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# Precalc

0
105
5
+287

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

### 5+0 Answers

#1
+1

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

Jan 1, 2020
#2
+287
+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
+108734
+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
+287
+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
+108734
0

Good work EpicWater.

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

Melody  Jan 7, 2020