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Complex Numbers

a + bi, the sum of a real and an imaginary number.

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Defining Complex Numbers

The coldest possible temperature, known as absolute zero is almost –460 degrees Fahrenheit. What is the square root of this number?

Watch This

First, watch this video.

Khan Academy: Introduction to i and Imaginary Numbers

Then, watch this video.

Khan Academy: Complex Numbers


Before this concept, all numbers have been real numbers. 2, -5, 11 , and 13 are all examples of real numbers. Look at #1 from the Review Queue. With what we have previously learned, we cannot find 25 because you cannot take the square root of a negative number. There is no real number that, when multiplied by itself, equals -25. Let’s simplify 25.


In order to take the square root of a negative number we are going to assign 1 a variable, i. i represents an imaginary number. Now, we can use i to take the square root of a negative number.


All complex numbers have the form a+bi, where a and b are real numbers. a is the real part of the complex number and b is the imaginary part. If b=0, then a is left and the number is a real number. If a=0, then the number is only bi and called a pure imaginary number. If b0 and a0, the number will be an imaginary number.

Example A

Find 162 .

Solution: First pull out the i. Then, simplify 162 .


Investigation: Powers of i

In addition to now being able to take the square root of a negative number, i also has some interesting properties. Try to find i2,i3, and i4.

1. Write out i2 and simplify. i2=ii=11=12=1

2. Write out i3 and simplify. i3=i2i=1i=i

3. Write out i4 and simplify. i4=i2i2=11=1

4. Write out i5 and simplify. i5=i4i=1i=i

5. Write out i6 and simplify. i6=i4i2=11=1

6. Do you see a pattern? Describe it and try to find i19.

You should see that the powers of i repeat every 4 powers. So, all the powers that are divisible by 4 will be equal to 1. To find i19, divide 19 by 4 and determine the remainder. That will tell you what power it is the same as.


Example B


a) i32

b) i50

c) i7


a) 32 is divisible by 4, so i32=1.

b) 50÷4=12, with a remainder of 2. Therefore, i50=i2=1.

c) 7÷4=1, with a remainder of 3. Therefore, i7=i3=i

Example C

Simplify the complex expressions.

a) \begin{align*}(6-4i)+(5+8i)\end{align*}

b) \begin{align*}9-(4+i)+(2-7i)\end{align*}

Solution: To add or subtract complex numbers, you need to combine like terms. Be careful with negatives and properly distributing them. Your answer should always be in standard form, which is \begin{align*}a + bi\end{align*}.

a) \begin{align*}(6-4i)+(5+8i)={\color{red}6}{\color{blue}-4i}+{\color{red}5}+{\color{blue}8i}={\color{red}11}+{\color{blue}4i}\end{align*}

b) \begin{align*}9-(4+i)+(2-7i)={\color{red}9-4}{\color{blue}-i}+{\color{red}2}{\color{blue}-7i}={\color{red}7}{\color{blue}-8i}\end{align*}

Intro Problem Revisit We're looking for \begin{align*}\sqrt{-460}\end{align*} .

First we need to pull out the \begin{align*}i\end{align*}. Then, we need to simplify \begin{align*}\sqrt{460}\end{align*} .

\begin{align*}\sqrt{-460}=\sqrt{-1} \cdot \sqrt{460}=i\sqrt{460}=i\sqrt{4 \cdot 115}=2i\sqrt{115}\end{align*}

Guided Practice


1. \begin{align*}\sqrt{-49}\end{align*}

2. \begin{align*}\sqrt{-125}\end{align*}

3. \begin{align*}i^{210}\end{align*}

4. \begin{align*}(8-3i)-(12-i)\end{align*}


1. Rewrite \begin{align*}\sqrt{-49}\end{align*} in terms of \begin{align*}i\end{align*} and simplify the radical.


2. Rewrite \begin{align*}\sqrt{-125}\end{align*} in terms of \begin{align*}i\end{align*} and simplify the radical.

\begin{align*}\sqrt{-125}=i\sqrt{125}=i\sqrt{25 \cdot 5}=5i\sqrt{5}\end{align*}

3. \begin{align*}210 \div 4=52\end{align*}, with a remainder of 2. Therefore, \begin{align*}i^{210}=i^2=-1\end{align*}.

4. Distribute the negative and combine like terms.



Imaginary Numbers
Any number with an \begin{align*}i\end{align*} associated with it. Imaginary numbers have the form \begin{align*}a + bi\end{align*} or \begin{align*}bi\end{align*}.
Complex Numbers
All real and imaginary numbers. Complex numbers have the standard form \begin{align*}a + bi\end{align*}, where \begin{align*}a\end{align*} or \begin{align*}b\end{align*} can be zero. \begin{align*}a\end{align*} is the real part and \begin{align*}bi\end{align*} is the imaginary part.
Pure Imaginary Numbers
An imaginary number without a real part, only \begin{align*}bi\end{align*}.


Simplify each expression and write in standard form.

  1. \begin{align*}\sqrt{-9}\end{align*}
  2. \begin{align*}\sqrt{-242}\end{align*}
  3. \begin{align*}6\sqrt{-45}\end{align*}
  4. \begin{align*}-12i\sqrt{98}\end{align*}
  5. \begin{align*}\sqrt{-32} \cdot \sqrt{-27}\end{align*}
  6. \begin{align*}7i\sqrt{-126}\end{align*}
  7. \begin{align*}i^8\end{align*}
  8. \begin{align*}16i^{22}\end{align*}
  9. \begin{align*}-9i^{65}\end{align*}
  10. \begin{align*}i^{365}\end{align*}
  11. \begin{align*}2i^{91}\end{align*}
  12. \begin{align*}\sqrt{-\frac{16}{80}}\end{align*}
  13. \begin{align*}(11-5i)+(6-7i)\end{align*}
  14. \begin{align*}(14+2i)-(20+9i)\end{align*}
  15. \begin{align*}(8-i)-(3+4i)+15i\end{align*}
  16. \begin{align*}-10i-(1-4i)\end{align*}
  17. \begin{align*}(0.2+1.5i)-(-0.6+i)\end{align*}
  18. \begin{align*}6+(18-i)-(2+12i)\end{align*}
  19. \begin{align*}-i+(19+22i)-(8-14i)\end{align*}
  20. \begin{align*}18-(4+6i)+(17-9i)+24i\end{align*}

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