4-1.Complex numbers
hard

જો ${\left( { - \,2\, - \,\frac{1}{3}\,i} \right)^3} = \frac{{x \,+ \,iy}}{{27}}(i\, = \,\sqrt { - 1} ),$ જ્યાં $x$ અને $y$ વાસ્તવિક સંખ્યા છે તો $y -x$ ની કિમત મેળવો. 

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

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