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Problem 2

# Discuss continuity and uniform continuity for the following functions. (a) $$f(z)=\frac{1}{1-z} \quad(|z|<1)$$ (b) $$f(z)=\frac{1}{z} \quad(|z| \geq 1)$$ (c) $f(z)=\left\\{\begin{array}{ll}\frac{|z|}{z} & \text { if } 0<|z| \leq 1 \\\ 0 & \text { if } z=0\end{array}\right.$ (d) $f(z)=\left\\{\begin{array}{ll}\frac{\operatorname{Re} z}{z} & \text { if } 0<|z|<1 \\ 1 & \text { if } z=0 .\end{array}\right.$

Expert verified
(a) $$f(z)=\frac{1}{1-z}$$ is continuous but not uniformly continuous. (b) $$f(z)=\frac{1}{z}$$ is continuous and uniformly continuous. (c) $$f(z)=\frac{\left|z\right|}{z}$$ when $$0<\left|z\right|\leq 1$$ and $$f(z) = 0$$ when $$z=0$$ is continuous but not uniformly continuous. (d) $$f(z)=\frac{Re\,z}{z}$$ when $$0<\left|z\right|<1$$ and $$f(z) = 1$$ when $$z=0$$ is neither continuous nor uniformly continuous.
See the step by step solution

## Step 1: Function a - Analysis

Function $$f$$($$z$$) = $$\frac{1}{1-z}$$, where $$\left| z \right| < 1$$. This function is continuous because it is a quotient of two polynomial functions and the denominator is not equal to zero. To show this, suppose $$z_0$$ is in the interval (-1,1). Then, for any $$\epsilon > 0$$, there exists some $$\delta > 0$$ such that $$\left| \frac{1}{1-z} - \frac{1}{1-z_0} \right| < \epsilon$$ whenever $$\left| z - z_0 \right| < \delta$$. However, the function is not uniformly continuous because it is not bounded.

## Step 2: Function b - Analysis

Function $$f$$($$z$$) = $$\frac{1}{z}$$, where $$\left| z \right| \geq 1$$. This function is continuous and uniformly continuous. Intuitively, it is possible to draw the graph of the function without lifting the pen, and the rate of change of the function does not accelerate infinitely. To show this, consider any two points $$z, z_0$$ in the domain (that is, $$\left| z \right| \geq 1$$ and $$\left| z_0 \right| \geq 1$$), then for any $$\epsilon > 0$$, there exists some $$\delta > 0$$ such that $$\left| \frac{1}{z} - \frac{1}{z_0} \right| < \epsilon$$ whenever $$\left| z - z_0 \right| < \delta$$, and this $$\delta$$ can be made independent of the choice of $$z_0$$.

## Step 3: Function c - Analysis

Function $$f$$($$z$$) = $$\frac{\left|z\right|}{z}$$, for $$0<\left|z\right| \leq 1$$, and $$f$$($$z$$) = 0, for $$z=0$$. This function is continuous but not uniformly continuous. Although at each point in the domain, for any $$\epsilon > 0$$, there exists some $$\delta > 0$$ can be found such that $$\left| \frac{\left|z\right|}{z} - \frac{\left|z_0\right|}{z_0} \right| < \epsilon$$ whenever $$\left| z - z_0 \right| < \delta$$, this $$\delta$$ depends on the choice of $$z_0$$.

## Step 4: Function d - Analysis

Function $$f$$($$z$$) = $$\frac{Re\,z}{z}$$, for $$0<\left|z\right|<1$$, and $$f$$($$z$$) = 1, for $$z=0$$. This function is neither continuous nor uniformly continuous. In the vicinity of 0, the value of the function jumps from 1 to -1, and so continuity at $$z=0$$ does not hold.

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