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4-1.Complex numbers
hard
माना $\left(-2-\frac{1}{3} i\right)^{3}=\frac{x+i y}{27}(i=\sqrt{-1})$, जहाँ $x$ तथा $y$ वास्तविक संख्यायें हैं, तो $y - x$ बराबर है
A
$91$
B
$-85$
C
$85$
D
$-91$
(JEE MAIN-2019)
Solution
$\left(-2-\frac{i}{3}\right)^{3}=\left(\frac{x+i y}{27}\right)$
$(-1)^{3}\left(2^{3}+\frac{i^{3}}{27}+3(2) \frac{i^{2}}{9}+3(2)^{2} \cdot \frac{i}{3}\right)=\frac{x-i y}{27}$
$-\left[8-\frac{i}{27}-\frac{2}{3}+4 i\right]=\frac{x+i y}{27}$
$\Rightarrow \frac{x}{27}=-8+\frac{2}{3}$
$\text { and } \frac{y}{27}=\frac{1}{27}-4$
$\frac{y-x}{27}=\frac{1}{27}-4+8-\frac{2}{3}$
$=\frac{1+27 \times 4-18}{27}$
$=\frac{109-18}{27}$
$=\frac{91}{27}$
$\Rightarrow y-x=91$
Standard 11
Mathematics
