Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Prove It

Math Help - Help with Sketching Region in Complex Plane

  1. #1
    Newbie
    Joined
    May 2012
    From
    Land down under
    Posts
    13

    Help with Sketching Region in Complex Plane

    Hey guys,

    I am unsure about the steps needed to sketch the following region in a complex plane. It's the only question regarding the sketching of complex planes which I am unsure of the methodology use to approach.

    |z -3| = |z + 2i|

    I have done questions like |z - 2i| = 2 etc, never done a question with 2 modulus equal to each other. Any help would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,487
    Thanks
    1391

    Re: Help with Sketching Region in Complex Plane

    I would convert everything to Cartesians:

    $\displaystyle \begin{align*} \left| z - 3 \right| &= \left| z + 2i \right| \\ \left| \left( x - 3 \right) + y\,i \right| &= \left| x + \left( y + 2 \right)\, i \right| \\ \sqrt{ \left( x - 3 \right) ^2 + y^2 } &= \sqrt{ x^2 + \left( y + 2 \right) ^2 } \\ \left( x - 3 \right) ^2 + y^2 &= x^2 + \left( y + 2 \right) ^2 \\ x^2 - 6x + 9 + y^2 &= x^2 + y^2 + 4y + 4 \\ -6x + 9 &= 4y + 4 \\ -6x + 5 &= 4y \\ y &= -\frac{3}{2}x + \frac{5}{4} \end{align*}$

    So the graph will be the line with gradient $\displaystyle \begin{align*} -\frac{3}{2} \end{align*}$ and y intercept $\displaystyle \begin{align*} \left( 0, \frac{5}{4} \right) \end{align*}$.
    Thanks from TehNewbie
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2012
    From
    Land down under
    Posts
    13

    Re: Help with Sketching Region in Complex Plane

    Quote Originally Posted by Prove It View Post
    I would convert everything to Cartesians:

    $\displaystyle \begin{align*} \left| z - 3 \right| &= \left| z + 2i \right| \\ \left| \left( x - 3 \right) + y\,i \right| &= \left| x + \left( y + 2 \right)\, i \right| \\ \sqrt{ \left( x - 3 \right) ^2 + y^2 } &= \sqrt{ x^2 + \left( y + 2 \right) ^2 } \\ \left( x - 3 \right) ^2 + y^2 &= x^2 + \left( y + 2 \right) ^2 \\ x^2 - 6x + 9 + y^2 &= x^2 + y^2 + 4y + 4 \\ -6x + 9 &= 4y + 4 \\ -6x + 5 &= 4y \\ y &= -\frac{3}{2}x + \frac{5}{4} \end{align*}$

    So the graph will be the line with gradient $\displaystyle \begin{align*} -\frac{3}{2} \end{align*}$ and y intercept $\displaystyle \begin{align*} \left( 0, \frac{5}{4} \right) \end{align*}$.
    Thanks mate, I see how much easier it now becomes converting it into Cartesian form. Once again thank you for the in depth explanation.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 5th 2011, 08:25 PM
  2. Sketching regions in complex plane
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 9th 2010, 10:30 AM
  3. Sketching |z - 1| + |z + i| <= 2 in complex plane
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: October 28th 2010, 05:08 AM
  4. Sketching complex plane
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 19th 2010, 08:00 AM
  5. Sketching in the complex plane
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 9th 2009, 04:54 AM

Search Tags


/mathhelpforum @mathhelpforum