Arithmetic Operations on Complex Numbers

Solved Examples: Add, Subtract, Multiply, and Divide Complex Numbers

Samuel Dominic Chukwuemeka (SamDom For Peace)

Calculators: Calculators for Complex Numbers
Prerequisites:
(1.) Laws of Exponents
(2.) Polynomials

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

(1.) Perform the indicated operation.
(2.) Use at least two methods for each question as applicable.

Unless stated otherwise:
(3.) Express your answer in standard form.
(4.) Write the real part.
(5.) Write the imaginary part.

(1.) Perform the indicated operations.

$ (a.)\;\; (-2 - 4i) + (1 + 6i) \\[3ex] (b.)\;\; (2 - 3i) - (3 + 2i) \\[3ex] (c.)\;\; (-3 + 7i) - (-7i - 12) \\[3ex] (d.)\;\; (-3 + 2i)(-6 - 8i) \\[3ex] (e.)\;\; (7 - 6i)(-8 + 3i) \\[3ex] $

$ (a.) \\[3ex] \underline{First\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] (-2 - 4i) + (1 + 6i) \\[3ex] -2 - 4i + 1 + 6i \\[3ex] -2 + 1 - 4i + 6i \\[3ex] -1 + 2i \\[5ex] \underline{Second\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} -2 - 4i \\ +\hspace{5em} \\ 1 + 6i \\ \hline -1 + 2i \\ \hline \\[5ex] \end{array} \\[5ex] real\;\;part = -1 \\[3ex] imaginary\;\;part = 2 \\[5ex] (b.) \\[3ex] \underline{First\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] (2 - 3i) - (3 + 2i) \\[3ex] 2 - 3i - 3 - 2i \\[3ex] 2 - 3 - 3i - 2i \\[3ex] -1 - 5i \\[5ex] \underline{Second\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} 2 - 3i \\ -\hspace{5em} \\ 3 + 2i \\ \hline -1 - 5i \\ \hline \\[5ex] \end{array} \\[5ex] real\;\;part = -1 \\[3ex] imaginary\;\;part = -5 \\[5ex] (c.) \\[3ex] \underline{First\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] (-3 + 7i) - (-7i - 12) \\[3ex] -3 + 7i + 7i + 12 \\[3ex] -3 + 12 + 7i + 7i \\[3ex] 9 + 14i \\[5ex] \underline{Second\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} -3 + 7i \\ -\hspace{5em} \\ -12 - 7i \\ \hline 9 + 14i \\ \hline \\[5ex] \end{array} \\[5ex] real\;\;part = 9 \\[3ex] imaginary\;\;part = 14 \\[5ex] (d.) \\[3ex] \underline{First\;\;Approach:\;\;Horizontal\;\;(FOIL)\;\;Method} \\[3ex] (-3 + 2i)(-6 - 8i) \\[3ex] F:\;\; -3(-6) = 18 \\[3ex] O:\;\; -3(-8i) = 24i \\[3ex] I:\;\; 2i(-6) = -12i \\[3ex] L:\;\; 2i(-8i) = -16i^2 = -16(-1) = 16 \\[3ex] \implies \\[3ex] 18 + 24i - 12i + 16 \\[3ex] 34 + 12i \\[5ex] \underline{Second\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} -3 + 2i \\ * \hspace{7em} \\ -6 - 8i \\ \hline 24i - 16i^2 \\ 18 - 12i \hspace{5em} \\ \hline 18 + 12i - 16i^2 \hspace{2em} \\ \hline \\[5ex] \end{array} \\[5ex] 18 + 12i - 16i^2 \\[3ex] 18 + 12i - 16(-1) \\[3ex] 18 + 12i + 16 \\[3ex] 34 + 12i \\[5ex] \underline{Third\;\;Approach:\;\;Box\;\;Method} \\[3ex] \begin{array} {c|c} &-3 & +2i \\ \hline -6 & 18 & -12i \\ \hline -8i & +24i & -16i^2 \\ \hline \end{array} \\[5ex] 18 - 12i + 24i - 16i^2 \\[3ex] 18 + 12i - 16(-1) \\[3ex] 18 + 12i + 16 \\[3ex] 34 + 12i \\[3ex] real\;\;part = 34 \\[3ex] imaginary\;\;part = 12 \\[5ex] (e.) \\[3ex] \underline{First\;\;Approach:\;\;Horizontal\;\;(FOIL)\;\;Method} \\[3ex] (7 - 6i)(-8 + 3i) \\[3ex] F:\;\; 7(-8) = -56 \\[3ex] O:\;\; 7(3i) = 21i \\[3ex] I:\;\; -6i(-8) = 48i \\[3ex] L:\;\; (-6i)(3i) = -18i^2 = -18(-1) = 18 \\[3ex] \implies \\[3ex] -56 + 21i + 48i + 18 \\[3ex] -38 + 69i \\[5ex] \underline{Second\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} 7 - 6i \\ * \hspace{7em} \\ -8 + 3i \\ \hline 21i - 18i^2 \\ -56 + 48i \hspace{5em} \\ \hline -56 + 69i - 18i^2 \hspace{2em} \\ \hline \\[5ex] \end{array} \\[5ex] -56 + 69i - 18i^2 \\[3ex] -56 + 69i - 18(-1) \\[3ex] -56 + 69i + 18 \\[3ex] -38 + 69i \\[5ex] \underline{Third\;\;Approach:\;\;Box\;\;Method} \\[3ex] \begin{array} {c|c} &7 & -6i \\ \hline -8 & -56 & +48i \\ \hline +3i & +21i & -18i^2 \\ \hline \end{array} \\[5ex] -56 + 48i + 21i - 18i^2 \\[3ex] -56 + 69i - 18(-1) \\[3ex] -56 + 69i + 18 \\[3ex] -38 + 69i \\[3ex] real\;\;part = -38 \\[3ex] imaginary\;\;part = 69 $

