Results 1 to 6 of 6

Math Help - Finding complex roots

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    37

    Finding complex roots

    Hi.

    I'm a little stuck on the following problem:

    How many of the roots to the equation:

    z - e^z + 5 = 0

    lie in the region Re(z) < 0?


    I'm just not sure how to approach this. I've solved similar problems through Rouche's theorem, but in all those cases I'm finding the number of roots within a closed circular contour. Here I really don't know what to do! Any help will be greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Rewrite to z + 5 = e^{z} and ask, "Just how well-behaved is that exponential for Re(z) < 0?"
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    37
    Hi.

    Well on the real line, where we have z = x + 0i, the exponential will lie in the interval [0,1]. Am I onto something?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    Maybe a hint, but e^z is really well-behaved for y = 0. What happens as x and y head negative? What happens to |e^z|?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2009
    Posts
    37
    Quote Originally Posted by TKHunny View Post
    Maybe a hint, but e^z is really well-behaved for y = 0. What happens as x and y head negative? What happens to |e^z|?
    |e^z| is equal to e^x. At x = 0, this is equal to 1. As the value of x decreases from 0, e^x will also decrease. Hence, as mentioned in my above post, the expression will lie between 0 and 1 (as x approaches negative infinity, e^x will approach zero). Of course I also know that e^z = e^(x + iy) = e^(x)(cos(y) + isin(y)). This last expression will fluctuate between zero and 1. This we know that e^(z) is also less than or equal to 1. But I don't know how to proceed from here
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    You're not quite catching the vision.

    Expand z + 5 to real and imaginary parts.

    Expand e^z to real and imaginary parts.

    Equate Real Parts to obtain \frac{x+5}{e^{x}} = \cos(y)

    Graph left and right sides separately. You should see something about possible intersections.

    There are similar lessons to be learned from the Imaginary comparison.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding complex roots of polynomials
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: November 28th 2010, 02:33 PM
  2. Complex Number finding the roots
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: November 3rd 2010, 02:40 PM
  3. Finding complex roots?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 10th 2009, 10:58 AM
  4. Complex Analysis: Finding Roots
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 29th 2009, 08:50 PM
  5. Complex Numbers - Finding Roots
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: April 23rd 2007, 08:50 PM

Search Tags


/mathhelpforum @mathhelpforum