Jawab Soal Fungsi Kompleks Turunan

Matematika Universitas
#1

Diketahui f(z) = -3z^2

  1. Ubahlah ke bentuk f(z) = u(x,y)+iv(x,y) ; z=x+yi
  2. Buktikan f'(z) = -6z dengan f'(z) = ux(x,y) + ivx(x,y)

Penyelesaian:

  1. Ubahlah ke bentuk f(z) = u(x,y)+iv(x,y) ; z=x+yi
    f(z) = -3z^2 = -3 (x+yi)^2 = -3(x^2+2xyi-y^2)
    f(z) = -3x^2-6xyi-3y^2
    Maka:
    f(z) = u(x,y)+iv(x,y)
    f(z) = (-3x^2 +3y^2) + i(-6xy)
    f(z) = -3(x^2 -y^2) + i(-6xy)

  2. Buktikan f'(z) = -6z dengan f'(z) = ux(x,y) + ivx(x,y)
    f(z) = u(x,y)+iv(x,y)
    f(z) =-3x^2 +3y^2 + i(-6xy)

    Maka:
    u(x,y) = -3x^2 +3y^2; ux(x,y) = -6x
    v(x,y) =-6xy; vx(x,y) = -6y

    Sehingga didapat:
    f'(z) = ux(x,y) + ivx(x,y)
    f'(z) = (-6x) + i (-6y)
    f'(z) = -6 (x+yi)
    f'(z) = -6z

    Terbukti f'(z) = -6z dengan f'(z) = ux(x,y) + ivx(x,y)