Practice 2 Solutions

Complete the statement with always, sometimes, or never.

When the denominator contains an imaginary number the denominator always needs to be rationalized.
Real and imaginary numbers are always part of the complex number system.

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

$4 \sqrt{- 48} + \sqrt{- 3} (4 i - 7)$

$4 i \sqrt{4^{2} \cdot 3} + i \sqrt{3} (4 i - 7) 16 i \sqrt{3} + 4 i^{2} \sqrt{3} - 7 i \sqrt{3} 9 i \sqrt{3} + 4 (- 1) \sqrt{3}$

$- 4 \sqrt{3} + 9 i \sqrt{3} Complex$

$(6 + i \sqrt{10}) (12 - i \sqrt{10}) - \sqrt{- 360}$

$72 - 6 i \sqrt{10} + 12 i \sqrt{10} - i^{2} \sqrt{10^{2}} - 6 i \sqrt{10} 72 - 10 i^{2} 72 - 10 (- 1)$

$82 Real, complex$

Simplify.

$\frac{9}{7 - 11 i}$

$\frac{9}{7 - 11 i} (\frac{7 + 11 i}{7 + 11 i}) \frac{63 + 99 i}{49 - 121 i^{2}} \frac{63 + 99 i}{49 - 121 (- 1)} \frac{63 + 99 i}{49 + 121}$

$\frac{63 + 99 i}{170}$

$\frac{\sqrt{- 4}}{\sqrt{2} + \sqrt{- 3}}$

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

$\frac{2 \sqrt{3} + 2 i \sqrt{2}}{5}$

$\frac{13 - 2 i}{5 i}$

$\frac{13 - 2 i}{5 i} (\frac{i}{i}) \frac{13 i - 2 i^{2}}{5 i^{2}} \frac{13 i - 2 (- 1)}{5 (- 1)}$

$- \frac{2 + 13 i}{5}$

$8 i (x - 3 i) + 3 i (3 x - \sqrt{- 16})$

$8 i x - 24 i^{2} + 3 i (3 x - 4 i) 8 i x - 24 i^{2} + 9 i x - 12 i^{2} 17 i x - 36 i^{2} 17 i x - 36 (- 1)$

$17 i x + 36$

${(9 x + \sqrt{- 2})}^{2}$

$(9 x + i \sqrt{2}) (9 x + i \sqrt{2}) 81 x^{2} + 9 i x \sqrt{2} + 9 i x \sqrt{2} + i^{2} \sqrt{2^{2}} 81 x^{2} + 18 i x \sqrt{2} + 2 (- 1)$

$81 x^{2} + 18 i x \sqrt{2} - 2$

$(4 + \sqrt{- 49}) (2 - \sqrt{- 25})$

$(4 + 7 i) (2 - 5 i) 8 - 20 i + 14 i - 35 i^{2} 8 - 6 i - 35 (- 1) 8 - 6 i + 35$

$43 - 6 i$

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

$\frac{2 - 5 i}{1 + 3 i} (\frac{1 - 3 i}{1 - 3 i}) \frac{2 - 6 i - 5 i + 15 i^{2}}{1 - 9 i^{2}} \frac{2 - 11 i + 15 (- 1)}{1 - 9 (- 1)} \frac{2 - 11 i - 15}{1 + 9}$

$\frac{- 13 - 11 i}{10}$

$\frac{8 + \sqrt{- 12}}{6 i}$

$\frac{8 + 2 i \sqrt{3}}{6 i} \frac{4 + i \sqrt{3}}{3 i} (\frac{i}{i}) \frac{4 i + i^{2} \sqrt{3}}{3 i^{2}} \frac{4 i + (- 1) \sqrt{3}}{3 (- 1)}$

$- \frac{4 i - \sqrt{3}}{3}$

Determine if the following expressions form an identity.

$4 (\sqrt{- 200} - 3 i) + \sqrt{- 32} and 20 (\sqrt{- 18} - i) + 12 \sqrt{- 8} + \sqrt{- 64}$

$\begin{matrix} 4 (10 i \sqrt{2} - 3 i) + 4 i \sqrt{2} 40 i \sqrt{2} - 12 i + 4 i \sqrt{2} 44 i \sqrt{2} - 12 i & 20 (3 i \sqrt{2} - i) + 24 i \sqrt{2} + 8 i 60 i \sqrt{2} - 20 i + 24 i \sqrt{2} + 8 i 84 i \sqrt{2} - 12 i \end{matrix}$

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

$(7 - 2 i) (21 + 6 i) and (3 + i) (49 - 12 i)$

$\begin{matrix} 147 + 42 i - 42 i - 12 i^{2} 147 - 12 (- 1) 159 & 147 - 36 i + 49 i - 12 i^{2} 147 + 13 i - 12 (- 1) 159 + 13 i \end{matrix}$

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

Determine the value of Q that makes the identity true.

$\frac{7 i}{2 + i} = \frac{Q}{1 - 2 i}$

$7 i (1 - 2 i) = Q (2 + i) 7 i - 14 i^{2} = 2 Q + Q i 7 i - 14 (- 1) = 2 Q + Q i 14 + 7 i = 2 Q + Q i 14 = 2 Q Q = 7$

$Q = 7$

$(10 - 3 i) (2 + Q i) = 23 + 4 i$

$20 + 10 Q i - 6 i - 3 Q i^{2} = 23 + 4 i 20 + 10 Q i - 6 i - 3 Q (- 1) = 23 + 4 i 20 + 3 Q + 10 Q i - 6 i = 23 + 4 i \begin{array}{lcc} 20 + 3 Q = 23 & 10 Q i - 6 i = 4 i \\ 3 Q = 3 & 10 Q i = 10 i \\ Q = 1 & Q = 1 \end{array}$

$Q = 1$