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Imaginary Numbers

i = sqrt (-1)

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Imaginary Numbers

The solutions to a quadratic equation show up as the x-intercepts of the corresponding quadratic function. The parabola y=x22x+3\begin{align*}y=x^2-2x+3\end{align*} is shown below.

What does this graph tell you about the solutions to x22x+3=0\begin{align*}x^2-2x+3=0\end{align*}?

Guidance

When humans first created the concept of numbers, they only had the counting numbers (whole numbers) {1,2,3,...}\begin{align*}\{1,2,3,...\}\end{align*} because numbers were meant to count physical objects. It took a long time (more than 700 years!) before even the concept of 0\begin{align*}0\end{align*} was invented in the year 500 AD in India. Since then, humans have slowly been adding to our number system so that it can work for us. Fractions, decimals, and negative numbers have become an important part of our world. For a long time, it was accepted that the square root of negative numbers did not exist. In order for a solution to exist to the equation x2=1\begin{align*}x^2=-1\end{align*}, mathematicians invented a solution. This solution is called the imaginary number and is noted by the letter i\begin{align*}i\end{align*}:

1=i\begin{align*}\sqrt{-1}=i\end{align*} and i2=1\begin{align*}i^2=-1\end{align*}

Imaginary numbers were not commonly accepted in mathematics until the 1700s, but since then they have become an important part of our number system and are especially important in physics. They are called imaginary because they cannot be found on a traditional number line of real numbers. The square root of any negative number can be written in terms of the imaginary number i\begin{align*}i\end{align*}:

• 4=41=41=2i\begin{align*}\sqrt{-4}=\sqrt{4\cdot -1}=\sqrt{4}\sqrt{-1}=2i\end{align*}
• 5=51=51=5i\begin{align*}\sqrt{-5}=\sqrt{5\cdot -1}=\sqrt{5}\sqrt{-1}=\sqrt{5}i\end{align*}
• 16=161=161=4i\begin{align*}\sqrt{-16}=\sqrt{16\cdot -1}=\sqrt{16}\sqrt{-1}=4i\end{align*}

You can perform addition, subtraction, and multiplication with imaginary numbers just like regular numbers (you can also do division, but that is a bit more complicated and won't be considered here). When performing multiplication, remember that i2=1\begin{align*}i^2=-1\end{align*}. You should always express answers in such a way that the i\begin{align*}i\end{align*} does not have an exponent.

• 2i+13i+4\begin{align*}2i+1-3i+4\end{align*} simplifies to i+5\begin{align*}-i+5\end{align*} or 5i\begin{align*}5-i\end{align*}
• 3i2i\begin{align*}3i\cdot 2i\end{align*} can be multiplied to 6i2=6(1)=6\begin{align*}6i^2=6(-1)=-6\end{align*}

Numbers that are a combination of imaginary and real numbers, such as 5i\begin{align*}5-i\end{align*}, are called complex numbers. You will study complex numbers in much more depth in future courses.

Example A

Express 49\begin{align*}\sqrt{-49}\end{align*} as a simplified imaginary number.

Solution: 49=491=491=7i\begin{align*}\sqrt{-49}=\sqrt{49\cdot -1}=\sqrt{49}\sqrt{-1}=7i\end{align*}

Example B

Express 40\begin{align*}\sqrt{-40}\end{align*} as a simplified imaginary number.

Solution: 40=4101=4101=210i\begin{align*}\sqrt{-40}=\sqrt{4\cdot 10\cdot -1}=\sqrt{4}\sqrt{10}\sqrt{-1}=2\sqrt{10}i\end{align*}

Example C

Simplify the following expression: (4+3i)+(65i)\begin{align*}(4+3i)+(6-5i)\end{align*}

Solution: Simplify by combining the real numbers with the real numbers and the imaginary numbers with the imaginary numbers:

(4+3i)+(65i)=102i\begin{align*}(4+3i)+(6-5i)=10-2i\end{align*}

Concept Problem Revisited

The parabola y=x22x+3\begin{align*}y=x^2-2x+3\end{align*} has no x-intercepts, as shown below.

This means that the solutions to the equation x22x+3=0\begin{align*}x^2-2x+3=0\end{align*} are not real numbers. The solutions are complex numbers. You can still find the solutions using the quadratic formula, but your result will be 2 complex number solutions.

Vocabulary

Complex Number
A complex number is a number in the form a+bi\begin{align*}a + bi\end{align*} where a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are real numbers and i2=1\begin{align*}i^2=-1\end{align*}.
Imaginary Number
An imaginary number is a number such that its square is a negative number.
16\begin{align*}\sqrt{-16}\end{align*} is an imaginary number because its square is –16.
16=i16=4i\begin{align*}\sqrt{-16}=i\sqrt{16} = 4i\end{align*}

Guided Practice

Simplify each of the following:

1. (53i)(2+4i)\begin{align*}(5-3i)-(2+4i)\end{align*}

2. 3i(4i25i+3)\begin{align*}3i(4i^2-5i+3)\end{align*}

3. (7+2i)(3i)\begin{align*}(7+2i)(3-i)\end{align*}

4. 12\begin{align*}\sqrt{-12}\end{align*}

1. (53i)(2+4i)=53i24i=37i\begin{align*}(5-3i)-(2+4i)=5-3i-2-4i=3-7i\end{align*}

2.

3i(4i25i+3)=3i(415i+3)=3i(45i+3)=3i(15i)=3i15i2=3i15(1)=3i+15

3.

(7+2i)(3i)=217i+6i2i2=21i2i2=21i2(1)=21i+2=23i

4. 12=431=2i3\begin{align*}\sqrt{-12}=\sqrt{4\cdot 3\cdot -1}=2i\sqrt{3}\end{align*}

Practice

Express each as a simplified imaginary number.

1. 300\begin{align*}\sqrt{-300}\end{align*}
2. 32\begin{align*}\sqrt{-32}\end{align*}
3. 418\begin{align*}4 \sqrt{-18}\end{align*}
4. 75\begin{align*}\sqrt{-75}\end{align*}
5. 98\begin{align*}\sqrt{-98}\end{align*}

Simplify each of the following:

1. (8+5i)(12+8i)\begin{align*}(8+5i) - (12+8i)\end{align*}
2. (7+3i)(45i)\begin{align*}(7+3i)(4-5i)\end{align*}
3. (2+i)(4i)\begin{align*}(2+i)(4-i)\end{align*}
4. 3(5i4)2(6i7)\begin{align*}3(5i-4) - 2(6i-7)\end{align*}
5. 5i(3i2i2+4)\begin{align*}5i(3i-2i^2+4)\end{align*}
6. (3+4i)+(11+6i)\begin{align*}(3+4i) + (11+6i)\end{align*}
7. (5+2i)(15i)\begin{align*}(5+2i)(1-5i)\end{align*}
8. (1+i)(1i)\begin{align*}(1+i)(1-i)\end{align*}
9. 2(6i3)4(2i+6)\begin{align*}2(6i-3) - 4(2i+6)\end{align*}
10. i3\begin{align*}i^3\end{align*}
11. i4\begin{align*}i^4\end{align*}
12. i6\begin{align*}i^6\end{align*}
13. 5i(3i2i2+4)\begin{align*}5i(3i-2i^2+4)\end{align*}

Vocabulary Language: English

Imaginary Number

Imaginary Number

An imaginary number is a number that can be written as the product of a real number and $i$.
Imaginary Numbers

Imaginary Numbers

An imaginary number is a number that can be written as the product of a real number and $i$.