+296 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.
+296 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!
+118687 \(z\bar{z}^3+\bar{z}z^3=350\\ (a+bi)(a-bi)^3+(a-bi)(a+bi)^3=350\\ (a^2-(bi)^2)[(a-bi)^2+(a+bi)^2]=350\\ (a^2+b^2)[(a^2-2abi-b^2)+(a^2+2abi-b^2)]=350\\ (a^2+b^2)[(2a^2-2b^2)]=350\\ (a^2+b^2)(a^2-b^2)=175\\ a^4-b^4=175 \)
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
