Problem 2 Let u(x, y) = x2 −y 2 −y. Find a real-valued function v(x, y) such that v(0, 0) = 1 and together, u and v satisfy the Cauchy-Riemann equations in the entire complex plane. Cauchy–Riemann equations are given by formulas: ∂v ∂u = ∂x ∂y ∂u ∂v =− ∂y ∂x We have given function: u(x, y) = x2 − y 2 − y Calculating partial derivatives respect x and y we have: ∂u = 2x ∂x ∂u = −2y − 1 ∂y We can calculate
∂v dv = (−2y − 1) ⇒ = −(2y + 1) ∂x dx v(x, y) = −(2y + 1)
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dx = −(2y + 1)x + C2
∂v dv = 2x ⇒ = 2x ∂y dy v(x, y) = 2x
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dy = 2xy + C1
v1 (x, y) = 2xy + C1 v2 (x, y) = −2xy − x + C2 Functions have to satisfy condition v(0, 0) = 1 ⇒ C1 = C2 = 1 thus we have: v1 (x, y) = 2xy + 1 v2 (x, y) = −2xy − x + 1 First function does not satisfy Cauchy–Riemann equations ∂u ∂v1 = ∂x ∂y ∂u ∂v1 =− ∂y ∂x 2x = 2y −2y − 1 6= −2y 1
the second one does
∂u ∂v2 = ∂x ∂y ∂v2 ∂u =− ∂y ∂x 2x = 2y −2y − 1 = −(2y + 1)
so a solution is: v(x, y) = −2xy − x + 1
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