Practice 1 Solutions

Complete the statement with always, sometimes, or never.

When expressions containing the imaginary unit are simplified, the result is sometimes a pure imaginary answer.
When the radicand is negative, i should always be simplified out of the radicand first.

Simplify. Then classify by all sets to which it belongs: real, pure imaginary, complex.

$(3 + 2 i) (4 - i) + \sqrt{- 81}$

$12 - 3 i + 8 i - 2 i^{2} + i \sqrt{9^{2}} 12 + 5 i - 2 (- 1) + 9 i 12 + 14 i + 2$

$14 + 14 i Complex$

$i^{2} (3 - \sqrt{2}) + \sqrt{18}$

$- 1 (3 - \sqrt{2}) + \sqrt{3^{2} \cdot 2} - 3 + \sqrt{2} + 3 \sqrt{2}$

$- 3 + 4 \sqrt{2} Real, complex$

Note

Q: Why are real numbers also complex?

A: Because real numbers are a subset of the complex number system.

$(- 7 + 4 i) (- 7 - 4 i)$

$49 + 28 i - 28 i - 16 i^{2} 49 - 16 (- 1) 49 + 16$

$65 Real, complex$

${(2 + 5 i)}^{2} - {(2 - 5 i)}^{2}$

$4 + 10 i + 25 i^{2} - (4 - 10 i + 25 i^{2}) 4 + 10 i + 25 i^{2} - 4 + 10 i - 25 i^{2}$

$20 i Pure imaginary, complex$

Simplify.

$\frac{3}{\sqrt{- 11}}$

$\frac{3}{i \sqrt{11}} (\frac{i \sqrt{11}}{i \sqrt{11}}) \frac{3 i \sqrt{11}}{11 i^{2}} \frac{3 i \sqrt{11}}{11 (- 1)}$

$- \frac{3 i \sqrt{11}}{11}$

Note

Q: What is the first step in this problem (and for any problem with a negative radicand)?

A: Simplify out –1 from the radical in the denominator.

$\frac{2 + i}{2 - i}$

$\frac{2 + i}{2 - i} (\frac{2 + i}{2 + i}) \frac{4 + 4 i + i^{2}}{4 - i^{2}} \frac{4 + 4 i + (- 1)}{4 - (- 1)}$

$\frac{3 + 4 i}{5}$

Note

Q: How do you rationalize the denominator with the imaginary unit?

A: Multiply by the complex conjugate.

$\frac{x - 3}{2 x + \sqrt{- 9}}$

$\frac{x - 3}{2 x + 3 i} (\frac{2 x - 3 i}{2 x - 3 i}) \frac{2 x^{2} - 3 i x - 6 x + 9 i}{4 x^{2} - 9 i^{2}} \frac{2 x^{2} - 3 i x - 6 x + 9 i}{4 x^{2} - 9 (- 1)}$

$\frac{2 x^{2} - 3 i x - 6 x + 9 i}{4 x^{2} + 9}$

$(x - \sqrt{- 4}) (3 x - 7 i)$

$(x - 2 i) (3 x - 7 i) 3 x^{2} - 7 i x - 6 i x + 14 i^{2} 3 x^{2} - 13 i x + 14 (- 1)$

$3 x^{2} - 13 i x - 14$

Note

Q: What property is used to multiply binomials?

A: The distributive property

$(5 y - 6 i) (5 y + 6 i)$

$25 y^{2} + 30 i y - 30 i y - 36 i^{2} 25 y^{2} - 36 (- 1)$

$25 y^{2} + 36$

$\frac{5 - \sqrt{- 3}}{1 + \sqrt{- 6}}$

$\frac{5 - i \sqrt{3}}{1 + i \sqrt{6}} (\frac{1 - i \sqrt{6}}{1 - i \sqrt{6}}) \frac{5 - 5 i \sqrt{6} - i \sqrt{3} + i^{2} \sqrt{18}}{1 - 6 i^{2}} \frac{5 - 5 i \sqrt{6} - i \sqrt{3} + (- 1) 3 \sqrt{2}}{1 - 6 (- 1)} \frac{5 - 5 i \sqrt{6} - i \sqrt{3} - 3 \sqrt{2}}{1 + 6}$

$\frac{5 - 5 i \sqrt{6} - i \sqrt{3} - 3 \sqrt{2}}{7}$

$\frac{7 i}{1 - 3 i}$

$\frac{7 i}{1 - 3 i} (\frac{1 + 3 i}{1 + 3 i}) \frac{7 i + 21 i^{2}}{1 - 9 i^{2}} \frac{7 i + 21 (- 1)}{1 - 9 (- 1)} \frac{- 21 + 7 i}{1 + 9}$

$\frac{- 21 + 7 i}{10}$

$(3 i - \sqrt{2}) (9 i + 3 \sqrt{2})$

$27 i^{2} + 9 i \sqrt{2} - 9 i \sqrt{2} - 3 \sqrt{2^{2}} 27 (- 1) - 6 - 27 - 6$

–33

Note

Q: If an expression starts with a complex number, will the simplified answer always be complex? If so, why?

A: Yes, because all numbers are complex.

Determine if the following expressions form an identity.

Note

Q: Why are the identities in this lesson not polynomial identities?

A: Because polynomial identities cannot have imaginary numbers as coefficients.

$5 \sqrt{- 12} + \sqrt{- 27} and 4 \sqrt{- 75} - 7 \sqrt{- 3}$

$\begin{matrix} 5 i \sqrt{2^{2} \cdot 3} + i \sqrt{3^{3}} 10 i \sqrt{3} + 3 i \sqrt{3} 13 i \sqrt{3} & 4 i \sqrt{5^{2} \cdot 3} - 7 i \sqrt{3} 20 i \sqrt{3} - 7 i \sqrt{3} 13 i \sqrt{3} \end{matrix}$

This is an identity because the expressions equal one another.

$10 i (5 + \sqrt{- 40}) and - 20 \sqrt{2} + 10 \sqrt{25}$

$\begin{matrix} 50 i + 10 i \sqrt{- 40} 50 i + 10 i^{2} \sqrt{2^{3} \cdot 5} 50 i - 20 \sqrt{10} & - 20 \sqrt{2} + 10 \sqrt{5^{2}} - 20 \sqrt{2} + 50 - 20 \sqrt{2} + 50 \end{matrix}$

This is NOT an identity because the expressions do NOT equal one another.

Determine the value of Q that makes the identity true.

$(8 + Q i) (3 - i) = 30 + 10 i$

$\begin{array}{rcl} 24 - 8 i + 3 Q i - Q i^{2} & = & 30 + 10 i \\ 24 + Q - 8 i + 3 Q i & = & 30 + 10 i \\ \begin{matrix} 24 + Q = 30 & - 8 i + 3 Q i = 10 i \\ Q = 6 & 3 Q i = 18 i \\ Q = 6 \end{matrix} \end{array}$

$Q = 6$

Note

The value Q can be solved for in more than one way. It is not necessary to solve both ways but may be helpful to show that the answer is correct.

$\frac{Q}{5 - 2 i} = \frac{4 + i}{11}$

$11 Q = (5 - 2 i) (4 + i) 11 Q = 20 + 5 i - 8 i - 2 i^{2} 11 Q = 20 - 3 i + 2 11 Q = 22 - 3 i 11 Q = 22$

$Q = 2$