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In class we have spoken a lot about imaginary numbers. I have learned that an imaginary number or the use of an "i" in a problem is simply the square root of negative one. An imaginary number is just a number that is on different plane than real numbers. This in turn is why it is given its name of "imaginary" for the reason being that it does not fall under the set of real numbers.

Example:
(8+10i^2) (12-3i)+ (2+7i^2) ( 30+4i)

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In class, we have discussed the ways we can solve and simplify mathematic expressions involving imaginary and complex numbers. In an equation, when the letter "i" is present in an expression, it represents the square root of -1. This number being imaginary, is able to be shown on a graph when different planes of the graph are observed like we saw in the video.

Example Problem:
(6-7i^2)^2+(14+6i)^3

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For the past couple of classes we have been studying the use of imaginary numbers in Algebra, and what their basic function is. We have done several problems using "i" --> the square root of negative one. If a number is "imaginary", it means it does not cross on the x-axis, therefore its roots are "imaginary" and it cannot be seen on the grid. To represent imaginary numbers, we use "i", equivalent to the square root of negative one, as I stated earlier. Complex numbers are written out in the form -- a + bi, where a and b are both real numbers, and i is the imaginary number.

The equation I created is:
(3+2i)/(7+2i)

:)

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In class, we have been learning about imaginary numbers. I have learned that i is the imaginary number that is equal to the square root of negative one. A complex number is a solution with an imaginary number in it. Thus, it cannot be graphed on a Cartesian plane. A complex number is normally written in in a+bi form. We also learned that multiplying i to an exponent can give you four different products. i^2 is equivalent to -1, i^3 is equivalent to -i, i^4 is equivalent to 1,i^5 is equivalent to i. This works out to put complex numbers into the simplistic a+bi form.
Problem:
(6+2i^2) (3+4i^3)+3

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Part 1:
Throughout our studies of imaginary numbers in class, we have gained the understanding of some very key concepts related to imaginary numbers. i is the term used to represent imaginary numbers, and is equal to the square root of negative 1. The term "imaginary number" is actually quite the misnomer, as imaginary numbers are not in fact imaginary. What caused them to get this name is the fact that they cannot be seen on a Cartesian plane, unlike real numbers. An imaginary number, by itself, is one that is expressed in terms of the square root of a negative number. In laymen's terms, a number times i is just that number times the square root of negative 1. When i is squared, or multiplied by itself, it becomes -1.
Part 2:
(4+i^4) (5+i^2) (2+2i)

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We have been learning about imaginary numbers and complex numbers in class.  We have learnt that we can have a solution for an equation that is a complex number, which can be written in a+bi form. i (which is an imaginary number) is equal to the square root of negative one, so we use it if we have a negative under a square root. One can multiple, add, subtract and do any other operation on an imaginary  number. Complex solutions have a imaginary number in them (the i in the a+bi), so can not be traditionally graphed on the Cartesian plane as one would do with a real number solution. Hence why they are complex solutions. Not all equations that have imaginary numbers in them necessarily have complex solutions, but many a time the do.

An equation to try:
(3+3i)/2(1-3i)/3

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Part 1:
In class, I have learned and gained a better understanding of imaginary numbers and complex numbers. Imaginary and complex numbers are related to each other because a complex number is written as a+bi. The "i" part of the equation is an imaginary number. The "a" and "b" part of the equation are real numbers. Equations using imaginary numbers do not need to result in one whole or real number. It can be left in the form a+bi (complex number). "i" is equal to the square root of -1.
Part 2:
(4+6i^4)(3+2i)(1+5i)

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Part 1:
The class has been studying the imaginary number and complex numbers in depth. A complex number and an imaginary are related because a complex number is usually written in the form a+bi. i is the imaginary unit, and it is also equal to to the square root of -1. In the equation a+bi, a and b are real numbers. Real numbers include all rational and irrational, basically numbers along a number line. Whereas imaginary number are not in that category. While i= the square root of -1, i²=-1. This helps to solve equations so that the answer can be put into the complex number a+bi. When encountering the square root of any negative number you can simplify it, or make it more manageable by taking out the square root of -1 and replacing it with "i." An example of this is the square root  of -9. First you can take out the square root of -1 to get the square root of 9 i (the i is outside the square root) Then if the number inside the square root can be simplified you can do so. In this case 9 is a perfect square, the square root of 9 is 3. So the answer is 3i.

Part 2:
Below is my challenge problem for people to solve. We have worked with add, subtracting, multiplying and dividing complex numbers.
Tips:
For adding and subtracting two complex numbers only like terms will interact. For example both a's will add or subtract and both bi's will add or subtract. In multiplication or division both a's and both bi's will multiply and divide amongst each other.

Good Luck!
(6i+7i¹⁰⁰)(3i+4)

# data[].attributes.body
In class we have been working on imaginary numbers and complex numbers. An imaginary number is the square root of -(number), where i would be written in place of that because you can't have a negative number underneath a radical. Imaginary numbers are numbers represented as i and in the formula a+bi, a and b are real numbers. Real numbers include all rational numbers and all rational numbers, whole numbers, natural numbers, decimals or fractions. While finding the roots on a Cartesian plane imaginary numbers can not be graphed. i^2 is equal to -1.

