Results 1 to 6 of 6

Math Help - Inverse of z

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    221

    Inverse of z

    Using z^{-1} = \frac{z*}{|z|^2}, geometrically show how you would construct z^{-1}.

    NOTE z* means the conjugate of z.

    I have no idea. I know how to show it algebraically, since we'd have:

    \frac{1}{(a+ib)} = \frac{(a-ib)}{a^2+b^2}

    Then, multiply the left side by the complex conjugate and we get the right hand side...

    How'd I show this geometrically?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,803
    Thanks
    1692
    Awards
    1
    Plot the point z on a graph.
    Reflect that point in the real axis; that is z conjugate.
    Now divide by the absolute value of z squared.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by Plato View Post
    Plot the point z on a graph.
    Reflect that point in the real axis; that is z conjugate.
    Now divide by the absolute value of z squared.
    How do you "divide" something on the graph...

    So if we have our axis', we'd have:

    ib on the vertical axis (above 0)
    -ib on the vertical axis (below 0)

    And we'd have a somewhere on the traditional x-axis (greater than 0)

    So then a - ib will put us in the 4th quadrant, and that is our conjugate...

    How do I graph a^2 + b^2 and then divide this by the point I already created from the conjugate?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    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 Ideasman View Post
    How do you "divide" something on the graph...

    So if we have our axis', we'd have:

    ib on the vertical axis (above 0)
    -ib on the vertical axis (below 0)

    And we'd have a somewhere on the traditional x-axis (greater than 0)

    So then a - ib will put us in the 4th quadrant, and that is our conjugate...

    How do I graph a^2 + b^2 and then divide this by the point I already created from the conjugate?
    i suppose you divide the co-ordinates of the reflected point by the quantity a^2 + b^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2006
    Posts
    221
    No idea how. I don't even know what my "b" is. It'd have to be "ib," no? A diagram would be helpful, but probably too hard to do on here.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    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 Ideasman View Post
    No idea how. I don't even know what my "b" is. It'd have to be "ib," no? A diagram would be helpful, but probably too hard to do on here.
    well, plotting the complex number on an Argand diagram, you will get a point (a,b). reflect in the real axis to obtain the point (a,-b). then divide by the quantity a^2 + b^2 to get the point \left( \frac a{a^2 + b^2}, - \frac b{a^2 + b^2}\right). this will be the geometric representation of the complex number z^{-1}

    of course, this works if a and b are not both zero. if a = b = 0, then z = z^{-1}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. left inverse equivalent to inverse
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 30th 2011, 03:58 PM
  2. Inverse tan and inverse of tanh
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 31st 2010, 06:20 AM
  3. secant inverse in terms of cosine inverse?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: August 30th 2010, 08:08 PM
  4. Replies: 1
    Last Post: April 9th 2010, 05:51 PM
  5. inverse trig values and finding inverse
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 6th 2009, 12:04 AM

Search Tags


/mathhelpforum @mathhelpforum