# Understanding the Origin of Complex Numbers
Complex numbers, an extension of the real numbers, have captivated mathematicians for centuries. The journey to our current understanding of complex numbers has been a long and sometimes controversial one. In this article, we will explore the origins of complex numbers and their development over time.
## A Brief History
The concept of complex numbers first arose in the context of solving polynomial equations. The need for complex numbers became apparent with cubic equations, which sometimes led to the disturbing occurrence of square roots of negative numbers, something that was not understood at the time.
The Italian mathematician Gerolamo Cardano is often credited with introducing complex numbers in his 1545 work “Ars Magna,” while trying to solve cubic equations. However, these numbers were not fully accepted until much later. Rene Descartes coined the term “imaginary” for these puzzling square roots of negative numbers, as they did not correspond to any known number.
It was not until the 18th century that complex numbers gained broader acceptance. Leonhard Euler introduced the notation \( i \) for the square root of -1, and his formula \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \) linked complex numbers to trigonometry.
The geometric interpretation of complex numbers as points on a plane (now known as the complex plane or Argand plane) was developed by Caspar Wessel and Jean-Robert Argand. This view allowed for a more intuitive understanding and acceptance of complex numbers.
Today, complex numbers are essential in various fields of science and engineering, including electrical engineering, quantum physics, applied mathematics, and more.
## Mathematical Foundation of Complex Numbers
A complex number is defined as a number of the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit with the property that \( i^2 = -1 \). The real part of the complex number is \( a \), and the imaginary part is \( b \).
### Problems and Solutions
Let’s now dive into some problems that will deepen our understanding of complex numbers and their properties.
#### Problem 1:
Evaluate \( i^4 \) and justify why \( i^4 = 1 \).
**Solution:**
Given \( i^2 = -1 \), if we square \( i^2 \) again, we get \( i^4 = (i^2)^2 = (-1)^2 = 1 \).
#### Problem 2:
Find the real and imaginary parts of \( 3 + 4i \).
**Solution:**
The real part is 3, and the imaginary part is 4.
#### Problem 3:
Determine the complex conjugate of \( 6 – 2i \).
**Solution:**
The complex conjugate is obtained by changing the sign of the imaginary part: \( 6 + 2i \).
#### Problem 4:
Simplify \( (1 + 2i)(3 – 4i) \).
**Solution:**
\[
(1 + 2i)(3 – 4i) = 3 – 4i + 6i – 8i^2 = 3 + 2i + 8 = 11 + 2i
\]
#### Problem 5:
Calculate the magnitude (absolute value) of \( 3 + 4i \).
**Solution:**
The magnitude is \( \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).
#### Problem 6:
Divide \( 1 + i \) by \( 1 – i \) and express the result in standard form.
**Solution:**
\[
\frac{1 + i}{1 – i} = \frac{(1 + i)(1 + i)}{(1 – i)(1 + i)} = \frac{1 + 2i + i^2}{1 – i^2} = \frac{1 + 2i – 1}{1 + 1} = \frac{2i}{2} = i
\]
#### Problem 7:
Solve the equation \( z^2 + 1 = 0 \) for \( z \).
**Solution:**
\[
z^2 = -1 \Rightarrow z = \pm i
\]
#### Problem 8:
Convert \( 5e^{i\pi/3} \) to rectangular form.
**Solution:**
Using Euler’s formula, \( 5e^{i\pi/3} = 5(\cos(\pi/3) + i\sin(\pi/3)) = 5\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{5}{2} + \frac{5\sqrt{3}}{2}i \).
#### Problem 9:
Find the product of \( i \) and its complex conjugate.
**Solution:**
The complex conjugate of \( i \) is \( -i \). Thus, \( i \cdot (-i) = -i^2 = 1 \).
#### Problem 10:
Express \( \sqrt{-25} \) as a complex number.
**Solution:**
\[
\sqrt{-25} = \sqrt{25} \cdot \sqrt{-1} = 5i
\]
#### Problem 11:
If \( z = a + bi \) and \( z^2 = 4 – 3i \), find \( a \) and \( b \).
**Solution:**
\[
(a + bi)^2 = a^2 – b^2 + 2abi = 4 – 3i
\]
Matching real and imaginary parts, we get:
\[
a^2 – b^2 = 4
\]
\[
2ab = -3
\]
Without loss of generality, by solving these equations simultaneously, we get \( a = 1, b = -\frac{3}{2} \) or \( a = -1, b = \frac{3}{2} \).
#### Problem 12:
Find the polar form of \( -1 – i \).
**Solution:**
Magnitude: \( r = \sqrt{1^2 + 1^2} = \sqrt{2} \)
Argument: \( \theta = \arctan\left(\frac{-1}{-1}\right) + \pi = \arctan(1) + \pi = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \)
Polar form: \( \sqrt{2} \left( \cos\left(\frac{5\pi}{4}\right) + i\sin\left(\frac{5\pi}{4}\right) \right) \)
#### Problem 13:
Prove that \( (a + bi)(a – bi) \) is always a real number.
**Solution:**
\[
(a + bi)(a – bi) = a^2 – abi + abi – b^2i^2 = a^2 + b^2
\]
Since \( a \) and \( b \) are real, \( a^2 + b^2 \) is also real.
#### Problem 14:
Given \( z_1 = 1 + i \) and \( z_2 = 2 – i \), find \( z_1 + z_2 \).
**Solution:**
\( z_1 + z_2 = (1 + i) + (2 – i) = 1 + 2 + i – i = 3 \)
#### Problem 15:
Express \( 8i \) as a complex number in polar form.
**Solution:**
Magn
