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10.1: Imaginary Numbers

Difficulty Level: At Grade Created by: CK-12

Name: __________________

Imaginary Numbers

1. Explain, using words and equations, why the equation $x^2=-1$ has no answer, but $x^3=-1$ does.

OK, so now we are going to use our imaginations. (Didn’t think we were allowed to do that in math class, did you?) Suppose there were an answer to $x^2=-1?$ Obviously it wouldn’t be a number that we are familiar with (such as $5,-\frac{3}{4}$, or $\pi$). So, let’s just give it a new name: $i$, because it’s imaginary. What would it be like?

The definition of the imaginary number $i$ is that it is the square root of $-1$:

$i = \sqrt{-1}$ or, equivalently, $i^2=-1$

Based on that definition, answer the following questions. In each case, don’t just guess—give a good mathematical reason why the answer should be what you say it is!

2. What is $i(-i)?$ ($^*$Remember that $-i$ means $-1 \times i.$)

3. What is $(-i)^2?$

4. What is $(3i)^2?$

5. What is $(-3i)^2?$

6. What is $\left (\sqrt{2}i\right )^2?$

7. What is $\left (\sqrt{2i}\right )^2?$

8. What is $\sqrt{-25}?$

9. What is $\sqrt{-3}?$

10. What is $\sqrt{-8}?$

11. Fill in the following table.

$&i^1\\&i^2\\&i^3\\&i^4\\&i^5\\&i^6\\&i^7\\&i^8\\&i^9\\&i^{10}\\&i^{11}\\&i^{12}$

12. Fill in the following table.

$&1^{100}\\&i^{101}\\&i^{102}\\&i^{103}\\&i^{104}$

Now let’s have some more fun!

13. $(3 + 4i)^2 =$

14. $(3 + 4i) (3 - 4i) =$

15. $\left (\frac{1}{i}\right )^2 =$

16. Simplify the fraction $\frac{1}{i}$. (Hint: Multiply the top and bottom by $i$.)

17. Square your answer to $^\#16$. Did you get the same answer you got to $^\#15?$ Why or why not?

18. Simplify the fraction $\frac{1}{3+2i}$. (Hint: Multiply the top and bottom by $3-2i$.)

Name: __________________

Homework: Imaginary Numbers

We began our in-class assignment by talking about why $x^3=-1$ does have a solution, whereas $x^2=-1$ does not. Let’s talk about the same thing graphically.

1. On the graph below, do a quick sketch of $y=x^3$.

a. Draw, on your graph, all the points on the curve where $y=1$. How many are there?

b. Draw, on your graph, all the points on the curve where $y=0$. How many are there?

c. Draw, on your graph, all the points on the curve where $y=-1$. How many are there?

2. On the graph below, do a quick sketch of $y=x^2$.

a. Draw, on your graph, all the points on the curve where $y=1$. How many are there?

b. Draw, on your graph, all the points on the curve where $y=0$. How many are there?

c. Draw, on your graph, all the points on the curve where $y=-1$. How many are there?

3. Based on your sketch in number $2\ldots$

a. If $a$ is some number such that $a>0$, how many solutions are there to the equation $x^2=a?$

b. If $a$ is some number such that $a=0$, how many solutions are there to the equation $x^2=a?$

c. If $a$ is some number such that $a<0$, how many solutions are there to the equation $x^2=a?$

d. If $i$ is defined by the equation $i^2=-1$, where the heck is it on the graph?

OK, let’s get a bit more practice with $i$.

4. In class, we made a table of powers of $i$, and found that there was a repeating pattern.

Make that table again quickly below, to see the pattern.

$&i^1\\&i^2\\&i^3\\&i^4\\&i^5\\&i^6\\&i^7\\&i^8\\&i^9\\&i^{10}\\&i^{11}\\&i^{12}$

5. Now let’s walk that table backward. Assuming the pattern keeps up as you back up, fill in the following table. (Start at the bottom.)

