System Of Complex Equations

For 1, I'm going to use the same argument as before.

Without loss of generality, let $z_1$, $z_2$, $z_3$ and $z_4$ be arranged counter-clockwise around the origin.

The resultant of two equal [in magnitude] vectors lies on the bisector of the angle between them.

So, $z_1+z_2=a_1$ must lie on the bisector of the angle between $z_1$ and $z_2$.

Similarly, $z_3+z_4=a_2$ must lie on the bisector of the angle between $z_3$ and $z_4$.

Since $a_1+a_2=0$, we can conclude that they lie on the same line.

That means [you can see this visually, I don't know if it's rigorous] $z_1$ and $z_4$ lies on the same line and $z_3$ and $z_2$ lies on the same line.

And $|z_1|+|z_4|$ and $|z_2|+|z_3|$ are the diagonals of a quadrilateral and since they are equal and bisect one another, they are the diagonals of a rectangle.

You should add that the complex numbers are non-zero. I will work on 2 now.

I think we cannot conclude about both (1) and (2) . If we take z1,z3 as diametrical end points of unit circle and z2 , z4 as diametrical end points of same circle but a different diameter. It will satisfy both the conditions. And it is not necessary that both diameters will be perpendicular to each other. hence the 2 conditions are not sufficient to conclude that z1,z2,z3,z4 will be vertices of square

THE ARE THE ROOTS OF THE EQUATION Z^5=1

1) If you take any two numbers as diametrical points on a circle of radius 1. You get two diameters and if you rotate these two diameters about the origin independently you will get all the possible values of z1 z2 z3 z4 and they will not necessarily form a square 2) Similarly, If you take any three points as vertices of an equilateral triangle circumscribed by the circle of radius 1 and the remaining 2 points as diametrical points of the same circle you can rotate these two figures independently about the origin to get all the 5 points.

In general, for a system with n points, you can form subsets of regular polygon(s) and diameter(s) and rotate them to get the values.

$(Z_1*Z_2) / (Z_3*Z_4) =1$ ////no

According to me I think it the four complex numbers can lie on any rectangle

1) Let $z_1 = a+bi$. Then, $z_2 = \pm c \pm \sqrt{a^2+b^2-c^2}i$, $z_3 = -a-bi$, $z_4 = \mp c \mp \sqrt{a^2+b^2-c^2}i$, $c \in \mathbb{R}$ (or some permutation of this; WLOG $z_1$, $z_2$, $z_3$ and $z_4$ are arranged clockwise or counter-clockwise around the origin), i.e. we have that $z_1$ and $z_3$ are reflections of each other over the origin, as are $z_2$ and $z_4$. Geometrically, the object $z_1 z_2 z_3 z_4$ is a rectangle centered around the origin. I'll come back to you on 2)!

@Finn Hulse @Rajsuryan Singh @Adrian Neacșu @Harsh Depal @Vishal Sharma @Jack D'Aurizio @Mursalin Habib @Michael Mendrin Thanks for working on the 3 variable case. Care to tackle the 4 variable and 5 variable case?

that z1=z3=-1 and z2=z4=1

No, It is not necessary that they should lie on corners of square. Second condition gives points are collinear and first gives they lie on a circle with radius 1. example (1,0) (-1,0) (1/2,root3/2) (-1/2,-root3/2)

THEY ARE THE ROOTS OF THE EQUATION Z^5=0

well, since $\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =\left| { z }_{ 4 } \right| =1$ is true, ${ z }_{ 1 },{ z }_{ 2 },{ z }_{ 3 },{ z }_{ 4 }$ all lie on the unit circle with radius one and center as origin which is one obvious conclusion you make...so they make a cyclic quadrilateral when joined..and one of the properties of a cyclic quadrilateral states that the sum of the opposite angles is always $\quad \pi$ radians... from the statement 2 given ${ z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }+{ z }_{ 4 }=0$ using ${ z }_{ i }=\cos { { \theta }_{ i } } +i\sin { { \theta }_{ i } }$ i get $\cos { { \theta }_{ 1 } } +\cos { { \theta }_{ 2 } } +\cos { { \theta }_{ 3 } } +\cos { { \theta }_{ 4 } } =0$ and similarly $\sin { { \theta }_{ 1 } } +\sin { { \theta }_{ 2 } } +\sin { { \theta }_{ 3 } } +\sin { { \theta }_{ 4 } } =0$ and i know ${ \theta }_{ 1 }+{ \theta }_{ 2 }+{ \theta }_{ 3 }+{ \theta }_{ 4 }=2\pi$ if that cyclic quadrilateral were to be a square,,then i would have the sum of any two angles as $\quad \pi$..but from the two conditions we just got.. $2\cos { \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } } \cos { \frac { { \theta }_{ 1 }-{ \theta }_{ 2 } }{ 2 } } =(-2)\cos { \frac { { \theta }_{ 3 }+{ \theta }_{ 4 } }{ 2 } } \cos { \frac { { \theta }_{ 3 }-{ \theta }_{ 4 } }{ 2 } }$

$2\sin { \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } } \cos { \frac { { \theta }_{ 1 }-{ \theta }_{ 2 } }{ 2 } } =(-2)\sin { \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } } \cos { \frac { { \theta }_{ 3 }-{ \theta }_{ 4 } }{ 2 } }$ we get ${ \tan { \frac { { \theta }_{ 1 }+{ \theta }_{ 2 } }{ 2 } } }^{ 2 }=1$ if it were a square i should get the sum of any two angles as $\pi$ but that doesn't satisfy the condition we just ended on..so..i conclude the points don't have to make a square...

this is my thinking....i might be wrong..if so,then i would be glad to know the flaw in my approach...

$z_{1}, z_{2}, z_{3},$ and $z_{4}$ are vertices of a rectangle ABCD. Here is the proof(rigorous):

In general, we can prove it for any $|z| = a(>0)$ rather than 1.

First, you should know that the value of $|p+q|^2 = 2|p^2|(1 + \cos \theta)$, where $\theta$ is theta is the angle that complex numbers $p$ and $q$ lying on a circle centered about origin subtend at origin.

Let $\alpha$ be the angle between lines OA and OB, and $\beta$ be the angle between lines OC and OD,

$z_{1}+ z_{2} = -(z_{3}+z_{4})$

Taking modulus and squaring, we get:

$2|z^2|(1+ \cos \alpha|) = 2|z^2| (1 + \cos \beta)$

$\cos \alpha = \cos \beta$ , or $\alpha = 2 \pi - \beta$ (not possible)

Similarly, if $\gamma$ and $\delta$ are angles between $OA$ and $OD$ , and $OB$ and $OC$ respectively, we can prove that

$\gamma = \delta$

Now, as $\alpha, \beta$, and $\gamma, \delta$ are pairs of vertically opposite angles, we get that

Lines AOC , BOD are straight, i.e. the diagonals intersect at O.

Thus, all angles of the quadrilateral are angle in semicircle, i.e. $90^{\circ}$ , and the quadrilateral is rectangle.

Also, it is not necessiraly a square. As a counter example, $z , \bar{z},-z , -\bar{z}$ always satisfy the given relations, but don't necessarily form a square.

i think it's not possible .if condition 1 is true ,then 2 must be false and vice-verse