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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.

 Jan 1, 2020
 #1
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+1

The four complex nubmers are \(\pm 5 \pm 4i\), so the area is 2*5*2*4 = 80.

 Jan 1, 2020
 #2
avatar+163 
+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!

 Jan 1, 2020
 #3
avatar+106887 
+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

 Jan 6, 2020
 #4
avatar+163 
+1

Good Job!!! Melody, you are correct. I also got 48. A bit late on your response, but I answered the problem before seeing you post this. 

 Jan 7, 2020
 #5
avatar+106887 
0

Good work EpicWater.

It is always good when you can solve it by yourself!   laugh

Melody  Jan 7, 2020

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