Question

Prove |Z|^2 = Z. Z bar .

Answer 1

1) Fix w in C, and let ε > 0 be given. 

Then, note that 
|f(z) - f(w)| = |z^2 - w^2| 
...............= |z + w| |z - w| 
...............= |(z - w) + 2w| |z - w| 
...............≤ (|z - w| + 2|w|) |z - w|, by triangle inequality 
...............= |z - w|^2 + 2|w| * |z - w| 
...............< |z - w| + 2|w| * |z - w|, assuming that |z - w| < 1 
...............< (1 + 2|w|) |z - w|. 

So given ε > 0, let δ = min {1, ε/(1 + 2|w|)}. 
Then, |z - w| < δ ==> |f(z) - f(w)| < (1 + 2|w|) |z - w| < ε, as required. 
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2) These are much easier. 

a) Given ε > 0, let δ = ε. 
Then, |z - w| < δ ==> |f(z) - f(w)| = |z_bar - w_bar| = |(z - w)_bar| = |z - w| < ε. 

b) Given ε > 0, let δ = ε. 

Writing z = x + iy and w = a + bi for some x, y, a, b in R: 
Then, |z - w| < δ ==> |f(z) - f(w)| = |y - b| = √(y - b)^2 ≤ √[(x - a)^2 + (y - b)^2] = |z - w| < ε. 
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3) Assuming that you mean {Z_n} → z: 

Suppose that {Z_n} → w as well. 
Given ε > 0, there exists a positive integer N such that 
|z_n - z| < ε/2 and |z_n - w| < ε/2 for all n > N. 

So, |z - w| ≤ |z_n - z| + |z_n - w| < ε/2 + ε/2 = ε for all n > N. 
Since ε > 0 is arbitrary, we conclude that w = z.