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Exercise 1.1 · Q49

Q.If x+iy=(a+ib)3x+iy = (a+ib)^3, show that xa+yb=4(a2−b2)\dfrac{x}{a}+\dfrac{y}{b} = 4(a^2-b^2)

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Expand (a+ib)3=a3+3a2(ib)+3a(ib)2+(ib)3=a3+3a2bi−3ab2−ib3=(a3−3ab2)+i(3a2b−b3)(a+ib)^3=a^3+3a^2(ib)+3a(ib)^2+(ib)^3=a^3+3a^2bi-3ab^2-ib^3=(a^3-3ab^2)+i(3a^2b-b^3). So x=a3−3ab2x=a^3-3ab^2 and y=3a2b−b3y=3a^2b-b^3. Then xa=a2−3b2\dfrac{x}{a}=a^2-3b^2 (dividing every term by aa, valid for a≠0a\neq0) and yb=3a2−b2\dfrac{y}{b}=3a^2-b^2 (dividing by bb, valid for b≠0b\neq0). Adding: $ …

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