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Miscellaneous Exercise 1 (Subjective) · Q193

Q.Find the real numbers xx and yy such that x1+2i+y3+2i=5+6i−1+8i\dfrac{x}{1+2i}+\dfrac{y}{3+2i} = \dfrac{5+6i}{-1+8i}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
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Rationalise: dfracx1+2i=dfracx(1−2i)5\\dfrac{x}{1+2i}=\\dfrac{x(1-2i)}{5} and dfracy3+2i=dfracy(3−2i)13\\dfrac{y}{3+2i}=\\dfrac{y(3-2i)}{13}. Combine over the denominator 6565: dfrac13x(1−2i)+5y(3−2i)65=dfrac(13x+15y)−(26x+10y)i65\\dfrac{13x(1-2i)+5y(3-2i)}{65}=\\dfrac{(13x+15y)-(26x+10y)i}{65}. On the right, dfrac5+6i−1+8i=dfrac(5+6i)(−1−8i)(−1+8i)(−1−8i)=dfrac−5−40i−6i−48i21+64=dfrac−5−46i+4865=dfrac43−46i65\\dfrac{5+6i}{-1+8i}=\\dfrac{(5+6i)(-1-8i)}{(-1+8i)(-1-8i)}=\\dfrac{-5-40i-6i-48i^2}{1+64}=\\dfrac{-5-46i+48}{65}=\\dfrac{43-46i}{65}. Equating real and imaginary parts of the two sides: 13x+15y=4313x+15y=43 and $-(26x+10y)=-46\ …

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