(2.) ACT What is the sum of the complex numbers 3 − 4i and 5 + 3i?

$ F.\;\; 7 \\[3ex] G.\;\; 27 \\[3ex] H.\;\; -1 + 8i \\[3ex] J.\;\; 8 - i \\[3ex] K.\;\; 15 - 2i \\[3ex] $

$ \underline{First\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] (3 - 4i) + (5 + 3i) \\[3ex] 3 - 4i + 5 + 3i \\[3ex] 3 + 5 - 4i + 3i \\[3ex] 8 - i \\[5ex] \underline{Second\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} 3 - 4i \\ +\hspace{5em} \\ 5 + 3i \\ \hline 8 - i \\ \hline \end{array} $

(3.) Perform the indicated operations.

$ (a.)\;\; \dfrac{7}{5 - 12i} \\[5ex] (b.)\;\; \dfrac{5}{2 + 5i} \\[5ex] (c.)\;\; \dfrac{4 + i}{-3 - 2i} \\[5ex] (d.)\;\; \dfrac{33 + 10i}{4 - 5i} \\[5ex] $

$ (a.) \\[3ex] \dfrac{7}{5 - 12i} \\[5ex] Conjugate = 5 + 12i \\[3ex] \implies \\[3ex] \dfrac{7}{5 - 12i} * \dfrac{5 + 12i}{5 + 12i} \\[5ex] \dfrac{7(5 + 12i)}{(5 - 12i)(5 + 12i)} \\[5ex] \dfrac{35 + 84i}{5^2 - (12i)^2} \\[5ex] \dfrac{35 + 84i}{25 - 144i^2} \\[5ex] \dfrac{35 + 84i}{25 - 144(-1)} \\[5ex] \dfrac{35 + 84i}{25 + 144} \\[5ex] \dfrac{35 + 84i}{169} \\[5ex] \dfrac{35}{169} + \dfrac{84i}{169} \\[5ex] real\;\;part = \dfrac{35}{169} \\[5ex] imaginary\;\;part = \dfrac{84}{169} \\[5ex] $ (b.)
Conjugate of $2 + 5i = 2 - 5i$

