

Let’s solve Question 1 from the provided PDF document. The question asks to evaluate the limits of given functions.
Question 1:
Part (a)
[ \lim_{z \to 0} \frac{\overline{z}}{z} ]
Solution:
Here, (\overline{z}) is the complex conjugate of (z). Let (z = x + iy), then (\overline{z} = x – iy).
Thus,
[ \frac{\overline{z}}{z} = \frac{x – iy}{x + iy} ]
Multiplying the numerator and the denominator by the conjugate of the denominator:
[ \frac{x – iy}{x + iy} \cdot \frac{x – iy}{x – iy} = \frac{(x – iy)^2}{x^2 + y^2} = \frac{x^2 – 2ixy – y^2}{x^2 + y^2} ]
As (z \to 0), both (x) and (y) approach 0. Thus,
[ \frac{x^2 – y^2}{x^2 + y^2} ]
approaches 0 and
[ \frac{-2ixy}{x^2 + y^2} ]
also approaches 0. Therefore, the limit is:
[ \boxed{1} ]
Part (b)
[ \lim_{z \to -1} \frac{z^4 – 2z^2 + 1}{z + 1} ]
Solution:
First, let’s factorize the numerator. Notice that (z = -1) is a root of the numerator.
[ z^4 – 2z^2 + 1 = (z^2 – 1)^2 = (z + 1)^2(z – 1)^2 ]
Thus, the limit becomes:
[ \lim_{z \to -1} \frac{(z + 1)^2(z – 1)^2}{z + 1} ]
We can cancel one ((z + 1)) term from the numerator and the denominator:
[ \lim_{z \to -1} (z + 1)(z – 1)^2 ]
Now, substitute (z = -1):
[ (-1 + 1)(-1 – 1)^2 = 0 \cdot 4 = 0 ]
Therefore, the limit is:
[ \boxed{0} ]
Part (c)
[ \lim_{z \to 0} \frac{\Re(z^2)}{|z|^2} ]
Solution:
Let (z = x + iy), then:
[ z^2 = (x + iy)^2 = x^2 – y^2 + 2ixy ]
[ \Re(z^2) = x^2 – y^2 ]
[ |z|^2 = x^2 + y^2 ]
Thus, the limit becomes:
[ \lim_{z \to 0} \frac{x^2 – y^2}{x^2 + y^2} ]
As (z \to 0), both (x) and (y) approach 0. The limit is indeterminate, but we can analyze along different paths:
- If (y = 0): [ \frac{x^2}{x^2} = 1 ]
- If (x = 0): [ \frac{-y^2}{y^2} = -1 ]
Since the limit depends on the path, it does not exist in general.
Therefore, the limit is:
[ \boxed{\text{Does not exist}} ]
Part (d)
[ \lim_{z \to \infty} \frac{z \Re(z)}{|z|} ]
Solution:
Let (z = x + iy), then:
[ \Re(z) = x ]
[ |z| = \sqrt{x^2 + y^2} ]
Thus, the limit becomes:
[ \lim_{z \to \infty} \frac{z x}{\sqrt{x^2 + y^2}} ]
Express (z) in polar form (z = re^{i\theta}), where (r = |z|):
[ \frac{re^{i\theta} \cdot r\cos\theta}{r} = re^{i\theta}\cos\theta = r\cos\theta e^{i\theta} ]
As (z \to \infty), (r \to \infty):
[ r \cos\theta \cdot e^{i\theta} ]
However, for (z \to \infty), the term (\cos\theta) does not change the nature of the limit since it’s multiplied by (r), which goes to infinity.
Thus, the limit is:
[ \boxed{\infty} ]
These are the solutions to the first question’s parts (a) to (d). If you need further assistance with any other question, feel free to ask!


The function ( f(z) = \frac{1}{z} ) is a fundamental example in complex analysis, and it has several interesting properties:
Basic Properties
- Definition:
[ f(z) = \frac{1}{z} ]
Here, ( z ) is a complex number ( z = x + iy ), where ( x ) and ( y ) are real numbers. - Domain:
The function ( f(z) = \frac{1}{z} ) is defined for all complex numbers ( z ) except ( z = 0 ). The point ( z = 0 ) is a singularity, specifically a pole of order 1. - Behavior at ( z = 0 ):
As ( z ) approaches 0, the magnitude ( |f(z)| = \left|\frac{1}{z}\right| ) becomes very large. This indicates a pole at ( z = 0 ).