$&i^{-4}\\&i^{-3}\\&i^{-2}\\&i^{-1}\\&i^0$

6. Did it work? Let’s figure it out. What should $i^0$ be, according to our general rules of exponents?

7. What should $i^{-1}$ be, according to our general rules of exponents? Can you simplify it to look like the answer in your table?

8. What should $i^{-2}$ be, according to our general rules of exponents? Can you simplify it to look like the answer in your table?

9. What should $i^{-3}$ be, according to our general rules of exponents? Can you simplify it to look like the answer in your table?

10. Simplify the fraction $\frac{i}{4-3i}$.

Name: __________________

Complex Numbers

A complex number is written in the form $a+bi$ where $a$ and $b$ are real numbers. $a$ is the “real part” and $bi$ is the “imaginary part.”

Examples are: $3+4i$ ($a$ is $3$, $b$ is $4$) and $3-4i$ ($a$ is $3$, $b$ is $-4$).

1. Is $4 \ a$ complex number? If so, what are $a$ and $b$? If not, why not?

2. Is $i \ a$ complex number? If so, what are $a$ and $b$? If not, why not?

3. Is $0 \ a$ complex number? If so, what are $a$ and $b$? If not, why not?

All four operations—addition, subtraction, multiplication, and division—can be done to complex numbers, and the answer is always another complex number. So, for the following problems, let $X=3+4i$ and $Y=5-12i$. In each case, your answer should be a complex number, in the form $a+bi.$

4. Add: $X+Y$

5. Subtract: $X-Y$

6. Multiply: $XY$

7. Divide: $\frac{X}{Y}$ ($^*$To get the answer in $a+bi$ form, you will need to use a trick we learned yesterday.)

8. Square: $X^2$

The complex conjugate of a complex number $(a+bi)$ is defined as $(a-bi)$. That is, the real part stays the same, and the imaginary part switches sign.

9. What is the complex conjugate of $(5-12i)?$

10. What do you get when you multiply $(5-12i)$ by its complex conjugate?

11. Where have we used complex conjugates before?

For two complex numbers to be equal, there are two requirements: the real parts must be the same, and the imaginary parts must be the same. In other words, $2+3i$ is only equal to $2+3i.$ It is not equal to $2-3i$ or to $3+2i$ or to anything else. So it is very easy to see if two complex numbers are the same, as long as they are both written in $a+bi$ form: you just set the real parts equal, and the imaginary parts equal. (If they are not written in that form, it can be very tricky to tell: for instance, we saw earlier that $\frac{1}{i}$ is the same as $-i$ !)

12. If $2-3i=m+ni,$ what are $m$ and $n$?

13. Solve for $x$ and $y: (x-6y)+(x+2y)i = 1-3i$

Finally, remember...rational expressions? We can have some of those with complex numbers as well!

14. Simplify. As always, your answer should be in the form $a+bi. \ \frac{\left(\frac{4+2i}{3+2i}\right )}{\left (\frac{5-3i}{7-i}\right )}$

15. Simplify. $\frac{4+2i}{3+2i} - \frac{5-3i}{7-i}$

Name: __________________

Homework: Complex Numbers

1. $(3+7i) - (4+7i) =$
2. $(5-3i) + (5-3i) =$
3. $2(5-3i) =$
4. $(5-3i) (2+0i) =$
5. What is the complex conjugate of $(5-3i)?$
6. What do you get when you multiply $(5-3i)$ by its complex conjugate?
7. What is the complex conjugate of $7$?
8. What do you get when you multiply $7$ by its complex conjugate?
9. What is the complex conjugate of $2i$?
10. What do you get when you multiply $2i$ by its complex conjugate?
11. What is the complex conjugate of $(a+bi)?$
12. What do you get when you multiply $(a+bi)$ by its complex conjugate?
13. I’m thinking of a complex number $z$. When I multiply it by its complex conjugate (designated as $z^*$) the answer is $25$.
1. What might $z$ be?
2. Test it, and make sure it works—that is, that $(z) \ (z^*)=25!$
14. I’m thinking of a different complex number $z$. When I multiply it by its complex conjugate, the answer is $3+2i.$
1. What might $z$ be?
2. Test it, and make sure it works—that is, that $(z) \ (z^*)=3+2i!$
15. Solve for $x$ and $y: x^2+2x^2i+4y+40yi = 7-2i$
16. Finally, a bit more exercise with rational expressions. We’re going to take one problem and solve it two different ways. The problem is $\frac{3}{2+i}-\frac{7i}{3+4i}$. The final answer, of course, must be in the form $a+bi$.
1. Here is one way to solve it: the common denominator is $(2+i) \ (3+4i)$. Put both fractions over the common denominator and combine them. Then, take the resulting fraction, and simplify it into $a+bi$ form.
2. Here is a completely different way to solve the same problem. Take the two fractions we are subtracting and simplify them both into $a+bi$ form, and then subtract.
3. Did you get the same answer? (If not, something went wrong...) Which way was easier?

Name: __________________

Me, Myself, and the Square Root of i

We have already seen how to take a number such as $\frac{3}{2-i}$ and rewrite it in $a+bi$ format. There are many other numbers—such as $2^i$ and $\log(i)$—that do not look like $a+bi$, but all of them can be turned into $a+bi$ form. In this assignment, we are going to find $\sqrt{i}$—that is, we are going to rewrite $\sqrt{i}$ so that we can clearly see its real part and its imaginary part.

How do we do that? Well, we want to find some number $z$ such that $z^2=i$. And we want to express $z$ in terms of its real and imaginary parts—that is, in the form $a+bi$. So what we want to solve is the following equation:

$(a+bi)^2 = i$

You are going to solve that equation now. When you find $a$ and $b$, you will have found the answers.

Stop now and make sure you understand how I have set up this problem, before you go on to solve it.

1. What is $(a+bi)^2?$ Multiply it out.

2. Now, rearrange your answer so that you have collected all the real terms together and all the imaginary terms together.

Now, we are trying to solve the equation $(a+bi)^2=i.$ So take the formula you just generated in number $2$, and set it equal to $i.$ This will give you two equations: one where you set the real part on the left equal to the real part on the right, and one where you set the imaginary part on the left equal to the imaginary part on the right.

3. Write down both equations.

4. Solve the two equations for $a$ and $b$. (Back to “simultaneous equations,” remember?) In the end, you should have two $(a,b)$ pairs that work in both equations.

5. So... now that you know $a$ and $b$, write down two complex answers to the problem $x^2=i.$

6. Did all that work? Well, let’s find out. Take your answers in $^\#5$ and test them: that is, square it, and see if you get $i.$ If you don’t, something went wrong!

7. OK, we’re done! Did you get it all? Let’s find out. Using a very similar technique to the one that we used here, find $\sqrt{-i}$: that is, find the two different solutions to the problem $z^2=i.$ Check them!

Name: __________________

The Many Merry Cube Roots of -1

When you work with real numbers, $x^2=1$ has two different solutions ($1$ and $-1$). But $x^2=-1$ has no solutions at all. When you allow for complex numbers, things are much more consistent: $x^2=-1$ has two solutions, just like $x^2=1$. In fact, $x^2=n$ will have two solutions for any number $n$— positive or negative, real or imaginary or complex. There is only one exception to this rule.

1. What is the one exception?

You might suspect that $x^3=n$ should have three solutions in general—and you would be right! Let’s take an example. We know that when we are working with real numbers, $x^3=-1$ has only one solution.

2. What is the one solution?

But if we allow for complex answers, $x^3=-1$ has three possible solutions. We are going to find the other two.

How do we do that? Well, we know that every complex number can be written as $(a+bi)$, where $a$ and $b$ are real numbers. So if there is some complex number that solves $x^3=-1$, then we can find it by solving the $a$ and $b$ that will make the following equation true:

$(a + bi)^3 = -1$

You are going to solve that equation now. When you find $a$ and $b$, you will have found the answers.