$ \dfrac{5}{2 + 5i} \\[5ex] \rightarrow \dfrac{5}{2 + 5i} * \dfrac{2 + 5i}{2 - 5i} \\[5ex] \rightarrow \dfrac{5(2 - 5i)}{(2 + 5i)(2 - 5i)} \\[5ex] 5(2 - 5i) = 10 - 25i \\[3ex] (2 + 5i)(2 - 5i) = (2)^2 - (5i)^2...Difference\:\: of\:\: two\:\: squares \\[3ex] = 4 - (5^2 * i^2) \\[3ex] = 4 - (25 * -1) \\[3ex] = 4 - (-25) \\[3ex] = 4 + 25 = 29 \\[3ex] \rightarrow \dfrac{10 - 25i}{29} \\[5ex] \rightarrow \dfrac{10}{29} - \dfrac{25}{29}i \\[5ex] Real\:\: part = \dfrac{10}{29} \\[5ex] Imaginary\:\: part = -\dfrac{25}{29} \\[5ex] $ (c.)
Conjugate of $-3 - 2i = -3 + 2i$

$ \dfrac{4 + i}{-3 - 2i} \\[5ex] = \dfrac{4 + i}{-3 - 2i} * \dfrac{-3 + 2i}{-3 + 2i} \\[5ex] = \dfrac{(4 + i)(-3 + 2i)}{(-3 - 2i)(-3 + 2i)} \\[5ex] \underline{Numerator} \\[3ex] (4 + i)(-3 + 2i) \\[3ex] = -12 + 8i - 3i + 2i^2 \\[3ex] = -12 + 5i + 2(-1) \\[3ex] = -12 + 5i - 2 \\[3ex] = -14 + 5i \\[5ex] \underline{Denominator} \\[3ex] (-3 - 2i)(-3 + 2i) = (-3)^2 - (2i)^2...Difference\:\: of\:\: two\:\: squares \\[3ex] = 9 - (2^2 * i^2) \\[3ex] = 9 - (4 * -1) \\[3ex] = 9 - (-4) \\[3ex] = 9 + 4 \\[3ex] = 13 \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{-14 + 5i}{13} \\[5ex] = -\dfrac{14}{13} + \dfrac{5}{13}i \\[5ex] Real\:\: part = -\dfrac{14}{13} \\[3ex] Imaginary\:\: part = \dfrac{5}{13} \\[5ex] $ (d.)
Conjugate of $4 - 5i = 4 + 5i$

$ \dfrac{33 + 10i}{4 - 5i} \\[5ex] = \dfrac{33 + 10i}{4 - 5i} * \dfrac{4 + 5i}{4 + 5i} \\[5ex] = \dfrac{(33 + 10i)(4 + 5i)}{(4 - 5i)(4 + 5i)} \\[5ex] \underline{Numerator} \\[3ex] (33 + 10i)(4 + 5i) \\[3ex] = 132 + 165i + 40i + 50i^2 \\[3ex] = 132 + 205i + 50(-1) \\[3ex] = 132 + 205i - 50 \\[3ex] = 82 + 205i \\[5ex] \underline{Denominator} \\[3ex] (4 - 5i)(4 + 5i) = 4^2 - (5i)^2...Difference\:\: of\:\: two\:\: squares \\[3ex] = 16 - (5^2 * i^2) \\[3ex] = 16 - (25 * -1) \\[3ex] = 16 - (-25) \\[3ex] = 16 + 25 \\[3ex] = 41 \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{82 + 205i}{41} \\[5ex] = \dfrac{82}{41} + \dfrac{205i}{41} \\[5ex] = 2 + 5i \\[3ex] Real\:\: part = 2 \\[3ex] Imaginary\:\: part = 5 $

(4.) ACT What is the product of the complex numbers (-2i + 5) and (2i + 5)?

