Consider complex numbers $z,z^2,z^3,z^4$ in that order which form a cyclic quadrilateral . If $\arg z=\alpha$ and $\alpha$ lies in $[0,2\pi]$.Find the values $\alpha$ can take.

I encountered this question in one competitive exam.i tried using the property of cyclic quadrilateral to get $$\arg\left(\frac{z^3-z^4}{z-z^4}\right)+\arg\left(\frac{z-z^2}{z^3-z^2}\right)=\pi$$ This can be simplfied furher but does not help.

I also tried using coni theorem but of no use. Answer given is alpha lies in $(0,\frac{2\pi}{3})and(\frac{4\pi}{3},2\pi)$

  • $\begingroup$ Rats! The question edit ruins my answer! $\endgroup$ Aug 18 '20 at 13:54
  • 1
    $\begingroup$ oh i am sorry i now only managed to look at the question properly. $\endgroup$ Aug 18 '20 at 13:56

Here is an alternative way to render $|z|=1$ -- by mathematical induction, of all things.

Suppose that $z,z^2,z^3,z^4$ lie on a circle for nonzero $z$. Then through multiplying all elements by $z$ we infer that $z^2,z^3,z^4,z^5$ also lie on a circle, which must be the same as the first circle because of the overlapping three points $z^2,z^3,z^4$. Similarly $z^6,z^7,...$ lie on the same circle.

Now go the other way. Given $z,z^2,z^3,z^4$ on a circle divide by $z$, then $1,z,z^2,z^3$ also lie on a circle which is again the same as the initial one. Repeating this process we find $z^{-1},z^{-2},...$ also lie on this circle.

Thus the same circle contains all points with the form $z^n$ for all integers $n$, positive, negative and zero. But the circle has to be bounded and the set of powers just identified is bounded only for $|z|=1$.

Given $|z|=1$, how the argument is restricted is a matter of definition. If we require the points $z,z^2,z^3,z^4$ to be in rotational order in the quadrilateral, then we must have one of two cases:

  • If the order is counterlockwise, then $0<\alpha<2\pi/3$ because to preserve the rotational order we have to have $\arg z^4-\arg z=3\alpha<2\pi$.

  • If the order is clockwise, then the inverse powers $z^{-1},z^{-2},z^{-3},z^{-4}$ are in counterclockwise order and we now require $\arg z^{-4}-\arg z^{-1}=3\alpha<2\pi$. This gives the second set $4\pi/3<\alpha<2\pi$ if arguments are taken to be in $[0,2\pi)$.

But, arguably, the points still lie on the circle even if they are not in this rotational order, so the cyclic quadrilateral exists unless it is degenerated by pairs of vertices coinciding. Such a coincidence occurs only if $n\alpha$ is multiple of $2\pi$ for $n\in\{1,2,3\}$. So from this point of view $\alpha$ can be anything at all in $[0,2\pi]$ except $0,2\pi/3,\pi,4\pi/3,2\pi$.

  • $\begingroup$ ok but how do you find the range of the values that $\alpha$ can take $\endgroup$ Aug 18 '20 at 13:28
  • $\begingroup$ Actually I disagree with the other answers about that. My edit is coming ... . $\endgroup$ Aug 18 '20 at 13:32
  • $\begingroup$ the expression $\arg z^4-\arg z=3\alpha<2\pi$.made all the difference i got it now thanks! $\endgroup$ Aug 18 '20 at 14:09

By Ptolemy we obtain: $$|z-z^2|\cdot|z^3-z^4|+|z-z^4|\cdot|z^2-z^3|=|z-z^3|\cdot|z^2-z^4|$$ or $$|z|+|z^2+z+1|=|(z+1)^2|.$$ Now, we can use a triangle inequality.

Id est, for $|z|=r$ we obtain: $$(\cos\alpha,\sin\alpha)||(r^2\cos2\alpha+r\cos\alpha+1,r^2\sin2\alpha+r\sin\alpha),$$ which gives $$\sin\alpha(r^2\cos2\alpha+r\cos\alpha+1)=\cos\alpha(r^2\sin2\alpha+r\sin\alpha)$$ or $$\sin\alpha=r^2\sin\alpha$$ and since $\sin\alpha\neq0$, we obtain $r=1$.

  • $\begingroup$ okay so $argz=arg(z^2+z+1).$ but how can we get range $\alpha$ $\endgroup$ Aug 18 '20 at 5:02
  • $\begingroup$ The argument is equal as long as $-\frac{2\pi}{3}\leq\alpha\leq\frac{2\pi}{3} $ $\endgroup$ Aug 18 '20 at 5:08

Like in Michael’s solution, use Ptolemy to obtain $|z|+|z^{2}+z+1|=|z^{2}+2z+1|$.

Refer to the picture, it is obvious that $|z^{2}|=1$ and consequentlly $|z|=1$. For $-\frac{2\pi}{3}\leq\alpha\leq\frac{2\pi}{3}$ the equation is valid. Hint: at which angle $\alpha$ does the direction of $z^{2}+z+1$ become opposite of $z$?

Argand plane


For the modulus issue, let us use the classical equivalence (see here):

$$a,b,c,d \ \text{constitute a cyclic quadrilateral} \ \iff \ $$ $$\underbrace{[a,c;b,d]}_{\text{cross ratio}}=\frac{(b-a)}{(b-c)} /\frac{(d-a)}{(d-c)} \ \text{is real}\tag{1}$$

In our case, (1) becomes:

$$[z,z^3;z^2,z^4]=\left(\frac{z^2-z}{z^2-z^3}\right) \times \left(\frac{z^4-z^3}{z^4-z}\right) \in \mathbb{R}\tag{2}$$

Taking into account different simplifications coming in particular from $z^3-1=(z-1)(z^2+z+1)$, (2) is equivalent to :

$$z+1+\tfrac{1}{z} \in \mathbb{R} \ \iff \ Im\left(z+1+\tfrac{1}{z}\right)=0$$

otherwise said, with $z=re^{i\theta}$,

$$(r-\tfrac1r) \sin(\theta)=0$$

as $\theta \ne k \pi$ (such values would give degenerated quadrilaterals), we have necessarily $r-\tfrac1r=0$, giving $r=1$.

For the angle issue, let us assume that $z=re^{i \theta}$ with $0<\theta<\pi$ without loss of generality (this is up to a symmetry with respect to the $x$-axis). It is equivalent to make a reasoning on $1,z,z^2,z^3$ which are points obtained from $z,z^2,z^3,z^4$ by a $-\theta$ rotation. It is geometrically clear that a necessary condition is that $z^3$ has an argument less than $2 \pi$ (otherwise, the order of points $1$ and $z^3$ wouldn't be respected). This condition $arg(z^3)<2 \pi$ gives

$$0<3\alpha<2\pi \ \iff \ 0<\alpha<2\pi/3\tag{3}$$

Moreover, this condition is in fact sufficient: all $\alpha$s verifying (3) give an adequate solution.


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