Complex Numbers 1 Question 4
4. Let $-2-\frac{1}{3} i^{3}=\frac{x+i y}{27}(i=\sqrt{-1})$, where $x$ and $y$ are real numbers, then $y-x$ equals
(2019 Main, 11 Jan I)
(a) 91
(b) 85
(c) -85
(d) -91
Show Answer
Answer:
Correct Answer: 4. (a)
Solution:
- We have, $\frac{x+i y}{27}=-2-\frac{1}{3} i^{3}=\frac{-1}{3}(6+i)^{3}$
$$ \begin{aligned} \Rightarrow \frac{x+i y}{27} & =-\frac{1}{27}\left(216+108 i+18 i^{2}+i^{3}\right) \\ & =-\frac{1}{27}(198+107 i) \end{aligned} $$
$\left[\because(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}, i^{2}=-1, i^{3}=-i\right]$
On equating real and imaginary part, we get
$$ \begin{gathered} x=-198 \text { and } y=-107 \\ \Rightarrow \quad y-x=-107+198=91 \end{gathered} $$
