There are four complex numbers such that

\(z \overline{z}^3 + \overline{z} z^3 = 350,\)

and both the real and imaginary parts of \(z\) are integers. These four complex numbers are plotted in the complex plane. Find the area of the quadrilateral formed by the four complex numbers as vertices.

EpicWater Jan 1, 2020

#2**+1 **

The complex numbers are all different though

So I don't think this is correct. Not that I can do better or anything. Keep up the good work!

EpicWater Jan 1, 2020

#3**+1 **

\(z\bar{z}^3+\bar{z}z^3=350\\ (a+bi)(a-bi)^3+(a-bi)(a+bi)^3=350\\ (a^2-(bi)^2)[(a-bi)^2+(a+bi)^2]=350\\ (a^2+b^2)[(a^2-2abi-b^2)+(a^2+2abi-b^2)]=350\\ (a^2+b^2)[(2a^2-2b^2)]=350\\ (a^2+b^2)(a^2-b^2)=175\\ a^4-b^4=175 \)

So the solution are (

\(z=+4\pm3i\\ z=-4\pm3i\)

So assuming i have made no silly mistakes the area is 48 units squared

coding:

z\bar{z}^3+\bar{z}z^3=350\\

(a+bi)(a-bi)^3+(a-bi)(a+bi)^3=350\\

(a^2-(bi)^2)[(a-bi)^2+(a+bi)^2]=350\\

(a^2+b^2)[(a^2-2abi-b^2)+(a^2+2abi-b^2)]=350\\

(a^2+b^2)[(2a^2-2b^2)]=350\\

(a^2+b^2)(a^2-b^2)=175\\

a^4-b^4=175

Melody Jan 6, 2020