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$\DeclareMathOperator{\aff}{Aff}$ Here is a funny problem I stumbled upon recently

I call the affix of a complex number the corresponding point in the argand plane, that is

$$\aff(\mathrm{i})=(0;1)$$

Find all $z \in \mathbf{C}$ such that the affixes of $z,z^2\ \& \ z^3$ form a right triangle.

I decided to Set $A=\aff(z),B=\aff(z^2),C=\aff(z^3)$

I have proven that :

  1. If $z$ is a solution, then $z\neq 0,\ z\neq 1, \ \& \ z\neq-1$
  2. If the triangle is right angled at A, $z$ must be a non zero imaginary number.
  3. If the triangle is rigth angled at B, $\aff(z)$ has to lie on the vertical line defined by $x=(-1)$, except for $z=(-1)$.

(I'll post my proof when I get some time to write it in $\mathrm{\LaTeX}$ soon.)

I'm down to the third case : I have proven that if the triangle is right angled at $C$, then : $$\Re \left(\frac{1+z}{z}\right)=0$$

Thanks to some experimentation, I know that $\aff(z)$ should lie on the circle centered at $(-\frac12;0)$ with radius $\frac12$, with $z$ again respecting the first condition above.

Now,how do I prove this ?

Thanks for the help.

Here is an animation.

Third case

T.D
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    The use of the word affix is uncommon, i think it's because is outdated, and this may confuse some readers. In the modern use of complex numbers one simply think the complex numbers as points in the complex plane. – jjagmath Dec 07 '20 at 15:59
  • yes, I am well-aware, but I have to teach it as it is a mandatory term in french exams, so I stuck to the old vocabulary. – T.D Dec 07 '20 at 16:02
  • I first come across the term "affix" maybe a week or so ago (I'm born-and-raised in the US), but now I seem to be noticing it more on Math.SE. ("Frequency illusion", I suppose.) Interestingly, a site search reveals sporadic uses here going back a decade. I'm surprised I'd never picked up on it in all that time. ... Anyway, I appreciate the context. – Blue Mar 21 '21 at 13:10

3 Answers3

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Partial Solution:

If $z=\cos t+i\sin t,$

$$m_{AB}=\dfrac{\sin2t-\sin t}{\cos 2t-\cos t}=-\cot\dfrac{3t}2\text{ if }\sin\dfrac t2\ne0$$

Similarly, $$ m_{BC}=-\cot\dfrac{5t}2, m_{CA}=-\cot2t$$

Now we need at exactly one of the products $m_{AB}\cdot m_{BC},m_{BC}\cdot m_{CA},m_{CA}\cdot m_{AB}$ to be $=-1$

lab bhattacharjee
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HINT

Angle between vectors $\vec{KM}, \vec{LM},$ where the points have complex coordinates $z_K, z_L, z_M,$ is equal to $$\arg{\frac{z_M-z_L}{z_M-z_K}}.$$ In the present exercise we want

$$\arg{\frac{z-z^2}{z-z^3}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$ or $$\arg{\frac{z^2-z}{z^2-z^3}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$ or $$\arg{\frac{z^3-z^2}{z^3-z}}=\frac{\pi}{2}+k\pi,\;k\in \mathbb{Z}$$

user376343
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  • yes this is exactly what is written there : if the real part is zero, the argument is +-pi/2 you forgot the minus sign :) – T.D Dec 07 '20 at 16:17
  • moreover it's easier to justify saying it's an imaginary number, because if there's an inverse (which happens here), the justification is trivial. – T.D Dec 07 '20 at 16:59
  • I did not forgotten the sign. My $k$ in $\frac{\pi}{2}+k\pi$ can be negative. – user376343 Dec 07 '20 at 19:49
  • yes I saw the Z after writing my comment sry – T.D Dec 07 '20 at 20:19
1

Oh dear,

It turns out I'm too tired to think, I found the following : let's set $z=a+ib$ as in the question

then : $$\frac{z+1}{z}=\frac{(a+1)+ib}{a+ib}=\frac{(a+1)+ib)(a-ib)}{a^2+b^2}$$ From which one gets $$\frac{z+1}{z}=\frac{a+a^2+b^2-ib}{a^2+b^2}=\frac{a+a^2+b^2}{a^2+b^2}-\frac{ib}{a^2+b^2}$$

now, because this is an imaginary number we get $$C(a,b)=a+a^2+b^2=0$$ which is the equation of an ellipse because it is a polynomial of two variables, of degree two. We see that there is no division, therefore, this is the equation of a circle

Let's rewrite it to see the circle's centre and radius : $$C(a,b)=a+a^2+(b-0)^2$$ That is with two canonical forms $$C(a,b)=\left(a+\frac{1}{2}\right)^2+(b-0)^2-\frac14=0$$ But then we get $$\left(a-\left(-\frac{1}{2}\right)\right)^2+(b-0)^2=\frac14$$ We just need a square after the equal sign to get the radius $$\left(a-\left(-\frac{1}{2}\right)\right)^2+(b-0)^2=\left(\frac12\right)^2$$

which is the equation of the circle mentioned in the question / OP.

Please note however that $(-1)$ and $0$ both belong to this circle.

T.D
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