$ A.\;\; 3 \\[3ex] B.\;\; 21 \\[3ex] C.\;\; 29 \\[3ex] D.\;\; 20i - 21 \\[3ex] E.\;\; 20i + 21 \\[3ex] $

$ \underline{First\;\;Approach:\;\;Horizontal\;\;(FOIL)\;\;Method} \\[3ex] (-2i + 5)(2i + 5) \\[3ex] F:\;\; -2i(2i) = -4i^2 = -4(-1) = 4 \\[3ex] O:\;\; -2i(5) = -10i \\[3ex] I:\;\; 5(2i) = 10i \\[3ex] L:\;\; 5(5) = 25 \\[3ex] \implies \\[3ex] 4 - 10i + 10i + 25 \\[3ex] 29 \\[5ex] \underline{Second\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} -2i + 5 \\ * \hspace{7em} \\ 2i + 5 \\ \hline -10i + 25 \\ -4i^2 + 10i \hspace{4em} \\ \hline -4i^2 \hspace{4em} + 25 \\ \hline \\[5ex] \end{array} \\[5ex] -4i^2 + 25 \\[3ex] -4(-1) + 25 \\[3ex] 4 + 25 \\[3ex] 29 \\[5ex] \underline{Third\;\;Approach:\;\;Box\;\;Method} \\[3ex] \begin{array} {c|c} &-2i & +5 \\ \hline 2i & -4i^2 & +10i \\ \hline +5 & -10i & +25 \\ \hline \end{array} \\[5ex] -4i^2 + 10i - 10i + 25 \\[3ex] -4(-1) + 25 \\[3ex] 4 + 25 \\[3ex] 29 $

(5.) ACT In the complex numbers, where i² = −1, $\dfrac{2 - i}{-3 + i}$ = ?

$ F.\;\; -\dfrac{2}{3} - i \\[5ex] G.\;\; -\dfrac{5}{8} + \dfrac{1}{8}i \\[5ex] H.\;\; -\dfrac{7}{8} + \dfrac{1}{8}i \\[5ex] J.\;\; -\dfrac{5}{10} + \dfrac{1}{10}i \\[5ex] K.\;\; -\dfrac{7}{10} + \dfrac{1}{10}i \\[5ex] $

$ \dfrac{2 - i}{-3 + i} \\[5ex] Denominator = -3 + i \\[3ex] Conjugate\;\;of\;\;Denominator = -3 - i \\[3ex] = \dfrac{2 - i}{-3 + i} * \dfrac{-3 - i}{-3 - i} \\[5ex] = \dfrac{(2 - i)(-3 - i)}{(-3 + i)(-3 - i)} \\[5ex] = \dfrac{-6 - 2i + 3i + i^2}{(-3)^2 - i^2} \\[5ex] = \dfrac{-6 + i + (-1)}{9 - (-1)} \\[5ex] = \dfrac{-6 + i - 1}{9 + 1} \\[5ex] = \dfrac{-7 + i}{10} \\[5ex] = -\dfrac{7}{10} + \dfrac{i}{10} \\[5ex] = -\dfrac{7}{10} + \dfrac{1}{10}i $

(6.) Simplify:

$ (a.)\;\; -3 + 6i - (-5 - 3i) - 8i \\[3ex] (b.)\;\; -3i \cdot 6i - 3(-7 + 6i) \\[3ex] (c.)\;\; -6i(8 - 6i)(-8 - 8i) \\[3ex] $

(a.)

$ -3 + 6i - (-5 - 3i) - 8i \\[3ex] -3 + 6i -1(-5 - 3i) - 8i \\[3ex] -3 + 6i + 5 + 3i - 8i \\[3ex] -3 + 5 + 6i + 3i - 8i \\[3ex] 2 + i \\[3ex] real\;\;part = 2 \\[3ex] imaginary\;\;part = 1 \\[5ex] $ (b.)

$ -3i \cdot 6i - 3(-7 + 6i) \\[3ex] -18i^2 + 21 - 18i \\[3ex] -18(-1) + 21 - 18i \\[3ex] 18 + 21 - 18i \\[3ex] 39 - 18i \\[3ex] real\;\;part = 39 \\[3ex] imaginary\;\;part = -18 \\[5ex] $ (c.)

