Complex numbers
(This page is not yet finished since it is so large, but a (VERY) large portion of it is finished.)
(This page is not yet finished since it is so large, but a (VERY) large portion of it is finished.)
A complex number is a number with a real part and an imaginary part. In other words, all complex numbers are in the form a+bi, where a and b are real numbers, and i is the imaginary number that represents the square root of -1. In a complex number, a is called the real part of the number and bi is called the imaginary part. There are many things to study about complex and imaginary numbers. In a complex number, if a=0, then it would just be an imaginary number, and if b=0, it would be a real number. Essentially, every number, real or imaginary, is a complex number. An example of a complex number would be 5+i, and another one would be 57.
First, let's find out what happens after we raise i to an exponent. Since i is the square root of -1, i^2=-1. Then what do you think i^3 is?
That's right, i^3 is -i, because i^3=i^2*i=-1*i=-i, so i^3=-i. Next, try to calculate what i^4 is.
i^4 is just i^2*i^2=-1*-1, which equals 1, so i^4 is 1. Finally, let us calculate i^999. Take a minute to think about this problem, and if you want to give up or you think you have the right answer, scroll down for the solution.
ANSWER:
Alright, I assume that you have thought about the previous problem, so here is the solution. Starting off, the high power 999 may be intimidating for you, so start with a few more calculations from the previous one. i^5 is just i*i^4, and now we know that i^4 is 1, it is simply i*1, which just yields i. Note how i^5 is the same as i^1. That is an interesting phenomenon. Let's calculate a few more powers to see if this continues.
i^6=i^5*i=i*i=-1, which is the same as i^2
i^7=i^6*i=-1*i=-i, which is the same as i^3
i^8=i^7*i=-i*i=1, which is the same as i^4
i^9=i^8*i=1*i=i, which is the same as i^1 or i^5
We could go on and on until we reach i^999, but hopefully, you are beginning to see a pattern in the powers. It seems like every four powers, the result of the power repeats itself. This might be a coincidence and in higher powers this will not be true, but here is a way to prove that it is.
Every power of i with an integer exponent can be written as either i^4x, i^4x*i, i^4x*i^2, or i^4x*i^3 where x is an integer. This is significant because i^4 is one, so i^4x can be rewritten as 1^x due to exponent rules (i^4x=(i^4)^x=1^x). Because 1 to any real power is 1, all the powers of i can be written as 1, 1*i, 1*i^2, or 1*i^3, which shows that there is a cycle. Now, back to the problem. i^999 can be dissected as
i^996*i^3
=1^249*i^3
=1*(-i)
=-i
So, through looking at patterns of powers of i we can find out that i^999 is -i. Now, you can use this to solve any power of i. Here is a handy set of rules. These rules are derived from the short exploration we did previously:
If i's exponent has a remainder of 1 when dividing by 4, the result will always be i
If i's exponent has a remainder of 2 when dividing by 4, the result will always be -1
If i's exponent has a remainder of 3 when dividing by 4, the result will always be -i
If i's exponent has a remainder of 0 when dividing by 4, the result will always be 1
For negative numbers, the remained is counted as the distance from the negative power of four directly smaller than it. For example, -11 would have a remainder of 1, as it is 1 higher than the smaller multiple, -12. Using calculations, we can confirm i^(-11) = 1/(i^11) = 1/(-i) = i, which get the same method as the remainder one.
Here are some practice problems for this. Try to solve them all before scrolling down and advancing to the next part.
What is i^256?
What is i^183?
What is i^394?
What is i^49394?
What is i^8047?
What is i^(-3847)?
What is i^50*i^39?
Now that we have explored the properties of purely imaginary numbers, let us now talk about fully complex numbers-numbers that both have a real part and an imaginary part. All complex numbers - regardless of if they are real, imaginary, or a mix - can be graphed on the complex plane.
The picture belongs to jillwillianms.github.io, which does not exist anymore.
In the complex plane, the x axis represents the real part of the complex number, and the y axis represents the imaginary part. For example, if you wanted to graph the complex number 5+6i, you would graph it here
at the purple dot. The dot has x of 5 and y of 6, which signifies its real part is 5 and the imaginary part is 6i. For any complex number a+bi, the x is a, and the y is b, not bi. Therefore, for this complex number, the correct way to graph it is start at the origin, (0,0), go 5 lines to the right, and go 6 lines up, and there is the point. Below is a picture explanation for the sequence of doing the problem.
It is now your turn to try graphing. For the following complex numbers, find the point on the complex plane above. You can copy-paste it and draw on it, you can print it out, or you can just plot it mentally. This is very straightforward and does not require much thought, so solutions will not be provided.
1+3i
5-2i
-3+3i
-10-5i
Next, new topic, the conjugate of a complex number, a+bi, is a-bi. For example, the conjugate of 1+3i is 1-3i , the conjugate of 5-2i is 5+2i, the conjugate of -3+3i is -3-3i, and the conjugate of -10-5i is -10+5i.
Find the conjugates of the following complex numbers:
30+32i
39-58i
-39
-67i
Okay, if you answered 30-32i, 39+59i, -39, and 67i, then you are correct. Here is a quick solution for all of these problems (I am very lazy, and I assume you understand all this. If you don't, don't worry, you will from all the applications of conjugates.)
30+32i=a+bi a=30, b=32 a-bi=30-32i
39-58i=a+bi a=39, b=-58 a-bi=39-(-58)i=39+58i
-39=a+bi a=-39, b=0 a-bi=-39-0i=-39
-67i=a+bi a=0, b=-67 a-bi=0-(-32)i
The main trick here is to find out what a and b are in the complex number, and then turn b into its negative form and attach it onto a.
Now, here are some more questions on a new topic, using basic operations on these new complex numbers. It's okay to not know how to answer them, but you should try to use your algebraic intuition as you did with variable to solve these. If you are inexperienced with variables and algebra and haven't read the other algebra sections, just try your best on these problems.
What is (5+3i)+(-4+i)?
What is (-3+i)+(2-7i)?
What is (5+6i)-(-10-9i)?
What is (2+2i)*(1+i)
For arithmetic, just combine like terms. The solution to #1 is to use the associative property and change it to (5+(-4))+(3i+i). As for multiplication, the trick here is to realize the i^2=-1, then multiply out all the pairs of terms. For example, here's how to solve #4.
2*1+2*i+2i*1+2i*i=2-2+2*i+2i*1=0+4i=4i
Polar form
Polar form is when you write