Results 1 to 10 of 10

Math Help - Complex numbers and Derivatives

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    130

    Complex numbers and Derivatives

    Hey guys, I've done more questions that I need to check.

    1) a) Simplify (z-1) (z^4 + z^3 + z^2 + z + 1)

    my solution: z^5 -1

    b) use part a to find all the roots of the polynomial P(z) = z^4 + z^3 + z^2 + z + 1 (you may keep your answer in exponential polar form

    my solution: I converted z^5 - 1 = 0 to exponential polar form and got the following roots: z = 1e^i0, 1e^i(2pi/5), 1e^i(-2pi/5), 1e^i(4pi/5), 1e^i(-4pi/5)

    2) Find dy/dx if root(y) + x loge(y) = cos(xy^2)

    my solution: using implicit diff and multiplication rule, i got dy/dx = [-log(y) - y^2 sin(xy^2)] / [ (x^2/2y) + 2x sin(xy^2) + (x^2/y)]

    please check my solutions to these problems

    thank you in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49

    Re: Complex numbers and Derivatives

    Your roots are good: 5th roots of unity - Wolfram|Alpha

    Simplifying the implicit differentiation is a bit messy but I'm pretty sure I don't like your (x^2)'s in the denominator.

    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    This is also embedded in the legs-uncrossed version of ...



    ... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.



    Larger

    I hope that helps.



    __________________________________________________ __________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; October 1st 2011 at 03:22 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2008
    Posts
    130

    Re: Complex numbers and Derivatives

    Hello Tom,

    Thanks for the prompt reply,

    I have revamped my solution fixing all silly mistakes - here it is:

    dy/dx = [ -y^2sin(xy^2) - (ln(y)/2root(y)) - 1/(2root(y))] / [ x/y + 2xysin(xy^2)]

    Am I correct?

    Regards
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49

    Re: Complex numbers and Derivatives

    No, your numerator was ok before. I suspect your differentiation is at fault.

    When you differentiate the equation implicitly, do you get the bottom row of my diagram (or equivalent)? You should.

    But if you do, then it's just an algebra problem. Use latex to show your steps. Click on 'Reply With Quote' and you can copy and paste the portion inside (and including) the [ tex ][ /tex ] tags, and then extend the following.

    {\displaystyle \frac{1}{2} \frac{1}{\sqrt{y}} \frac{dy}{dx} + \ln y + \frac{x}{y} \frac{dy}{dx} = -\sin(xy^2) [y^2 + 2xy \frac{dy}{dx}}]
    Last edited by tom@ballooncalculus; October 2nd 2011 at 03:20 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2008
    Posts
    130

    Re: Complex numbers and Derivatives

    {\displaystyle \frac{1}{2} \frac{1}{\sqrt{y}} \frac{dy}{dx} + \ln y + \frac{x}{y} \frac{dy}{dx} = \sin(xy^2) [y^2 + 2xy \frac{dy}{dx}}]

    Hello Balloon, I don't know how to use latex fluently, it would take me far too long to write it all out.

    I will work through them in words.

    First I multiplied out the brackets on the right hand side and took all the terms with a dy/dx to the left and the terms without a dy/dx to the right.
    I then took out the dy/dx as a common factor and divided what was left.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2008
    Posts
    130

    Re: Complex numbers and Derivatives

    Quote Originally Posted by tom@ballooncalculus View Post
    No, your numerator was ok before. I suspect your differentiation is at fault.

    When you differentiate the equation implicitly, do you get the bottom row of my diagram (or equivalent)? You should.

    But if you do, then it's just an algebra problem. Use latex to show your steps. Click on 'Reply With Quote' and you can copy and paste the portion inside (and including) the [ tex ][ /tex ] tags, and then extend the following.

    {\displaystyle \frac{1}{2} \frac{1}{\sqrt{y}} \frac{dy}{dx} + \ln y + \frac{x}{y} \frac{dy}{dx} = \sin(xy^2) [y^2 + 2xy \frac{dy}{dx}}]
    Is that supposed to be -sin(xy^2)?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jan 2008
    Posts
    130

    Re: Complex numbers and Derivatives

    I have redone the calculations, i now get

    [-y^2sin(xy^2) - loge(y)] / [ (1/(2root(y)) + (x/y) +2xy sin(xy^2)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49

    Re: Complex numbers and Derivatives

    Quote Originally Posted by andrew2322 View Post
    Is that supposed to be -sin(xy^2)?
    Oh, yes it is.

    Quote Originally Posted by andrew2322 View Post
    I have redone the calculations, i now get
    [-y^2sin(xy^2) - loge(y)] / [ (1/(2root(y)) + (x/y) +2xy sin(xy^2) ]
    Correct!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jan 2008
    Posts
    130

    Re: Complex numbers and Derivatives

    Thanks mate, have a great night!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49

    Re: Complex numbers and Derivatives

    And you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. raising complex numbers to complex exponents?
    Posted in the Advanced Math Topics Forum
    Replies: 10
    Last Post: March 25th 2011, 10:02 PM
  2. Replies: 1
    Last Post: September 27th 2010, 03:14 PM
  3. Replies: 2
    Last Post: February 7th 2009, 06:12 PM
  4. complex numbers and derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 31st 2008, 04:43 AM
  5. Replies: 1
    Last Post: May 24th 2007, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum