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Quiz about Complex Math For Kids
Quiz about Complex Math For Kids

Complex Math For Kids! Trivia Quiz


An introduction to imaginary numbers and complex mathematics based on algebra.

A multiple-choice quiz by Sedho7. Estimated time: 4 mins.
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Author
Sedho7
Time
4 mins
Type
Multiple Choice
Quiz #
378,030
Updated
Dec 03 21
# Qns
10
Difficulty
Average
Avg Score
7 / 10
Plays
221
- -
Question 1 of 10
1. Any elementary student who knows how to multiply will eventually be able to understand that a square root of a number, when multiplied to itself, will result in the number. For example, the square root of 16 is 4, because 4 x 4 = 16. However, the square root of a negative number is far more difficult to visualize. What specific letter is often assigned to represent the square root of -1? Hint


Question 2 of 10
2. Numbers that contain i are called imaginary, while numbers that don't contain i are called real. What term is used to refer to an expression that combines both real and imaginary terms? Hint


Question 3 of 10
3. Whenever you learn a new branch of mathematics (or of most things, actually), a good way to start is to try to look for similarities with things you already know. For example, the first thing we learn to do with (real) numbers is to add and subtract them. As it happens, the process is exactly the same for imaginary numbers. Just as 3 + 4 = 7, so does 3i + 4i = 7i. Now, what is 9i - 7i? Hint


Question 4 of 10
4. If adding and subtracting imaginary numbers is similar to real numbers, what happens when the numbers to be added or subtracted are complex? What is (2 + i) + (1 + 3i)? Hint


Question 5 of 10
5. Now that we've covered addition and subtraction, let's move on to multiplication of imaginary numbers. What would one get multiplying 3i x 5i? Remember that if i is the square root of -1, i x i = -1. Hint


Question 6 of 10
6. There is one interesting pattern that can be seen when i is multiplied to itself several times. By definition, i^2 = i x i = -1. Meanwhile, i^3 = i^2 x i = -i. Finally, i^4 = i^3 x i = 1. If one were to continue, i^5 = i, i^6 = -1, and one would soon realize that this sequence of i, -1, -i, 1, repeats again and again. Knowing this, what is the value of i^400? Hint


Question 7 of 10
7. We've talked about multiplying imaginary numbers, but what happens when one of the factors is complex? Simplify (1 + 2i)*(3i) by multiplying the imaginary factor 3i to 1, then multiplying 3i to 2i, then adding the results. Hint


Question 8 of 10
8. Now we move on to the last of the four basic operations. Similar to multiplication, the real and imaginary components of a complex number can be individually divided by the given divisor. Knowing this, what is (8 + 6i)/2? Hint


Question 9 of 10
9. Now that you know the basic operations for complex numbers, try combining the knowledge you've gleaned from multiple questions. Simplify this expression (-5 + 3i)*2i + (4 - 7i)? Hint


Question 10 of 10
10. When adding long series of numbers, one technique is to group them in such a way that they cancel out. Give it a try: i + i^2 + i^3 + i^4 + .... + i^40 + i^41? Hint



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Quiz Answer Key and Fun Facts
1. Any elementary student who knows how to multiply will eventually be able to understand that a square root of a number, when multiplied to itself, will result in the number. For example, the square root of 16 is 4, because 4 x 4 = 16. However, the square root of a negative number is far more difficult to visualize. What specific letter is often assigned to represent the square root of -1?

Answer: i

Square roots of negative numbers are called imaginary, and thus, the letter i is used to represent this. Also, if the square root of -1 is i, then this means that i x i = -1.
2. Numbers that contain i are called imaginary, while numbers that don't contain i are called real. What term is used to refer to an expression that combines both real and imaginary terms?

Answer: Complex

The expression 5 + 3i is a complex number. Its real component is 5, while its imaginary component is 3i. Note that this expression cannot be simplified further. Do NOT try to add 5 + 3 and "simplify" this as 8i.
3. Whenever you learn a new branch of mathematics (or of most things, actually), a good way to start is to try to look for similarities with things you already know. For example, the first thing we learn to do with (real) numbers is to add and subtract them. As it happens, the process is exactly the same for imaginary numbers. Just as 3 + 4 = 7, so does 3i + 4i = 7i. Now, what is 9i - 7i?

Answer: 2i

This can be extended to negative numbers as well. For example, 4i - 6i = -2i. Note is that the imaginary number "1i" is not usually written down with the number "1", and is simply written as "i".
4. If adding and subtracting imaginary numbers is similar to real numbers, what happens when the numbers to be added or subtracted are complex? What is (2 + i) + (1 + 3i)?

Answer: 3 + 4i

To add these complex numbers together, consider their real and imaginary components separately. The real components: 2 + 1 = 3; and the imaginary components i + 3i = 4i; and the final answer is 3 + 4i.
5. Now that we've covered addition and subtraction, let's move on to multiplication of imaginary numbers. What would one get multiplying 3i x 5i? Remember that if i is the square root of -1, i x i = -1.

Answer: -15

Multiply 3 x 5 = 15, and multiply i x i = -1, and multiply these results together to get -15.
6. There is one interesting pattern that can be seen when i is multiplied to itself several times. By definition, i^2 = i x i = -1. Meanwhile, i^3 = i^2 x i = -i. Finally, i^4 = i^3 x i = 1. If one were to continue, i^5 = i, i^6 = -1, and one would soon realize that this sequence of i, -1, -i, 1, repeats again and again. Knowing this, what is the value of i^400?

Answer: 1

Remembering that i^4 = 1 allows you to determine the value of i raised to any whole number. Every 4th value in the sequence (when the exponent is 4,8,12,16...400) has a value of 1.
7. We've talked about multiplying imaginary numbers, but what happens when one of the factors is complex? Simplify (1 + 2i)*(3i) by multiplying the imaginary factor 3i to 1, then multiplying 3i to 2i, then adding the results.

Answer: -6 + 3i

When multiplying two complex numbers, one would have to multiply all the possible combinations. For example, (1 + 2i)*(3 + 4i) = 1*3 + 1*4i + 2i*3 + 2i*4i = -5 + 10i. A common mnemonic used to ensure that all combinations have been covered is FOIL, which stands, for First (1*3), Outer (1*4i), Inner (2i*3), and Last (2i*4i).
8. Now we move on to the last of the four basic operations. Similar to multiplication, the real and imaginary components of a complex number can be individually divided by the given divisor. Knowing this, what is (8 + 6i)/2?

Answer: 4 + 3i

Dividing by a complex number requires significantly more effort, because the first step is to turn a complex denominator into a real number, and this is accomplished by multiplying BOTH the numerator and the denominator by a complex number. For example, to simplify (3 - i)/(1 + i), both complex numbers would first be multiplied by what is called the denominator's conjugate (1 - i), and the resulting expression would be (4 - 4i)/2, and this is equal to 2 - 2i.
9. Now that you know the basic operations for complex numbers, try combining the knowledge you've gleaned from multiple questions. Simplify this expression (-5 + 3i)*2i + (4 - 7i)?

Answer: -2 - 17i

Following the same MDAS rule in basic mathematics, (-5 +3i)*2i should be simplified first to yield -6 - 10i. Then adding this to (4 - 7i) gives the final result of -2 - 17i. If you've learned algebra, you can think of the letter i as a variable, and the only difference is that i^2 (and higher powers of i) should be replaced (i^2 = -1, i^4 = 1, etc).
10. When adding long series of numbers, one technique is to group them in such a way that they cancel out. Give it a try: i + i^2 + i^3 + i^4 + .... + i^40 + i^41?

Answer: i

Remember that i^2 = -1, i^3 = -i, and i^4 = 1. This means that i + i^2 + i^3 + i^4 = 0. Furthermore, because i^4 = 1, this means that i^5 = i, i^6 = i^2, i^7 = i^3, i^8 = i^4, and so the sum of these four (i^5 up to i^8) is still equal to 0. This goes on for every set of 4 consecutive numbers, which means that the sum up to i^40 is still 0. This leaves i^41, which is equal to i^1 = i.
Source: Author Sedho7

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