be the same for each value of $z$ such that $|z|=\frac{1}{\sqrt3}$. So just replace $z$ with the real number $\frac{1}{\sqrt3}$ and see what $$ \left|\frac{-3\frac{1}{\sqrt3}+2i}{2i\frac{1}{\sqrt3}+1}\right| $$ is.

Warning: your prof might not have liked this solution. But in case it is a multiple-choice question, of course it doesn't matter.

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$$z=a+bi\rightarrow ||z||=\sqrt{a^{2}+b^{2}}=\dfrac{1}{\sqrt{3}}\rightarrow a^{2}+b^{2}=\dfrac{1}{3}$$ $$2iz+1=(-2b+1)+2ai\rightarrow$$ $$||2iz+1||=\sqrt{(-2b+1)^{2}+(2a)^{2}}=\sqrt{4(a^{2}+b^{2})+1-4b}=\sqrt{\dfrac{7}{3}-4b}$$ $$-3z+2i=(-3a)+(2-3b)i\rightarrow$$ $$||-3z+2i||=\sqrt{(-3a)^{2}+(2-3b)^{2}}=\sqrt{9(a^{2}+b^{2})+4-12b}=\sqrt{7-12b}$$ $$\left|\left|\dfrac{-3z+2i}{2iz+1}\right|\right|=\dfrac{|-3z+2i|}{|2iz+1|}=\dfrac{\sqrt{7-12b}}{\sqrt{\frac{7}{3}-4b}}=\sqrt{\dfrac{7-12b}{\frac{7-12b}{3}}}=\sqrt{3}$$

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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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