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Prove limit of complex function

  1. Jul 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that lim_{z -> 1 - i} [x + i(2x+y)] = 1 + i

    where z = x + iy

    2. Relevant equations

    Prove using definition of complex limit.

    3. The attempt at a solution

    Start from |x + i(2x + y) - (1 + i)| < epsilon, need to transform LHS to an expression that includes |(x + iy) - (1 - i)|.

    |x + i 2x + i y - 1 - i| < epsilon
    |(x + iy) + i 2x - 1 - i| < epsilon
     
  2. jcsd
  3. Jul 15, 2009 #2
    Eh, OK I am assuming you handle these kind of limits the same way you do in one-variable calculus. So basically we need to find bounds for |x-1| and |y+1|. Note

    |x + i(2x + y) - (1 + i)| = |x + 2xi - 1 - i + iy| = |x + 2xi -1 - i + (-i) + i + iy| = |(x-1)(1+2i) + i i(y+1)|.

    So yeah, you should be able to take it from here.
     
  4. Jul 15, 2009 #3
    snipez90, thanks for your reply.

    Continue from your RHS, I get
    [tex]
    \begin{align*}
    |(x-1)(1+2i) + i (y + 1)| =& \left|(1+2i) [(x-1) + \frac{i(y+1)}{1 +
    2i}] \right| \\
    =& |1+2i| \left| (x-1) + \frac{i(y+1) (1 - 2i)}{(1 + 2i) (1 - 2i)} \right| \\
    =& |1+2i| \left| (x-1) + \frac{i(y+1) + 2(y+1)}{1 + 2i - 2i + 4}
    \right| \\
    =& |1+2i| \left| (x-1) + \frac{i + 2}{5} (y+1) \right| \\
    =& \sqrt{5} \left| (x-1) + \frac{i + 2}{5} (y+1) \right|
    \end{align*}
    [/tex]

    However, this is still not the same as required expression consisting of [tex]|(x-1) + i(y+1)|[/tex]. What should I do from here?
     
  5. Jul 15, 2009 #4
    Hmm, what I was thinking was that you could apply the triangle inequality to
    [tex]|(x-1)(1+2i) + i(y+1)|[/tex] (sorry about typo above) so that
    [tex]|(x-1)(1+2i) + i(y+1)| \le \sqrt{5}|x-1| + |y +1|.[/tex]
    Then require each of the two terms on the right to be less than [tex]\varepsilon/2,[/tex] and then choose delta accordingly.
     
  6. Jul 15, 2009 #5
    Thank you for your reply to both of my threads. I get the idea now; i didn't think of triangle inequality at all.
     
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