Geometric interpretation of complex number.

Question

Let $z \in \mathbb{C}, z^2, z^3$ be the verticies of a right triangle. Find the geometric images of $z$.

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I did not understand the question but I guess it want me to find the figure formed by $z$ under these constraints.

Let $z = x + iy$, then $z^2 = x^2 - y^2 +2xyi$ and $z^3 = x^3 - iy^3 + 3x^2iy - 3xy^2$

It forms a triangle, So, $|z - z^2|^2 + |z^3 - z^2|^2 = |z - z^3|^2$ by Pythagoras theorem.

But when I put the values in I get polynomial in two degree with 6th degree with many extra terms, in short a complete mess.

How can I do this question without actually explanding the Pythogoras theorem ?

Answer 1

Complex number multiplication (and exponentiation) has a geometric interpretation. It is described for instance in this video. When you know that, the problem becomes just a problem of euclidean geometry and you should try to avoid algebra as much as possible.

In general for euclidean geometry problems, I can recommend tools for playing with them, especially GeoGebra. For this case choose point A and select "complex number". Create it by clicking to the plane. Activate "Input" on the left. Write "z_1"right arrow"^2"enter. Then similarly create z^3 and join them with the polygon tool. Then use the "mouse cursor tool" to move the original point $z_1$ and see what happens. Then you can add other geometric constructions and search for patterns.

You can solve the first part ($z-z^3$ being the hypotenuse) just by a very simple angle chasing:

By $\angle pq$ I denote the (oriented) angle between lines $p,q$, namely $\angle((A,B),(A,C))$ is the angle at the point $A$ in a triangle $ABC$.

By mentioned properties of complex numbers we know that $\angle((z^2,z)(z^2,0)) = \angle((z,1)(z,0))$ and $\angle((z^2,0)(z^2,z^3)$ = $\angle((1,0)(1,z))$. Now we can sum it up: $$ \angle((z^2,z),(z^2,z^3)) = \angle((z^2,z)(z^2,0)) + \angle((z^2,0)(z^2,z^3) $$$$ = \angle((0,1)(1,z)) + \angle((1,z)(0,z)) = \angle((0,1)(0,z)) $$ So the angle is right iff $z$ lies on the imaginary axis.

The other two cases are a bit more tricky in planimetry -- they lead to determining the angle $\angle((z,0)(z,z^3)) = \angle((1,0)(1,z^2))$. Now you can return back to the algebra for a while and use the identity $z^2-1 = (z-1)(z+1)$. Everything else is easily geometrically solvable ;-)

Answer 2

If the vertex at the right angle is $z^2$, then $\dfrac{z^3-z^2}{z-z^2}$ must be purely imaginary. Thus $\dfrac{z^2-z}{1-z} = -z$ is purely imaginary. Thus $z = ri$ for some real $r$. Let $z = re^{i\theta}$ be the vertex at the right angle. From the figure we have \begin{align*} AB^2 &= r^2 + r^4 -2r^3 \cos \theta \\ BC^2 &= r^4 + r^6 - 2r^5 \cos \theta\\ AC^2 &= r^2+r^6 - 2r^4 \cos 2\theta \end{align*} Using $BC^2 = AB^2 + AC^2$, we gave $$ (r^2 + r^4 -2r^3 \cos \theta)+ (r^2+r^6 - 2r^4 \cos 2\theta) = r^4 + r^6 - 2r^5 \cos \theta$$ Simplifying we get $$1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta = 0$$ When $\theta = 90^\circ$, we have $1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta = 1+r^2 > 0$ and when $\theta = 180^\circ$, we have $1 - r\cos\theta -r^2\cos 2\theta + r^3 \cos \theta = (1+r)(1-r^2)$ When $r > 1$, this quantity is negative and hence there is a $\theta$ that $\dfrac{\pi}{2} < \theta < \pi $ for which the above equation is satisfied. Hence there is a right angled triangle with $z$ as the vertex at the right angle. An example is shown below. Similar situation happens when the vertex at right angle is $z^3$.

Answer 3

Hint: let $r = |z| \in \mathbb{R}^+\,$, so that $\bar z = r^2 / z\,$.

It forms a triangle, So, $|z - z^2|^2 + |z^3 - z^2|^2 = |z - z^3|^2$ by Pythagoras theorem.

This assumes that the hypotenuse is $z\,z_3\,$, and the following works on this assumption. The other cases would follow similarly.

Rewriting it in terms of conjugates:

$$ \require{cancel} \begin{align} 0 & = |z - z^2|^2 + |z^3 - z^2|^2 - |z - z^3|^2 \\ & = (z - z^2)(\bar z - \bar z^2) + (z^3 - z^2)(\bar z^3-\bar z^2) - (z - z^3)(\bar z -\bar z^3) \\ & = \bcancel{r^2} + r^4 - z \bar z^2 - \bar z z^2 + \cancel{r^6} + r^4 - z^3 \bar z^2 - \bar z^2 z^3 - \bcancel{r^2} - \cancel{r^6} + z \bar z^3 + \bar z z^3 \\ & = r^2 (z^2 + \bar z^2) - (r^2+r^4) (z+\bar z) +2 r^4 \\ & = r^2 \left(z^2+ \frac{r^4}{z^2}\right) -r^2(r^2+1)\left(z + \frac{r^2}{z}\right) + 2 r^4 \\ & = r^2\left( \left( z + \frac{r^2}{z}\right)^2 - \bcancel{2 r^2} \right) -r^2(r^2+1)\left(z + \frac{r^2}{z}\right) + \bcancel{2 r^4} \\ & = r^2 \left( z + \frac{r^2}{z}\right) \left( z + \frac{r^2}{z} - r^2 -1 \right) \\ & = r^2 \left( z + \frac{r^2}{z}\right)\,\frac{(z-1)(z-r^2)}{z} \end{align} $$

The latter can be easily solved and, discarding the degenerate triangles given by $z=0,1\,$, the solutions are $z=\pm i \,r\,$ with $r \in \mathbb{R}^+\,$, which is the same as $z = \lambda i$ with $\lambda \in \mathbb{R} \setminus \{0\}\,$.