Proving a Complex Analysis Limit: lim_{z \rightarrow 1+2i} [ix - (x+y)] = -3 + i

[Dunkle]

Homework Statement

Show that $$lim_{z \rightarrow 1+2i} [ix - (x+y)] = -3 + i$$.

Homework Equations

$$lim_{z \rightarrow z_0} f(z) = w_0$$ if and only if given $$\epsilon > 0$$ there exists a $$\delta > 0$$ such that $$0 < |z-z_0| < \delta \Rightarrow |f(z)-w_0| < \epsilon$$

The Attempt at a Solution

$$f(z) = ix-(x+y), w_0 = -3+i, z = x+iy, z_0 = 1+2i$$

I calculated the following:

$$|z-z_0| = \sqrt{(x-1)^2+(y-2)^2}$$ and

$$|f(z)-w_0| = \sqrt{(3-x-y)^2+(x-1)^2}$$

I need to somehow find a relationship between these, and this is where I'm struggling. Any help would be appreciated!

 

[mighty2000]

First, we need to rewrite the function in terms of z instead of x and y:

f(z) = ix-(x+y) = ix - (x+iy) = (i-1)x - iy

Next, we can use the definition of the limit to find a relationship between |z-z_0| and |f(z)-w_0|:

|f(z)-w_0| = |(i-1)x - iy - (-3+i)| = |(i-1)x + (1-i)y + 3-i|

We can then use the triangle inequality to split up the absolute value:

|f(z)-w_0| \leq |(i-1)x| + |(1-i)y| + |3-i|

Since we are trying to find a relationship between |z-z_0| and |f(z)-w_0|, we can rewrite the first two terms using the definition of |z-z_0|:

|(i-1)x| = |(i-1)(x-1+1)| = |(i-1)(z-z_0)| = |i-1||z-z_0| = \sqrt{2}|z-z_0|

|(1-i)y| = |(1-i)(y-2+2)| = |(1-i)(z-z_0)| = |1-i||z-z_0| = \sqrt{2}|z-z_0|

Substituting these back into the original equation, we get:

|f(z)-w_0| \leq \sqrt{2}|z-z_0| + \sqrt{2}|z-z_0| + |3-i| = 2\sqrt{2}|z-z_0| + |3-i|

Now, we can use this relationship to find a value for \delta that satisfies the definition of the limit. Let's choose \delta = \frac{\epsilon}{2\sqrt{2}+1}. Then, if 0 < |z-z_0| < \delta, we have:

|f(z)-w_0| \leq 2\sqrt{2}|z-z_0| + |3-i| < 2\sqrt{2}\frac{\epsilon}{2\sqrt{2}+1} + |3-i| = \epsilon

Therefore, we have shown that lim_{z