Fun with Complex and Imaginary Numbers

What is a complex number? Well, to get into that, we need to first understand what an imaginary number is. Okay, so you probably know that you can always have a square root of a positive number right? But what is the square root of a negative number??? For example, what in the world is sqrt(-1)??? This really doesn't make sense since for any real number n, n^2 = |n^2|, and since the absolute value of any real number is always positive, and 0^2 = 0. However, in 1637, Reni Descartes made an idea of an imaginary number, and let the value of sqrt(-1) equal to a variable i. Today, we call i an "imaginary number". A complex number is any number that can be expressed the form x + yi. In this forum, we will test your knowledge about complex numbers and see how good you are at it.

1) Let the value of (5i + 3)(2i + 4) be expressed in the form x + yi where x and y are real numbers. 

1a) What is x?

1b) What is y?

2) Let the value of (2i + 3)/(3i + 4) be expressed in the form a + bi where a and b are real number. 

2a) Find a.

2b) Find b.

BONUS: 3) Let there be quadratic be expressed in the form ax^2 + bx + c = 0. 

3a) What are the conditions of a and b such that the two roots of the quadratic are real?

3b) What are the conditions of a and b such that the two roots of the quadratic are imaginary?

prove that every prime 1 mod 4  can be expressed as a sum of squares (related to complex numbers)

x=-2, y=26; a=6, b=15. 

BONUS problem:

1. If b^2-4ac is non-negative (is great or equals 0) then the roots are real.

2. If b=0 and ac<0 then the roots are imaginary.  

You know what they should do is make a symbol like  ( i )  to represent imaginary numbers.

so if it is a negative 16 square root, it can now be -4i 

They should do that

Hamiltonian numbers or quaternions are a number system that extends the complex numbers.

Problem 1)

(5i+3)(2i+4)=x+yi

(5i+3)(2i+4)=2+26i

x=2, y=26

Problem 2)

(2i+3)/(3i+4)=a+bi

(2i+3)/(3i+4)=[(2i+3)(-3i+4)]/[(3i+4)(-3i+4)]=(18-i)/25

a=18/25, b=-1/25

Problem 3)

Let ax²+bx+c=0.

I will assume that a, b and c are real number constants with a≠0, and x is a complex number variable.

The two roots (or solutions) are different real numbers if b²-4ac>0.

The two roots (or solutions) are the same real number if b²-4ac=0.

The two roots (or solutions) are complex numbers but not real numbers if b²-4ac<0.

Wikipedia page: Quadratic equation

Good Job Everyone!