Results 1 to 2 of 2

Math Help - differentiation with complex variables

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    142

    differentiation with complex variables

    I need to find the values for which the following have derivatives and I need to find the derivatives at those points.

    (i) f(z) = x^2 + iy

    Then u(x,y) = x^2 and v(x,y) = y.
    du/dx = 2x
    du/dy = 0
    dv/dx = 0
    dv/dy = 1
    This means 2x = 1 and 0 = 0 and f(z) is only differentiable at x = 1/2.
    So it's differentiable at points z(1/2,y) and its derivative is df/dx = 2x = 1 at these points.

    (ii) f(z) = xy + xyi

    Then u(x,y) = xy and v(x,y) = xy.
    du/dx = y
    du/dy = x
    dv/dx = y
    dv/dy = y
    This means x = y and x = -y and f(z) is only differentiable along these lines. So it's differentiable at points z(x,x) and z(x,-x) and its derivative is df/dx = y + yi = x + xi and x - xi at these points.

    (iii) f(z) = y- 2xy + i(-x + x^2 - y^2).

    Then u(x,y) = y- 2xy and v(x,y) = -x + x^2 - y^2
    du/dx = -2y
    du/dy = 1-2x
    dv/dx = -1 + 2x
    dv/dy = -2y
    This means -2y = -2y and 1-2x = 1-2x and f(z) is differentiable everywhere.
    So its differentiable at points z(x,y) and its derivative is df/dx = -2y + i(-1 + 2x) at all points.

    Can someone just check that this work is accurate and stated correctly? I'm just learning this material.

    Thanks in advance for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by PvtBillPilgrim View Post
    I need to find the values for which the following have derivatives and I need to find the derivatives at those points.

    (i) f(z) = x^2 + iy

    Then u(x,y) = x^2 and v(x,y) = y.
    du/dx = 2x
    du/dy = 0
    dv/dx = 0
    dv/dy = 1
    This means 2x = 1 and 0 = 0 and f(z) is only differentiable at x = 1/2.
    So it's differentiable at points z(1/2,y) and its derivative is df/dx = 2x = 1 at these points.
    ok

    (ii) f(z) = xy + xyi

    Then u(x,y) = xy and v(x,y) = xy.
    du/dx = y
    du/dy = x
    dv/dx = y
    dv/dy = x
    This means x = y and x = -y and f(z) is only differentiable along these lines. So it's differentiable at points z(x,x) and z(x,-x) and its derivative is df/dx = y + yi = x + xi and x - xi at these points.
    yes, you made a mistake with dv/dy though, it is x

    (iii) f(z) = y- 2xy + i(-x + x^2 - y^2).

    Then u(x,y) = y- 2xy and v(x,y) = -x + x^2 - y^2
    du/dx = -2y
    du/dy = 1-2x
    dv/dx = -1 + 2x
    dv/dy = -2y
    This means -2y = -2y and 1-2x = 1-2x and f(z) is differentiable everywhere.
    So its differentiable at points z(x,y) and its derivative is df/dx = -2y + i(-1 + 2x) at all points.

    Can someone just check that this work is accurate and stated correctly? I'm just learning this material.

    Thanks in advance for your help.
    nice
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Implicit Differentiation with 3 Variables
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 30th 2010, 02:22 AM
  2. Complex Variables differentiation
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: October 11th 2009, 12:46 PM
  3. Numerical Differentiation - 2 variables
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 22nd 2009, 07:16 PM
  4. Replies: 3
    Last Post: November 2nd 2008, 06:13 AM
  5. differentiation, separation of variables???
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 11th 2008, 08:42 AM

Search Tags


/mathhelpforum @mathhelpforum