Ian H
Lv 7
Ian H asked in Science & MathematicsMathematics · 2 years ago

COMPLEX Suppose that f(z) is analytic, with f(0) = 3 - 2i and f(1) =6 - 5i . Find f(1 + i) if Im f (z) = 6x(2y - 1)?

2 Answers

Relevance
  • 2 years ago

    Im f(z)=6x(2y-1)i

    Let f(z)=U(x,y)+V(x,y)i, where V=(12xy-6x), then

    Ux=Vy=>

    Ux=12x=>

    U=6x^2+C(y)

    Uy=-Vx=>

    C'(y)=-12y+6=>

    C(y)=-6y^2+6y

    Thus,

    U(x,y)=6x^2-6y^2+6y=6(x^2-y^2+y)

    =>

    f(z)=6(x^2-y^2+y)+(12xy-6x)i

    z=1+i=>x=1 & y=1

    f(1+i)=6(1-1+1)+(12-6)i

    =>

    f(1+i)=6(1+i)

    Check:

    Ux=12x;

    Vy=12x

    =>

    Ux=Vy is valid

    Uy=-12y+6

    -Vx=-(12y-6)=-12y+6

    =>

    Uy=-Vx is valid

  • Ian H
    Lv 7
    2 years ago

    COMPLEX WAS SOLVED by KB (rewritten slightly)

    If f(z) = u(x,y) + i v(x,y), and f'(z) is taken to be derivative is wrt x,

    Im f'(z) = 12xy - 6x = ∂v/∂x …………………………………………...........(1)

    and integrating that wrt x

    v = 6x^2y - 3x^2 + g(y) ……………………………………….………......…(2)

    By one of the Cauchy-Riemann equations applied to (2)

    ∂v/∂y = 6x^2 + g'(y) = ∂u/∂x ………………………………………..........…(3)

    By the other Cauchy-Riemann equation used with (1)

    ∂u/∂y = -∂v/∂x = 6x – 12xy ………………………………………........……(4)

    From (4), ∂^u/∂y∂x = 6 – 12y …………………………………........……. (5)

    From (3), ∂^u/∂x∂y = g''(y)

    and since f(z) is analytic, these mixed partial derivatives are equal

    g''(y) = 6 - 12y, ……. and integrating wrt y

    g'(y) = 6y – 6y^2 + A, and again wrt y , ………………………..............…(6)

    g(y) = 3y^2 – 2y^3 + Ay + B ……………………………………….........…(7)

    v = 6x^2y - 3x^2 – 2y^3 +3y^2 + Ay + B …………………...................….(8)

    From (3), ∂u/∂x = 6x^2 + g'(y) = 6x^2 + 6y – 6y^2 + A

    and integrating that wrt x

    u = 2x^3 - 6xy^2 + 6xy + Ax + C ……………………………............……..(9)

    f(0) means f(0 + 0i) and substituting x = y = 0 into (9) and (8)

    f(0) = C + Bi = 3 – 2i, so B = -2, C = 3

    f(1) means f(1 + 0i) and substituting x = 1, y = 0 into (9) and (8)

    2 + A + C + (-3 + B)i = 2 + A + 3 + (-3 - 2)i = 6 – 5i

    A = 1 so, now we have f(z) = u(x,y) + i v(x,y), where

    u = 2x^3 - 6xy^2 + 6xy + x +3 ……………………………………................(10)

    v = 6x^2y - 3x^2 – 2y^3 + 3y^2 + y - 2 …………………….…...................(11)

    f(1 + i) = (2 - 6 + 6 + 1 + 3) + i (6 - 3 - 2 + 3 + 1 - 2) = 6 + 3i

    Thanks a lot KB

Still have questions? Get answers by asking now.