$ -6i(8 - 6i)(-8 - 8i) \\[3ex] \underline{1st\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] First\;\;two: \\[3ex] -6i(8 - 6i) \\[3ex] -48i + 36i^2 \\[3ex] -48i + 36(-1) \\[3ex] -48i - 36 \\[3ex] -36 - 48i \\[3ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (-36 - 48i)(-8 - 8i) \\[3ex] -36(-8) = 288 \\[3ex] -36(-8i) = 288i \\[3ex] -48i(-8) = 384i \\[3ex] -48i(-8i) = 384i^2 = 384(-1) = -384 \\[3ex] \implies \\[3ex] 288 + 288i + 384i - 384 \\[3ex] -96 + 672i \\[5ex] \underline{2nd\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] First\;\;two: \\[3ex] \begin{array}{c} 8 - 6i \\ * \hspace{4em} \\ \hspace{1em} -6i \\ \hline -48i + 36i^2 \\ \hline \\[5ex] \end{array} \\[5ex] \implies \\[3ex] -48i + 36i^2 \\[3ex] -48i + 36(-1) \\[3ex] -48i - 36 \\[3ex] -36 - 48i \\[3ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (-36 - 48i)(-8 - 8i) \\[3ex] \begin{array}{c} -36 - 48i \\ * \hspace{7em} \\ -8 - 8i \\ \hline 288i + 384i^2 \\ 288 + 384i \hspace{6em} \\ \hline 288 + 672i + 384i^2 \hspace{2.5em} \\ \hline \\[5ex] \end{array} \\[5ex] \implies \\[3ex] 288 + 672i + 384i^2 \\[3ex] 288 + 672i + 384(-1) \\[3ex] 288 + 672i - 384 \\[3ex] -96 + 672i \\[5ex] \underline{3rd\;\;Approach:\;\;Box\;\;Method} \\[3ex] First\;\;two: \\[3ex] -6i(8 - 6i) \\[3ex] \begin{array} {c|c|} &8 & -6i \hspace{3em} \\ \hline -6i & -48i & +36i^2 \\ \hline \end{array} \\[5ex] \implies \\[3ex] -48i + 36i^2 \\[3ex] -48i + 36(-1) \\[3ex] -48i - 36 \\[3ex] -36 - 48i \\[3ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (-36 - 48i)(-8 - 8i) \\[3ex] \begin{array} {c|c} &-8 & -8i \\ \hline -36 & 288 & +288i \\ \hline -48i & +384i & +384i^2 \\ \hline \end{array} \\[5ex] 288 + 288i + 384i + 384i^2 \\[3ex] 288 + 672i + 384(-1) \\[3ex] 288 + 672i - 384 \\[3ex] -96 + 672i \\[3ex] real\;\;part = -96 \\[3ex] imaginary\;\;part = 672 $

(7.) ACT In the complex plane, consider the segment whose endpoints are the points corresponding to $-6 + 3i$ and $2 - 7i$.
The midpoint of this segment corresponds to which of the following complex numbers?

$ F.\:\: -4 - 4i \\[3ex] G.\:\: -4 + 5i \\[3ex] H.\:\: -2 - 2i \\[3ex] J.\:\: -2 + 2i \\[3ex] K.\:\: 4 + 5i \\[3ex] $

$ Endpoints = -6 + 3i \;\;\;and\;\;\; 2 - 7i \\[3ex] Midpoint = \left(\dfrac{-6 + 2}{2}, \dfrac{3i + -7i}{2}\right) \\[5ex] = \left(\dfrac{-4}{2}, \dfrac{-4i}{2}\right) \\[5ex] = (-2, -2i) \\[3ex] = -2 - 2i $

(8.) ACT Given that i is the imaginary unit, which of the following complex numbers is equal to (7 + 6i)² ?

$ F.\;\; 13 \\[3ex] G.\;\; 85 \\[3ex] H.\;\; 13 + 84i \\[3ex] J.\;\; 14 + 12i \\[3ex] K.\;\; 85 + 84i \\[3ex] $

$ (7 + 6i)^2 \\[3ex] (7 + 6i)(7 + 6i) \\[3ex] 7(7) = 49 \\[3ex] 7(6i) = 42i \\[3ex] 6i(7) = 42i \\[3ex] 6i(6i) = 36i^2 = 36(-1) = -36 \\[3ex] \implies \\[3ex] 49 + 42i + 42i - 36 \\[3ex] 13 + 84i $

(9.) Simplify

$ (a.)\;\; (8 - 3i)^2 \\[3ex] (b.)\;\; 6(-7 + 6i)(-4 + 2i) \\[3ex] (c.)\;\; (-2 - 2i)(-4 - 3i)(7 + 8i) \\[3ex] $

(a.)

