Show that z/|z| is continuous by epsilon-delta definition where z=x+iy

To prove that the function \( f(z) = \frac{z}{|z|} \) is continuous using the epsilon-delta definition of continuity, where \( z = x + iy \), we'll follow these steps:

### 1. Statement of Continuity
We need to show that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( |z - z_0| < \delta \), then \( \left| \frac{z}{|z|} - \frac{z_0}{|z_0|} \right| < \epsilon \), where \( z \) and \( z_0 \) are complex numbers.

### 2. Express the Function
Let \( z = x + iy \) and \( z_0 = x_0 + iy_0 \), so:
\[
|z| = \sqrt{x^2 + y^2}, \quad |z_0| = \sqrt{x_0^2 + y_0^2}
\]
Thus, the function is:
\[
f(z) = \frac{z}{|z|} = \frac{x + iy}{\sqrt{x^2 + y^2}}
\]
Similarly:
\[
f(z_0) = \frac{z_0}{|z_0|} = \frac{x_0 + iy_0}{\sqrt{x_0^2 + y_0^2}}
\]

### 3. Difference Between the Functions
We need to estimate \( \left| \frac{z}{|z|} - \frac{z_0}{|z_0|} \right| \):
\[
\left| \frac{z}{|z|} - \frac{z_0}{|z_0|} \right| = \left| \frac{z |z_0| - z_0 |z|}{|z| |z_0|} \right| = \frac{|z |z_0| - z_0 |z||}{|z| |z_0|}
\]

### 4. Triangle Inequality
Use the triangle inequality to estimate the numerator:
\[
|z |z_0| - z_0 |z|| \leq |z| |z_0 - z_0| + |z_0| |z - z_0|
\]
This simplifies to:
\[
|z |z_0| - z_0 |z|| \leq |z| \cdot |z - z_0| + |z_0| \cdot |z - z_0| = (|z| + |z_0|) \cdot |z - z_0|
\]

Thus, we have:
\[
\left| \frac{z}{|z|} - \frac{z_0}{|z_0|} \right| \leq \frac{(|z| + |z_0|) \cdot |z - z_0|}{|z| \cdot |z_0|}
\]

### 5. Estimate the Ratio
As \( z \) approaches \( z_0 \), \( |z - z_0| \) is small. Since \( |z| \) and \( |z_0| \) are both close to each other, let's denote \( |z| \) and \( |z_0| \) by \( M \), where \( M \) is close to \( |z_0| \):
\[
\left| \frac{z}{|z|} - \frac{z_0}{|z_0|} \right| \leq \frac{2M \cdot |z - z_0|}{M^2} = \frac{2 |z - z_0|}{M}
\]

### 6. Epsilon-Delta Condition
Now, choose \( \delta \) such that \( \delta = \frac{\epsilon M}{2} \). Then:
\[
|z - z_0| < \delta \quad \Rightarrow \quad \frac{2 |z - z_0|}{M} < \epsilon
\]
Hence, \( \left| \frac{z}{|z|} - \frac{z_0}{|z_0|} \right| < \epsilon \).

### Conclusion
Therefore, by the epsilon-delta definition of continuity, \( f(z) = \frac{z}{|z|} \) is continuous for \( z \neq 0 \).

Do you have any questions or need further details? Here are some related questions:

1. How does the continuity of \( f(z) = \frac{z}{|z|} \) change at \( z = 0 \)?
2. How would you apply the epsilon-delta definition to prove the continuity of a real-valued function?
3. What is the significance of the triangle inequality in proving continuity?
4. Can the same method be applied to prove continuity of functions in higher dimensions?
5. How does the choice of \( \delta \) relate to the function's behavior near \( z_0 \)?

**Tip:** When proving continuity using epsilon-delta, focus on estimating the difference in function values and controlling it by manipulating the input difference.

Learn how to prove the continuity of the complex function f(z) = z/|z| using the epsilon-delta definition. This problem involves complex numbers, continuity, and the rigorous mathematical proof using epsilon and delta. Suitable for undergraduate students studying complex analysis.

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