Polar Form
In polar coordinates, where ( z = re^{i\theta} ):
[ f(z) = \frac{1}{z} = \frac{1}{re^{i\theta}} = \frac{1}{r} e^{-i\theta} ]
This shows that the magnitude ( |f(z)| = \frac{1}{r} ) and the argument ( \arg(f(z)) = -\theta ).
Transformation Properties
- Conformal Mapping:
The function ( f(z) = \frac{1}{z} ) is a conformal mapping except at ( z = 0 ). This means it preserves angles and the shape of infinitesimally small figures. - Inversion:
The function ( f(z) = \frac{1}{z} ) maps circles and lines in the complex plane to other circles and lines. Specifically:
- Circles passing through the origin are mapped to lines.
- Lines passing through the origin are mapped to circles passing through the origin.
- Circles not passing through the origin are mapped to other circles not passing through the origin.
- Lines not passing through the origin are mapped to circles passing through the origin.
- Symmetry:
The function ( f(z) = \frac{1}{z} ) has inversion symmetry. If ( z ) is reflected in the unit circle, the result is the complex conjugate of the reflection of ( \frac{1}{z} ).
Examples
- Unit Circle:
For ( |z| = 1 ), we have ( |f(z)| = \left|\frac{1}{z}\right| = 1 ). Thus, the unit circle is mapped onto itself under ( f(z) ). - Real Axis:
For ( z = x ) (where ( x ) is real), ( f(z) = \frac{1}{x} ) is also real. Thus, the real axis maps onto itself. - Imaginary Axis:
For ( z = iy ) (where ( y ) is real), ( f(z) = \frac{1}{iy} = -\frac{i}{y} ). Thus, the imaginary axis is mapped to itself with a rotation by ( -\pi/2 ).
Visualization
To visualize the transformation ( f(z) = \frac{1}{z} ), consider plotting various geometric shapes and observing their images under this mapping. Let’s plot a few key transformations.
- Unit Circle (unchanged).
- Lines through the origin (mapped to circles passing through the origin).
- Circles centered at the origin (mapped to themselves).
Let’s create some visualizations for these transformations.
Visualizing ( f(z) = \frac{1}{z} )
Let’s plot these transformations to see their effects on the complex plane.
Here are the visualizations of the transformation ( f(z) = \frac{1}{z} ):
- Left Plot: The original grid in the complex plane. The blue lines represent the real part of ( z ), and the red lines represent the imaginary part of ( z ).
- Right Plot: The transformed grid under the mapping ( f(z) = \frac{1}{z} ). The transformation inverts and rotates the grid lines, mapping straight lines through the origin to circles and vice versa, while also inverting distances.
This illustrates how the function ( \frac{1}{z} ) significantly alters the complex plane, showcasing its inversion properties and the preservation of angles due to its conformal nature.

import numpy as np
import matplotlib.pyplot as plt
Function to compute 1/z
def reciprocal(z):
return 1 / z
Define a range of points in the complex plane
x = np.linspace(-2, 2, 400)
y = np.linspace(-2, 2, 400)
X, Y = np.meshgrid(x, y)
Z = X + 1j * Y
Apply the transformation f(z) = 1/z
W = reciprocal(Z)
Plot the original and transformed grid
fig, ax = plt.subplots(1, 2, figsize=(14, 7))
Plot the original grid
ax[0].contour(X, Y, np.real(Z), levels=20, colors=’blue’, alpha=0.5)
ax[0].contour(X, Y, np.imag(Z), levels=20, colors=’red’, alpha=0.5)
ax[0].set_title(‘Original Grid in the Complex Plane’)
ax[0].set_xlabel(‘Re(z)’)
ax[0].set_ylabel(‘Im(z)’)
ax[0].grid(True)
Plot the transformed grid
ax[1].contour(np.real(W), np.imag(W), np.real(Z), levels=20, colors=’blue’, alpha=0.5)
ax[1].contour(np.real(W), np.imag(W), np.imag(Z), levels=20, colors=’red’, alpha=0.5)
ax[1].set_title(‘Transformed Grid under f(z) = 1/z’)
ax[1].set_xlabel(‘Re(w)’)
ax[1].set_ylabel(‘Im(w)’)
ax[1].grid(True)
plt.tight_layout()
plt.show()
بله، دقیقاً. تابع ( f(z) = \frac{1}{z} ) ویژگیهای جالبی در تبدیل اشکال در صفحه مختلط دارد. این تابع دایرهها و خطوط را به یکدیگر تبدیل میکند به صورت زیر:
- دایرهها و خطوط عبوری از مبدأ:
- دایرههایی که از مبدأ عبور میکنند، به خطوطی تبدیل میشوند که از مبدأ عبور میکنند.