Stop now and make sure you understand how I have set up this problem, before you go on to solve it.

3. What is $(a + bi)^3?$ Multiply it out.

4. Now, rearrange your answer so that you have collected all the real terms together and all the imaginary terms together.

Now, we are trying to solve the equation $(a+bi)^3=-1$. So take the formula you just generated in number $4$, and set it equal to $-1$. This will give you two equations: one where you set the real part on the left equal to the real part on the right, and one where you set the imaginary part on the left equal to the imaginary part on the right.

5. Write down both equations.

OK. If you did everything right, one of your two equations factors as $b(3a^2-b^2)=0$. If one of your two equations doesn’t factor that way, go back—something went wrong!

If it did, then let’s move on from there. As you know, we now have two things being multiplied to give $0$, which means one of them must be $0$. One possibility is that $b=0$: we’ll chase that down later. The other possibility is that $3a^2-b^2=0$, which means $3a^2=b^2$.

6. Solve the two equations for a and b by substituting $3a^2=b^2$ into the other equation

7. So... now that you know $a$ and $b$, write down two complex answers to the problem $x^3=-1$. If you don’t have two answers, look again!

8. But wait...shouldn’t there be a third answer? Oh, yeah... what about that $b=0$ business? Time to pick that one up. If $b=0$, what is a? Based on this $a$ and $b$, what is the third and final solution to $x^3=-1?$

9. Did all that work? Well, let’s find out. Take either of your answers in $^\#7$ and test it: that is, cube it, and see if you get $-1$. If you don’t, something went wrong!

Name: __________________

Homework: Quadratic Equations and Complex Numbers

I’m sure you remember the quadratic formula: $x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Back when we were doing quadratic equations, if we wound up with a negative number under that square root, we just gave up. But now we can solve these equations!

1. Use the quadratic formula to solve: $2x^2+6x+5=0$.
2. Use the quadratic formula to solve: $x^2-2x+5=0$.
3. Check one of your answers to $^\#2$.
4. Solve by completing the square: $2x^2+10x+17=0$.
5. In general, what has to be true for a quadratic equation to have two complex roots?
6. What is the relationship between the two complex roots?
7. Is it possible to have a quadratic equation with one complex root?

Name: __________________

Sample Test: Complex Numbers

1. Fill in the following table.

$&i^{-1}\\&i^0\\&i^1\\&i^2\\&i^3\\&i^4\\&i^5\\&i^6\\&i^7$

Simplify.

2. $(i)^{85} =$

3. $(5i)^2 =$

4. $(ni)^{103} =$

5. a. $\sqrt{-20} =$

b. Other than your answer to part (a), is there any other number that you can square to get $-20?$ If so, what is it?

6. $(3a - bi)^2 =$

7. a. Complex conjugate of $4+i=$

b. What do you get when you multiply $4+i$ by its complex conjugate?

If the following are simplified to the form $a+bi$, what are $a$ and $b$ in each case?

8. $-i$

a. $a=$

b. $b=$

9. $\frac{n}{i}$

a. $a=$

b. $b=$

10. $\frac{5x}{1+2i} - \frac{2i}{3-i}$

a. $a=$

b. $b=$

11. If $2x+3xi+2y = 28+9i$, what are $x$ and $y?$

12. Make up a quadratic equation (using all real numbers) that has two complex roots, and solve it.

13. a. Find the two complex numbers (of course in the form $z=a+bi$) that fill the condition $z^2=-2i.$

b. Check one of your answers to part (a), by squaring it to make sure you get $-2i.$

Extra credit: Complex numbers cannot be graphed on a number line. But they can be graphed on a $2-$ dimensional graph: you graph the point $x+iy$ at $(x,y)$.

a. If you graph the point $5+12i,$ how far is that point from the origin $(0,0)?$

b. If you graph the point $x+iy$, how far is that point from the origin $(0,0)?$

c. What do you get if you multiply the point $x+iy$ by its complex conjugate? How does this relate to your answer to part (b)?

Feb 23, 2012

Apr 29, 2014