Complex Numbers 2 Question 10
10. Let $z=x+i y$ be a complex number where, $x$ and $y$ are integers. Then, the area of the rectangle whose vertices are the root of the equation $z \bar{z}^{3}+\bar{z} z^{3}=350$, is
(2009)
(a) 48
(b) 32
(c) 40
(d) 80
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Answer:
Correct Answer: 10. (a)
Solution:
- Since,
$$ z \bar{z}\left(z^{2}+\bar{z}^{2}\right)=350 $$
$\Rightarrow \quad 2\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)=350$
$\Rightarrow \quad\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)=175$
Since, $x, y \in I$, the only possible case which gives integral solution, is
$$ \begin{aligned} x^{2}+y^{2} & =25 \\ x^{2}-y^{2} & =7 \end{aligned} $$
From Eqs. (i) and (ii),
$$ x^{2}=16, y^{2}=9 \Rightarrow x= \pm 4, y= \pm 3 $$
$\therefore$ Area of rectangle $=8 \times 6=48$
