If f (z) \( = \begin{cases} \frac {|z|}{Re(z)} &, \quad \text{if } {} \text{Re} (z)\neq 0 \\ 0 &, \quad \text{if } \text{ Re (z) = 0} \end{cases} \), then-
Understanding the Complex Function f(z)
The given function is defined for a complex number \( z = x + iy \), where \( x = Re(z) \) is the real part and \( y = Im(z) \) is the imaginary part. The function \( f(z) \) is defined piecewise:
- \( f(z) = \frac{|z|}{Re(z)} \) when \( Re(z) \neq 0 \)
- \( f(z) = 0 \) when \( Re(z) = 0 \)
We need to analyze the continuity of this function \( f(z) \). A function is continuous at a point \( z_0 \) if the limit of \( f(z) \) as \( z \to z_0 \) exists and is equal to \( f(z_0) \).
Analyzing Continuity at Re(z) ≠ 0
If \( Re(z_0) \neq 0 \), then in a small neighborhood around \( z_0 \), \( Re(z) \) remains non-zero. In this region, \( f(z) = \frac{|z|}{Re(z)} \). The functions \( |z| \) and \( Re(z) \) are continuous for \( z \neq 0 \). Since the denominator \( Re(z) \) is non-zero in this neighborhood, the quotient \( \frac{|z|}{Re(z)} \) is also continuous. Thus, \( f(z) \) is continuous at any point \( z_0 \) where \( Re(z_0) \neq 0 \).
Analyzing Continuity on the Imaginary Axis (Re(z) = 0), excluding z=0
Consider a point \( z_0 = iy_0 \) on the imaginary axis where \( y_0 \neq 0 \). At this point, \( Re(z_0) = 0 \), so \( f(z_0) = 0 \). To check continuity, we need to evaluate the limit \( \lim_{z \to z_0} f(z) \). Let's approach \( z_0 \) from a path where \( Re(z) \neq 0 \). Let \( z = x + iy \) approach \( z_0 = iy_0 \), meaning \( x \to 0 \) and \( y \to y_0 \). For \( x \neq 0 \), \( f(z) = \frac{|z|}{Re(z)} = \frac{\sqrt{x^2+y^2}}{x} \). As \( z \to z_0 \), \( x \to 0 \) and \( y \to y_0 \). The expression becomes \( \frac{\sqrt{x^2+y^2}}{x} \). As \( x \to 0 \), this expression tends to \( \frac{\sqrt{0^2+y_0^2}}{x} = \frac{|y_0|}{x} \). Since \( y_0 \neq 0 \), \( |y_0| > 0 \). The limit \( \lim_{x \to 0} \frac{|y_0|}{x} \) does not exist (it approaches \( +\infty \) or \( -\infty \) depending on the sign of \( x \)). Therefore, the limit \( \lim_{z \to z_0} f(z) \) does not exist for \( z_0 \neq 0 \) with \( Re(z_0)=0 \). Hence, \( f(z) \) is not continuous at any point on the imaginary axis except possibly at \( z=0 \).
Analyzing Continuity at z = 0
This is the point where the definition of \( f(z) \) changes. At \( z=0 \), \( Re(0)=0 \), so \( f(0) = 0 \). We need to evaluate the limit \( \lim_{z \to 0} f(z) \). Let's consider different paths approaching \( z=0 \).
- Approach along the real axis: Let \( z = x \), where \( x \) is real and \( x \neq 0 \). As \( z \to 0 \), \( x \to 0 \). For these points, \( Re(z) = x \neq 0 \). \( f(z) = f(x) = \frac{|x|}{x} \). As \( x \to 0 \) from the right (\( x > 0 \)), \( \frac{|x|}{x} = \frac{x}{x} = 1 \). So, the limit is 1. As \( x \to 0 \) from the left (\( x < 0 \)), \( \frac{|x|}{x} = \frac{-x}{x} = -1 \). So, the limit is -1. Since the left and right limits are different, the limit of \( f(z) \) along the real axis as \( z \to 0 \) does not exist.
- Approach along the imaginary axis: Let \( z = iy \), where \( y \) is real and \( y \neq 0 \). As \( z \to 0 \), \( y \to 0 \). For these points, \( Re(z) = 0 \). \( f(z) = f(iy) = 0 \) (by definition, since \( Re(iy)=0 \)). As \( y \to 0 \), the limit is 0.
Since the limit of \( f(z) \) depends on the path of approach (e.g., approaching along the real axis gives different values), the limit \( \lim_{z \to 0} f(z) \) does not exist. For a function to be continuous at a point, the limit must exist and be equal to the function value. Since the limit does not exist at \( z=0 \), the function \( f(z) \) is not continuous at \( z=0 \).
Conclusion on Continuity of f(z)
Based on the analysis:
- \( f(z) \) is continuous at all points where \( Re(z) \neq 0 \).
- \( f(z) \) is not continuous at any point where \( Re(z) = 0 \) and \( z \neq 0 \).
- \( f(z) \) is not continuous at \( z = 0 \).
Therefore, the function \( f(z) \) is not continuous on the entire imaginary axis (where \( Re(z)=0 \)), including the point \( z=0 \). The statement that \( f \) is not continuous at 0 is true.
Important Questions from Complex Analysis
For z ∈ ℂ let f(z) = \(\left\{ \begin{matrix} \rm \frac{\bar{z}^2}{z}\ if \ z \ne 0,\\\ \rm 0 \ \ otherwise. \end{matrix} \right.\)
Then which of the following statements is false?
Let f, g be entire functions such that \(\lim _{z \rightarrow \infty} \frac{f(z)}{z^n}=\lim _{z \rightarrow \infty} \frac{g(z)}{z^n}=1\) for some fixed positive integer n. Which of the following statements is true?
A function f : \(\mathbb{C}\)\(\longmapsto \mathbb{C}\) is said to be analytic at ∞, if the function g defined by \(g(w)=f\left(\frac{1}{w}\right)\) is analytic at 0 with an appropriate value given for g(0). Which of the following statements is true?
For z ∈ \(\mathbb{C}\), let ℜz denotes its real part. Let f be an entire function satisfying |f(z)| ≤ |z| |ℜz| on \(\mathbb{C}\). Which of the following statements are true?
For every n ≥ 1, consider the entire function \(p_n(z)=\sum_{k=0}^n \frac{z^k}{k !}\). Which of the following statements are true?
