Question

Suppose $z_1+z_2+z_3+z_4=0$ and $|z_1|=|z_2|=|z_3|=|z_4|=1if{z}_1,z_2,z_3,z_4$ are the vertices of a quadrilateral, then the quadrilateral must be a

• A. parallelogram
• B. rhombus
• C. rectangle
• D. square

Answer 1

Answer: (A), (C)

The correct options are
A rectangle
C parallelogram

$z_1+z_2+z_3+z_4=0$ .... (i)

${\overline{z}}_1+{\overline{z}}_2+{\overline{z}}_3+{\overline{z}}_4=0$

$\because |z_1|=1so{\overline{z}}_1=\frac{1}{z_1}$

$\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\frac{1}{z_4}=0$

$z_1{z}_2{z}_3+z_1{z}_2{z}_4+z_1{z}_3{z}_4+z_2{z}_3{z}_4=0$ .... (ii)

Now the equation where roots are $z_1,z_2,z_3,z_4$ is

$(z-z_1)(z-z_2)(z-z_3)(z-z_4)=0$ (using (i) and (ii) )

or $z^4+b{z}^2+d=0$ .... (iii)

Clearly if $\alpha$ is a root of (iii) then

$-\alpha$ is also a root of equation (iii)

so root of (iii) are of the form $\alpha ,\beta ,-\alpha ,-\beta$

$z_1+z_3=0z_2+z_4=0$

$\because z_1,z_2,z_3,z_4$ are vertices of parallelogram

Also $|z_1-z_3{|}^2$

$=|z_1{|}^2+|z_3{|}^2-z_1{\overline{z}}_3-{\overline{z}}_1{z}_3=2-z_1(-{\overline{z}}_1)-{\overline{z}}_1(-{\overline{z}}_1)=2+2=4$

$|z_1-z_3|=2similarly|z_2-z_4|=2$

Thus $z_1,z_2,z_3,z_4$ are vertices of a rectangle

($\because$ diagonals are equal length and bisect each other