$ (8 - 3i)^2 = (8 - 3i)(8 - 3i) \\[3ex] \underline{1st\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] (8 - 3i)(8 - 3i) \\[3ex] 8(8) = 64 \\[3ex] 8(-3i) = -24i \\[3ex] -3i(8) = -24i \\[3ex] -3i(-3i) = 9i^2 = 9(-1) = -9 \\[3ex] \implies \\[3ex] 64 - 24i - 24i - 9 \\[3ex] 55 - 48i \\[5ex] \underline{2nd\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] (8 - 3i)(8 - 3i) \\[3ex] \begin{array}{c} 8 - 3i \\ * \hspace{7em} \\ 8 - 3i \\ \hline -24i + 9i^2 \hspace{1em} \\ 64 - 24i \hspace{4.5em} \\ \hline 64 - 48i + 9i^2 \hspace{2em} \\ \hline \\[5ex] \end{array} \\[5ex] \implies \\[3ex] 64 - 48i + 9i^2 \\[3ex] 64 - 48i + 9(-1) \\[3ex] 64 - 48i - 9 \\[3ex] 55 - 48i \\[5ex] \underline{3rd\;\;Approach:\;\;Box\;\;Method} \\[3ex] (8 - 3i)(8 - 3i) \\[3ex] \begin{array} {c|c} &8 & -3i \\ \hline 8 & 64 & -24i \\ \hline -3i & -24i & +9i^2 \\ \hline \end{array} \\[5ex] 64 - 24i - 24i + 9i^2 \\[3ex] 64 - 48i + 9(-1) \\[3ex] 64 - 48i - 9 \\[3ex] 55 - 48i \\[3ex] real\;\;part = 55 \\[3ex] imaginary\;\;part = -48 \\[3ex] $ (b.)

$ 6(-7 + 6i)(-4 + 2i) \\[3ex] 6(-7 + 6i)(-4 + 2i) \\[3ex] \underline{1st\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] First\;\;two: \\[3ex] 6(-7 + 6i) \\[3ex] -42 + 36i \\[3ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (-42 + 36i)(-4 + 2i) \\[3ex] -42(-4) = 168 \\[3ex] -42(2i) = -84i \\[3ex] 36i(-4) = -144i \\[3ex] 36i(2i) = 72i^2 = 72(-1) = -72 \\[3ex] \implies \\[3ex] 168 - 84i - 144i - 72 \\[3ex] 96 - 228i \\[5ex] \underline{2nd\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] First\;\;two: \\[3ex] \begin{array}{c} -7 + 6i \\ * \hspace{4em} \\ \hspace{2.5em} 6 \\ \hline -42 + 36i \\ \hline \\[5ex] \end{array} \\[5ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (-42 + 36i)(-4 + 2i) \\[3ex] \begin{array}{c} -42 + 36i \\ * \hspace{7em} \\ -4 + 2i \\ \hline -84i + 72i^2 \\ 168 - 144i \hspace{5.5em} \\ \hline 168 - 228i + 72i^2 \hspace{2.5em} \\ \hline \\[5ex] \end{array} \\[5ex] \implies \\[3ex] 168 - 228i + 72i^2 \\[3ex] 168 - 228i + 72(-1) \\[3ex] 168 - 228i - 72 \\[3ex] 96 - 228i \\[5ex] \underline{3rd\;\;Approach:\;\;Box\;\;Method} \\[3ex] First\;\;two: \\[3ex] 6(-7 + 6i) \\[3ex] \begin{array} {c|c|} &-7 & +6i \hspace{3em} \\ \hline 6 & -42 & +36i \\ \hline \end{array} \\[5ex] \implies \\[3ex] -42 + 36i \\[3ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (-42 + 36i)(-4 + 2i) \\[3ex] \begin{array} {c|c} &-4 & +2i \\ \hline -42 & 168 & -84i \\ \hline +36i & -144i & +72i^2 \\ \hline \end{array} \\[5ex] 168 - 84i - 144i + 72i^2 \\[3ex] 168 - 228i + 72i^2 \\[3ex] 168 - 228i + 72(-1) \\[3ex] 168 - 228i - 72 \\[3ex] 96 - 228i \\[3ex] real\;\;part = 96 \\[3ex] imaginary\;\;part = -228 \\[3ex] $ (c.)

