JEE PYQ: Complex Numbers Question 40
Question 40 - 2019 (11 Jan Shift 1)
Let $\left(-2-\frac{1}{3}i\right)^3 = \frac{x+iy}{27}$ $(i = \sqrt{-1})$, where $x$ and $y$ are real numbers, then $y - x$ equals:
(1) $91$
(2) $-85$
(3) $85$
(4) $-91$
Show Answer
Answer: (1)
Solution
$(-2-\frac{i}{3})^3 = -8 - 3(4)(-\frac{i}{3}) + 3(-2)\frac{1}{9} + \frac{i}{27} \cdot i^2$… Actually: $(-2-\frac{i}{3})^3 = (-1)^3(2+\frac{i}{3})^3 = -(8 + 3 \cdot 4 \cdot \frac{i}{3} + 3 \cdot 2 \cdot \frac{-1}{9} + \frac{-i}{27}) = -(8+4i-\frac{2}{3}-\frac{i}{27})$. $= -\frac{22}{3} - \frac{107i}{27} = \frac{x+iy}{27}$. So $x = -198, y = -107$. $y - x = 91$.
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