Let the set of complex numbers (x1,y1),(x2,y2),(x3,y3),.......... denoting the points on the complex plane satisfying (xn+1,yn+1)=(3xn−yn,3yn+xn) for n=1,2,3,....... suppose (x100,y100)=(2,4) then the value of x1+y1 is equal to

The point (x1,y1),(x2,y2),...(xn+1,yn+1) Given (xn+1+i yn+1)=Zn+1(say)∴Zn+1=3xn−yn+i3yn+xn=3xn+iyn−yn+ixn =3xn+iyn+ixn−yniii=3xn+iyn+ix+iyn =3Zn+iZn=Zn(3+i) where Zn=xn+iyn =2Zncos⁡π6+isin⁡π6Zn+1=2Zneiπ6Zn+1Zn=2eiπ6=K(say)putting n=1,2,3....99, we getZ2Z1=K,Z3Z2=K,...Z100Z99=KMultiplying above all, we getZ100Z99=K99=299ei33π2=299i⇒Z100=299Z1i⇒Z1=Z100299i=2+4i299i⇒Z1−2−97−i2−98=x1+iy1⇒x1+y1=2−97−2−98=2−98(2−1)=2−98

Let the set of complex numbers (x1,y1),(x2,y2),(x3,y3),.......... denoting the points on the complex plane satisfying (xn+1,yn+1)=(3xn−yn,3yn+xn) for n=1,2,3,....... suppose (x100,y100)=(2,4) then the value of x1+y1 is equal to

The point (x1,y1),(x2,y2),...(xn+1,yn+1) Given (xn+1+i yn+1)=Zn+1(say)∴Zn+1=3xn−yn+i3yn+xn=3xn+iyn−yn+ixn =3xn+iyn+ixn−yniii=3xn+iyn+ix+iyn =3Zn+iZn=Zn(3+i) where Zn=xn+iyn =2Zncos⁡π6+isin⁡π6Zn+1=2Zneiπ6Zn+1Zn=2eiπ6=K(say)putting n=1,2,3....99, we getZ2Z1=K,Z3Z2=K,...Z100Z99=KMultiplying above all, we getZ100Z99=K99=299ei33π2=299i⇒Z100=299Z1i⇒Z1=Z100299i=2+4i299i⇒Z1−2−97−i2−98=x1+iy1⇒x1+y1=2−97−2−98=2−98(2−1)=2−98

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Let the set of complex numbers $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), . .........$ denoting the points on the complex plane satisfying $(x_{n + 1}, y_{n + 1}) = (\sqrt{3} x_{n} - y_{n}, \sqrt{3} y_{n} + x_{n})$ for $n = 1, 2, 3, . ......$ suppose $(x_{100}, y_{100}) = (2, 4)$ then the value of $x_{1} + y_{1}$ is equal to

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$2^{- 99}$

$2^{- 97}$

$2^{- 96}$

$2^{- 98}$

answer is B.

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Detailed Solution

The point $(x_{1}, y_{1}), (x_{2}, y_{2}), . .. (x_{n + 1}, y_{n + 1})$

Given $(x_{n + 1} + i y_{n + 1}) = Z_{n + 1} (say)$

$\begin{array}{l} ∴ Z_{n + 1} = (\sqrt{3} x_{n} - y_{n}) + i (\sqrt{3} y_{n} + x_{n}) \\ = \sqrt{3} (x_{n} + {iy}_{n}) - y_{n} + {ix}_{n} = \sqrt{3} (x_{n} + {iy}_{n}) + i (x_{n} - \frac{y_{n}}{i} \frac{i}{i}) \\ = \sqrt{3} (x_{n} + {iy}_{n}) + i (x + {iy}_{n}) = \sqrt{3} Z_{n} + {iZ}_{n} \\ = Z_{n} (\sqrt{3} + i) where Z_{n} = x_{n} + {iy}_{n} = 2 Z_{n} (\cos \frac{π}{6} + isin \frac{π}{6}) \\ Z_{n + 1} = 2 Z_{n} e^{i \frac{π}{6} \frac{Z_{n + 1}}{Z_{n}} = 2 e^{i \frac{π}{6}} = K (say)} \end{array}$

putting $n = 1, 2, 3 ....99, we get$

$\frac{Z_{2}}{Z_{1}} = K, \frac{Z_{3}}{Z_{2}} = K, . .. \frac{Z_{100}}{Z_{99}} = K$

Multiplying above all, we get

$\begin{array}{l} \frac{Z_{100}}{Z_{99}} = K^{99} = 2^{99} e^{i \frac{33 π}{2}} = 2^{99} i \Rightarrow Z_{100} = 2^{99} (Z_{1} i) \\ \Rightarrow Z_{1} = \frac{Z_{100}}{2^{99} i} = \frac{2 + 4 i}{2^{99} i} \Rightarrow Z_{1} - 2^{- 97} - i 2^{- 98} = x_{1} + i y_{1} \\ \Rightarrow x_{1} + y_{1} = 2^{- 97} - 2^{- 98} = 2^{- 98} (2 - 1) = 2^{- 98} \end{array}$