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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
- PinkgreenLv 73 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 HLv 73 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