Prove limit of complex function

[complexnumber]

Homework Statement

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

where z = x + iy

Homework Equations

Prove using definition of complex limit.

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

 

[snipez90]

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.

 

[complexnumber]

snipez90, thanks for your reply.

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

However, this is still not the same as required expression consisting of $$|(x-1) + i(y+1)|$$. What should I do from here?

 

[snipez90]

Hmm, what I was thinking was that you could apply the triangle inequality to
$$|(x-1)(1+2i) + i(y+1)|$$ (sorry about typo above) so that
$$|(x-1)(1+2i) + i(y+1)| \le \sqrt{5}|x-1| + |y +1|.$$
Then require each of the two terms on the right to be less than $$\varepsilon/2,$$ and then choose delta accordingly.

 

[complexnumber]

Thank you for your reply to both of my threads. I get the idea now; i didn't think of triangle inequality at all.