$ (-2 - 2i)(-4 - 3i)(7 + 8i) \\[3ex] \underline{1st\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] First\;\;two: \\[3ex] (-2 - 2i)(-4 - 3i) \\[3ex] -2(-4) = 8 \\[3ex] -2(-3i) = 6i \\[3ex] -2i(-4) = 8i \\[3ex] -2i(-3i) = 6i^2 = 6(-1) = -6 \\[3ex] \implies \\[3ex] 8 + 6i + 8i - 6 \\[3ex] 2 + 14i \\[3ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (2 + 14i)(7 + 8i) \\[3ex] 2(7) = 14 \\[3ex] 2(8i) = 16i \\[3ex] 14i(7) = 98i \\[3ex] 14i(8i) = 112i^2 = 112(-1) = -112 \\[3ex] \implies \\[3ex] 14 + 16i + 98i - 112 \\[3ex] -98 + 114i \\[5ex] \underline{2nd\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] First\;\;two: \\[3ex] \begin{array}{c} -2 - 2i \\ * \hspace{4em} \\ -4 - 3i \\ \hline 6i + 6i^2 \\ 8 + 8i \hspace{4.45em} \\ \hline 8 + 14i + 6i^2 \hspace{1.5em} \\ \hline \\[5ex] \end{array} \\[5ex] \implies \\[3ex] 8 + 14i + 6i^2 \\[3ex] 8 + 14i + 6(-1) \\[3ex] 8 + 14i - 6 \\[3ex] 2 + 14i \\[3ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (2 + 14i)(7 + 8i) \\[3ex] \begin{array}{c} 2 + 14i \\ * \hspace{5em} \\ 7 + 8i \\ \hline 16i + 112i^2 \\ 14 + 98i \hspace{5.5em} \\ \hline 14 + 114i + 112i^2 \hspace{2.5em} \\ \hline \\[5ex] \end{array} \\[5ex] \implies \\[3ex] 14 + 114i + 112i^2 \\[3ex] 14 + 114i + 112(-1) \\[3ex] 14 + 114i - 112 \\[3ex] -98 + 114i \\[5ex] \underline{3rd\;\;Approach:\;\;Box\;\;Method} \\[3ex] First\;\;two: \\[3ex] (-2 - 2i)(-4 - 3i) \\[3ex] \begin{array} {c|c} &-2 & -2i \\ \hline -4 & 8 & +8i \\ \hline -3i & +6i & +6i^2 \\ \hline \end{array} \\[5ex] \implies \\[3ex] 8 + 8i + 6i + 6i^2 \\[3ex] 8 + 14i + 6(-1) \\[3ex] 8 + 14i - 6 \\[3ex] 2 + 14i \\[3ex] Then\;\;multiply\;\;with\;\;the\;\;third \\[3ex] (2 + 14i)(7 + 8i) \\[3ex] \begin{array} {c|c} &2 & +14i \\ \hline 7 & 14 & +98i \\ \hline +8i & +16i & +112i^2 \\ \hline \end{array} \\[5ex] 14 + 98i + 16i + 112i^2 \\[3ex] 14 + 98i + 16i + 112(-1) \\[3ex] 14 + 114i - 112 \\[3ex] -98 + 114i \\[3ex] real\;\;part = -98 \\[3ex] imaginary\;\;part = 114 $

(10.) ACT Given that b is rational and $i = \sqrt{-1}$, the product of the expression $(3 + bi)$ and which of the following expressions must be a rational number?