Guide:
After learning about complex numbers and imaginary numbers I a able to apply my knowledge to other problems. For example multiplying, adding or subtracting complex numbers. There are several methods that can be used:
(2+3i)+(5+4i)
In solving this, because there is an operation in between the parenthesis as shown above, add the like terms. Both the real parts a (2+5) and the complex parts bi (3i +4i).
2+5+3i+4i=
7+7i
If the operation was a negative instead of a + the - would be distributed and then continue.

Another example is where multiplying is included in solving the problem.
(4+7i)(5+8i)
First use the F.O.I.L. method, which is also known as the distributive property.
20+32i+35i+56i^2
20+67i-56
*Remember i^2=-1*
-36+67i

Part two:
(3+6i)(9+4i) +(2+7i)^2

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Part One:

Our class is currently working on imaginary numbers and complex numbers. Complex numbers are typically written in the “a+bi” form, in which “a” and “b” are real numbers and “i” is an imaginary number. A number is considered “real” when it can be represented on a number line or a graph. The “real number” category includes whole numbers, natural numbers, integers, decimals, fractions, rational numbers, and irrational numbers. Unlike these real numbers, imaginary numbers cannot be physically graphed on a Cartesian plane (named after French mathematician Rene Descartes). An imaginary number is a number that is expressed by the square root of a negative number (usually the square root of –1, which is represented by the letter “i”). For example, an imaginary number may be 5i, which is the same as 5 times the square root of –1. It is important to know that i² = -1 and that i = the square root of -1.

Examples:

In addition to learning the definition of a complex number, I have also learned how to add, multiply, and subtract complex numbers. For example, if I am given (1+3i)(2+4i), I know that I can “F.O.I.L” (or use the distributive property) to multiply the complex numbers. Notice how the end result is also a complex number (a+bi). These are my steps:

(1+3i) (2+4i)

2 + 4i + 6i + 12i²

2 + 10i + 12(-1)

2 + 10i - 12

-10 + 10i

However, it might be different if I was given (1+3i) + (2+4i). With this equation, I am asked to add the complex numbers instead of multiplying them. This process is similar toward the process of subtracting two complex numbers. Instead of using the distributive property (as I would with multiplication), I will need to combine like terms. Notice how the end result is a complex number. These are my steps:

(1+3i) + (2+4i)

3 + 7i

There are many ways to combine complex numbers, but these are just two common ways.

Part Two:

Below is my own “challenging mathematics problem” that includes complex numbers. Good luck!

# data[].attributes.headline
While working with imaginary numbers, I have learned many things. First, the basic of what an imaginary number is. An imaginary number is a number that is expressed in terms of the square root of a negative number. This number is considered "imaginary" simply because it is not real-- it is not on the number line, and does not exist. When faced with a square root, where a negative number is under it, this is an imaginary number. You must remember that the square root of negative 1= i (imaginary). For example. the square root of -24 = the square root of 24 times the square root of -1. This can be represented as the square root of 24, with an 'i' before the radical. I have learned that when i is raised to different powers, the value of it differs. i to the 0 and i to the 4th is equal to 1. i to the 5th and i to the 1st are equal to i. i to the 2nd or i to the 6th is equal to -1. i to the 3rd or i to the 7th are equal to -i. Since you have for options, another way to do this is to divide the value (number before i) of i, by four. If there is no remainder the answer is 1, if there is a remainder of 1 or .25 the answer is i, if there is a remainder of 2 or .5 the answer is -1, and lastly, if there is a remainder of 3 or .75 the answer is -i.  

# data[].attributes.subject
While working with imaginary numbers, I have learned many things. First, the basic of what an imaginary number is. An imaginary number is a number that is expressed in terms of the square root of a negative number. This number is considered "imaginary" simply because it is not real-- it is not on the number line, and does not exist. When faced with a square root, where a negative number is under it, this is an imaginary number. You must remember that the square root of negative 1= i (imaginary). For example. the square root of -24 = the square root of 24 times the square root of -1. This can be represented as the square root of 24, with an 'i' before the radical. I have learned that when i is raised to different powers, the value of it differs. i to the 0 and i to the 4th is equal to 1. i to the 5th and i to the 1st are equal to i. i to the 2nd or i to the 6th is equal to -1. i to the 3rd or i to the 7th are equal to -i. Since you have for options, another way to do this is to divide the value (number before i) of i, by four. If there is no remainder the answer is 1, if there is a remainder of 1 or .25 the answer is i, if there is a remainder of 2 or .5 the answer is -1, and lastly, if there is a remainder of 3 or .75 the answer is -i.  

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Problem:
1/2i(15i+12)--> cubed.

It says anonymous, I don't know why. This is Maddy Barket.

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Part 1: Take 5 minutes to conduct a reflect analysis of your learning on a) the imaginary number or b) a complex number.   Your feedback should serve as a tool for someone to uses a tutorial on the concept.

Part 2: In supporting your learning you must provided a challenging mathematics problem on this topic.  *Do not answer the problem.  Allow someone else to visit your post to provide the answer.   You will then return to the post and let them know if the answer they provided is correct.