- خطوطی که از مبدأ عبور میکنند، به دایرههایی تبدیل میشوند که از مبدأ عبور میکنند.
- دایرهها و خطوطی که از مبدأ عبور نمیکنند:
- دایرههایی که از مبدأ عبور نمیکنند، به دایرههایی دیگر تبدیل میشوند که از مبدأ عبور نمیکنند.
- خطوطی که از مبدأ عبور نمیکنند، به دایرههایی تبدیل میشوند که از مبدأ عبور میکنند.
مثالها
- دایره واحد:
- دایره واحد (|z| = 1) تحت تبدیل (\frac{1}{z}) به خودش تبدیل میشود.
- خطوط افقی و عمودی:
- خطوط افقی و عمودی در صفحه مختلط به دایرههایی تبدیل میشوند که از مبدأ عبور میکنند.
بصریسازی
برای درک بهتر این تبدیلها، میتوانیم خطوط و دایرههای مختلف را در صفحه مختلط رسم کنیم و ببینیم که چگونه تحت این تبدیل تغییر میکنند.
کد برای رسم تبدیلهای دایرهها و خطوط
بیایید دایرهها و خطوط مختلف را رسم کنیم و مشاهده کنیم که چگونه تحت تبدیل ( \frac{1}{z} ) تغییر میکنند.
import numpy as np
import matplotlib.pyplot as plt
# Function to compute 1/z
def reciprocal(z):
return 1 / z
# Define a range of points for circles and lines
theta = np.linspace(0, 2 * np.pi, 400)
radii = [0.5, 1, 1.5]
lines = np.linspace(-2, 2, 400)
# Compute circles and lines
circles = [r * np.exp(1j * theta) for r in radii]
vertical_lines = [x + 1j * lines for x in [-1, 0, 1]]
horizontal_lines = [lines + 1j * y for y in [-1, 0, 1]]
# Plot the original and transformed shapes
fig, ax = plt.subplots(1, 2, figsize=(14, 7))
# Plot the original shapes
for circle in circles:
ax[0].plot(np.real(circle), np.imag(circle), label=f'Circle |z|={abs(circle[0]):.1f}')
for line in vertical_lines:
ax[0].plot(np.real(line), np.imag(line), label=f'Vertical Line x={line[0].real:.1f}')
for line in horizontal_lines:
ax[0].plot(np.real(line), np.imag(line), label=f'Horizontal Line y={line[0].imag:.1f}')
ax[0].set_title('Original Circles and Lines')
ax[0].set_xlabel('Re(z)')
ax[0].set_ylabel('Im(z)')
ax[0].legend()
ax[0].grid(True)
# Plot the transformed shapes
for circle in circles:
transformed = reciprocal(circle)
ax[1].plot(np.real(transformed), np.imag(transformed), label=f'Circle |z|={abs(circle[0]):.1f}')
for line in vertical_lines:
transformed = reciprocal(line)
ax[1].plot(np.real(transformed), np.imag(transformed), label=f'Vertical Line x={line[0].real:.1f}')
for line in horizontal_lines:
transformed = reciprocal(line)
ax[1].plot(np.real(transformed), np.imag(transformed), label=f'Horizontal Line y={line[0].imag:.1f}')
ax[1].set_title('Transformed Circles and Lines under f(z) = 1/z')
ax[1].set_xlabel('Re(w)')
ax[1].set_ylabel('Im(w)')
ax[1].legend()
ax[1].grid(True)
plt.tight_layout()
plt.show()
با این کد، میتوانیم مشاهده کنیم که چگونه دایرهها و خطوط در صفحه مختلط تحت تبدیل ( \frac{1}{z} ) تغییر شکل میدهند.
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