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Exercise 1.1 · Q54

Q.If (x+iy)3=u+iv(x+iy)^3 = u+iv, then show that ux+vy=4(x2−y2)\dfrac{u}{x}+\dfrac{v}{y} = 4(x^2-y^2)

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Expand (x+iy)3=x3+3x2(iy)+3x(iy)2+(iy)3=(x3−3xy2)+i(3x2y−y3)(x+iy)^3=x^3+3x^2(iy)+3x(iy)^2+(iy)^3=(x^3-3xy^2)+i(3x^2y-y^3). So u=x3−3xy2u=x^3-3xy^2 and v=3x2y−y3v=3x^2y-y^3. Then ux=x2−3y2\dfrac{u}{x}=x^2-3y^2 (for x≠0x\neq0) and vy=3x2−y2\dfrac{v}{y}=3x^2-y^2 (for y≠0y\neq0). Adding: $\dfrac{u}{x}+\dfrac{v}{ …

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