# 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 72 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 72 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