Question

Let \(z=x+iy\) be a complex number with \(x,y\in \mathbb{Z}\). Then the area (in square units) of the rectangle whose vertices are the roots of the equation \[ z\overline{z}^{\,3}+\overline{z}z^{3}=350 \] is:

• A. \(48\)
• B. \(32\)
• C. \(40\)
• D. \(44\)

Hint:

For equations involving both \(z\) and \(\overline z\), first convert them into expressions involving \(x\) and \(y\). Then use factorization and integer constraints to obtain the roots.

Answer 1

The Correct Option is A

Step 1: Express the equation in terms of \(x\) and \(y\).
Let \[ z=x+iy, \qquad \overline{z}=x-iy \] Then, \[ z\overline{z}=x^2+y^2 \] and \[ z\overline{z}^{\,3}+\overline{z}z^{3} = z\overline{z}\left(\overline{z}^{\,2}+z^2\right) \] Hence, \[ (x^2+y^2)\Big((x-iy)^2+(x+iy)^2\Big)=350 \]

Step 2: Simplify the expression.
Now, \[ (x-iy)^2+(x+iy)^2 = 2(x^2-y^2) \] Therefore, \[ 2(x^2+y^2)(x^2-y^2)=350 \] \[ (x^2+y^2)(x^2-y^2)=175 \] \[ x^4-y^4=175 \]

Step 3: Factorize.
\[ (x^2-y^2)(x^2+y^2)=175 \] Since \[ 175=1\times175=5\times35=7\times25 \] and \(x,y\) are integers, let \[ x^2-y^2=7, \qquad x^2+y^2=25 \] Adding, \[ 2x^2=32 \] \[ x^2=16 \] \[ x=\pm4 \] Then, \[ y^2=25-16=9 \] \[ y=\pm3 \]

Step 4: Determine the roots.
The roots are \[ 4+3i,\quad 4-3i,\quad -4+3i,\quad -4-3i \] These form a rectangle with \[ \text{length}=8 \] and \[ \text{breadth}=6 \]

Step 5: Find the area.
\[ \text{Area}=8\times6 \] \[ =48 \]

Step 6: Final conclusion.
Therefore, \[ \boxed{48} \]