Results 1 to 12 of 12
Like Tree1Thanks
  • 1 Post By SworD

Math Help - Complex variables - find an additional solution

  1. #1
    Member
    Joined
    Apr 2010
    Posts
    116
    Thanks
    1

    Complex variables - find an additional solution

    This is not really a calculus question but I am taking a course in basic complex analysis and this is from the first chapter.


    Show that the indicated numbers satisfy the given equation and find an additional solution, z_2.

    z^2+i=0, z_1=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i


    Obviously you can plug z_1 in and get a true statement, but how do you go about finding an additional solution?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    215
    Thanks
    56

    Re: Complex variables - find an additional solution

    Do you recall synthetic or polynomial division? One way is to factor out z - z_1 from z^2 + i and you will get a linear equation.

    Edit: you don't even need to go this complicated: just use the fact that z^2 + i is an even function, if z1 is a root, so is -z1.
    Last edited by SworD; January 13th 2013 at 01:22 PM.
    Thanks from Deveno
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2013
    From
    United States
    Posts
    5

    Re: Complex variables - find an additional solution

    Use the fact that complex zeros come in conjugate pairs. just conjugate z1 and call that z2, and plug it in and it also is a zero.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    215
    Thanks
    56

    Re: Complex variables - find an additional solution

    No, the above won't work because the polynomial does not have real coefficients.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2013
    From
    United States
    Posts
    5

    Re: Complex variables - find an additional solution

    Why won't the conjugate of z1 work? It is indeed a zero when I do plugged it in. Perhaps there is some theory I'm not seeing here, all the op's question asks is to find an additional solution. And so z1 conjugate is that solution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    215
    Thanks
    56

    Re: Complex variables - find an additional solution

    No it isn't. Recheck your calculations.

    The theory your not seeing here is that roots won't magically always be conjugate pairs. I could just create an arbitrary polynomial which clearly violates that rule, for example, by multiplying out the following:

    (z + i)(z + 2i)

    You are missing the fact that the coefficients of the polynomial have to be real in order for the roots to come in conjugate pairs.

    The other root in this case is actually the negative of z1, not the conjugate.
    Last edited by SworD; January 13th 2013 at 01:16 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2013
    From
    United States
    Posts
    5

    Re: Complex variables - find an additional solution

    We could probably just multiply z1 by e^(2kpi) for k in Z? Thanks sword for fixing my error, I didn't multiply correctly lol. Do you see an error with this new idea, using the fact that e^(2kpi)=1 is an identity.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Complex variables - find an additional solution

    Quote Originally Posted by farzero View Post
    We could probably just multiply z1 by e^(2kpi) for k in Z? Thanks sword for fixing my error, I didn't multiply correctly lol. Do you see an error with this new idea, using the fact that e^(2kpi)=1 is an identity.
    if it is an identity, this won't give you an additional answer.

    look suppose z_1 is a root (any root) of z^2 + c = 0.

    this means z_1^2 + c = 0.

    therefore: (-z_1)^2 + c = (-z_1)(-z_1) + c =

    (-1)(z_1)(-1)(z_1) + c = (-1)(-1)z_1^2 + c = z_1^2 + c = 0, that is: -z_1 is ALSO a root of z^2 + c = 0.

    note that we discovered this without any information about "c".
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,441
    Thanks
    1862

    Re: Complex variables - find an additional solution

    Quote Originally Posted by farzero View Post
    Why won't the conjugate of z1 work? It is indeed a zero when I do plugged it in. Perhaps there is some theory I'm not seeing here, all the op's question asks is to find an additional solution. And so z1 conjugate is that solution.
    Again, you are completely wrong. z_1= -\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i. Its complex conjugate is -\frac{\sqrt{2}}{2}- \frac{\sqrt{2}}{2}i. The square is \frac{2}{4}- 2\frac{2}{4}i- \frac{2}{4}= i so that if z is that complex conjugate, z^2+ i= 2i, not 0.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,441
    Thanks
    1862

    Re: Complex variables - find an additional solution

    Yes, having already determined one root, the simplest way to find the other is "division". If you wished to solve z^2+i= 0 from the beginning, use polar form. This equation is the same as z^2= -i and -i has modulus 1 and argument 3\pi/2 (or -\pi/4). To find the square root, take the square root of the modulus (1) and divide the argument by 2: (3\pi/2)/2= 3\pi/4 so that one square root of -i is cos(3\pi/4)+ i sin(3\pi/4)= -\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i which is the root we have.

    Since adding " 2\pi" to the argument does not change the number, we can also write -i as modulus 1, argument 2\pi+ 3\pi/2= 7\pi/2. That gives, as a second root, cos(7\pi/4)+ i sin(7\pi/4)= \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i.

    In other words, the second root is just the negative of the first root, just as you learned way back in basic algebra!
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Apr 2010
    Posts
    116
    Thanks
    1

    Re: Complex variables - find an additional solution

    Thank you everyone for your input!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Aug 2012
    From
    sanaa
    Posts
    18
    Thanks
    3

    Re: Complex variables - find an additional solution

    Quote Originally Posted by Ragnarok View Post
    This is not really a calculus question but I am taking a course in basic complex analysis and this is from the first chapter.


    Show that the indicated numbers satisfy the given equation and find an additional solution, z_2.

    z^2+i=0, z_1=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i


    Obviously you can plug z_1 in and get a true statement, but how do you go about finding an additional solution?
    \text{if }z_1 \text{ and } z_2 \text{ are roots then we can write the polynomial as : }

    (z-z_1)(z-z_2)=0 \Rightarrow \, \, z^2- z_1z-z_2z +z_1z_2 = 0  \, \Rightarrow \,\, z^2-(z_1+z_2)z +z_1z_2=0

    \text{so comparing this to the  original  : }z_1+z_2 =0 \, \,\text{then  }\fbox{ z_2=-z_1}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find a real solution for a complex root
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: September 16th 2012, 09:51 PM
  2. Find Particular and then General Solution of Complex Function
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: December 12th 2010, 05:10 PM
  3. Replies: 2
    Last Post: September 7th 2009, 03:01 PM
  4. Replies: 0
    Last Post: May 9th 2009, 12:36 PM
  5. solution using three variables
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 20th 2009, 03:26 PM

Search Tags


/mathhelpforum @mathhelpforum