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There are four complex numbers \(z\) 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.

Tanks :)

 Jun 20, 2019
 #1
avatar+101856 
+1

Some of our members may be able to do this in a better manner....but...here goes....!!!!

 

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

 

(a + bi)(a - bi)(a - bi)^2 + (a - bi)(a + bi) ( a + bi)^2  = 350

 

(a^2 + b^2) (a - bi)^2 + (a^2 + b^2)( a + bi)^2  = 350

 

(a^2 + b^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

 

2 (a^2 + b^2)(a^2 - b^2) = 350

 

(a^2 + b^2)(a^2 - b^2)  = 175

 

(4^2 + 3^2) (4^2 - 3^2)  = 175

 

(4^2 + 3^2) (4 + 3) (4 - 3)  = 175

 

(25) (7) (1) = 175

 

So..by a little trial and error...let the first  complex number be  4+ 3i     and its conjugate be  4 - 3i

And let the second be -4 - 3i    and its conjugate be  -4 + 3i

 

And we have that

 

(4 + 3i)(4 - 3i)^3   + (-4 + 3i)(-4 - 3i)^3  =

(4 + 3i)(4-3i)(4 - 3i)^2  + (-4 + 3i) (-4 -3i)(-4 - 3i)^2  =

(16 + 9) ( 16 - 24i - 9) + ( 16 + 9) (16 + 24i -9)  =

(16 + 9)  [ 7 - 24i + 7 + 24i ]  =

(25) [ 14]  = 350

 

So.....the four  points are    4 + 3i  , 4 - 3i , -4 -3i  and  - 4 + 3i

 

We can make the  translation to Cartesian co-ordinates   (4, 3) (4 - 3) (-4, - 3) (-4, 3)

 

See the graph here : 

 

 

These points form an 8 x 6  rectangle....so....the area  = 48 units^2

 

 

cool cool cool

 Jun 20, 2019
 #2
avatar+87 
+1

Wow... that's a great solution! Thanks :)

Pushy  Jun 20, 2019
 #3
avatar+101856 
0

THX, Pushy.....I have a feeling that at least one of our members, heureka, can present this in a WAY better manner than I can....you might want to check back later....he may give his method.....

 

 

 

cool coolcool

CPhill  Jun 20, 2019
 #4
avatar+22546 
+2

There are four complex numbers  such that
\(\large{z \overline{z}^3 + \overline{z} z^3 = 350}\),
and both the real and imaginary parts of  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.

 

\(\text{Let $z=a+bi$}\\ \text{Let $\overline{z}=a-bi$}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{z \overline{z}^3 + \overline{z} z^3} &=& \mathbf{350} \\\\ z \overline{z}\left( \overline{z}^2+z^2 \right) &=& 350 \\\\ && \boxed{z\overline{z} = |z|^2 =a^2+b^2\\ \overline{z}^2+z^2 =a^2-2abi-b^2 +a^2+2abi-b^2 =2(a^2-b^2)} \\\\ 2(a^2-b^2)(a^2+b^2) &=& 350 \\ (a^2-b^2)(a^2+b^2) &=& 175 \quad | \quad 175 = 7\cdot 25 \\ (a^2-b^2)(a^2+b^2) &=& 7\cdot 25 \\\\ \text{if } a>b \\ a^2-b^2 &=& 7 \quad (1) \\ a^2+b^2 &=& 25 \quad (2) \\ \hline (1)+(2): \quad 2a^2 &=& 32 \\ a^2 &=& 16 \\ \mathbf{ a } &=& \mathbf{\pm4} \\ \hline (2)-(1): \quad 2b^2 &=& 18 \\ b^2 &=& 9 \\ \mathbf{ b } &=& \mathbf{\pm3} \\ \hline \hline \end{array}\\ \begin{array}{|lcll|} \hline \text{The four complex numbers are: $\mathbf{(4+3i), (4-3i), (-4-3i), (-4+3i)} $} \\ \text{The four vertices are: $\mathbf{(4,3), (4,-3), (-4,-3), (-4,3)} $} \\ \text{The area of this rectangle is $8\cdot 6 = \mathbf{48} $} \\ \hline \end{array}\)

 

 

laugh

 Jun 21, 2019

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