$ A.\;\; i \\[3ex] B.\;\; bi \\[3ex] C.\;\; 3 + bi \\[3ex] D.\;\; 3 - bi \\[3ex] $

The product of a complex number and it's conjugate is a rational number.

$ complex\;\;number = 3 + bi \\[3ex] conjugate\;\;of\;\;the\;\;complex\;\;number = 3 - bi \\[3ex] $ Hence, the expression is $3 - bi$

(11.) ACT For all real numbers $x$ and the imaginary number $i$, which of the following expressions is equivalent to $(x - 3i)^3$

First Method: Pascal's Triangle (recommended for ACT because it is a timed test)
Here is a video on Pascal's Triangle
$ \:\:\:\: 1 \\[3ex] \:\:\:\:1\:\:\:\:2\:\:\:\:1 \\[3ex] \:1\:\:\:\:\:3\:\:\:\:\:3\:\:\:\:\:1 \\[3ex] 1(x)^3 + 3(x)^2(-3i) + 3(x)(-3i)^2 + 1(-3i)^3 \\[3ex] 1(x^3) - 9x^2i + 3x(-3)^2(i^2) + (-3)^3 (i^3) \\[3ex] x^3 - 9x^2i + 3x(9)(-1) + (-27)(-i) \\[3ex] x^3 - 9x^2i - 27x + 27i \\[3ex] $ Second Method: Normal Expansion
We shall multiply two terms first.
Then, multiply the product with the third term.

$ (x - 3i)(x - 3i) \\[3ex] x(x) = x^2 \\[3ex] x(-3i) = -3xi \\[3ex] (-3i)(x) = -3xi \\[3ex] (-3i)(-3i) = 9i^2 = 9(-1) = -9 \\[3ex] = x^2 - 3xi - 3xi - 9 \\[3ex] = x^2 - 6xi - 9 \\[5ex] (x^2 - 6xi - 9)(x - 3i) \\[3ex] x^2(x) = x^3 \\[3ex] x^2(-3i) = -3x^2i \\[3ex] -6xi(x) = -6x^2i \\[3ex] -6xi(-3i) = 18xi^2 = 18x(-1) = -18x \\[3ex] -9(x) = -9x \\[3ex] -9(-3i) = 27i \\[3ex] = x^3 - 3x^2i - 6x^2i - 18x - 9x + 27i \\[3ex] = x^3 - 9x^2i - 27x + 27i \\[3ex] = x^3 - 27x - 9x^2i + 27i \\[3ex] = x^3 - 27x + i(-9x^2 + 27) \\[3ex] Real\:\: part = x^3 - 27x \\[3ex] Imaginary\:\: part = -9x^2 + 27 $

(13.) Simplify $\left(\dfrac{1}{6} + \dfrac{i\sqrt{13}}{6}\right)^2$

$ \left(\dfrac{1}{6} + \dfrac{i\sqrt{13}}{6}\right)^2 \\[5ex] = \left(\dfrac{1}{6} + \dfrac{i\sqrt{13}}{6}\right)\left(\dfrac{1}{6} + \dfrac{i\sqrt{13}}{6}\right) \\[5ex] \dfrac{1}{6} * \dfrac{1}{6} = \dfrac{1}{36} \\[5ex] \dfrac{1}{6} * \dfrac{i\sqrt{13}}{6} = \dfrac{i\sqrt{13}}{36} \\[5ex] \dfrac{i\sqrt{13}}{6} * \dfrac{1}{6} = \dfrac{i\sqrt{13}}{36} \\[5ex] \dfrac{i\sqrt{13}}{6} * \dfrac{i\sqrt{13}}{6} = \dfrac{i^2(\sqrt{13})^2}{36} = \dfrac{(-1)(13)}{36} = -\dfrac{13}{36} \\[5ex] = \dfrac{1}{36} + \dfrac{i\sqrt{13}}{36} + \dfrac{i\sqrt{13}}{36} -\dfrac{13}{36} \\[5ex] = \dfrac{1}{36} -\dfrac{13}{36} + \dfrac{2i\sqrt{13}}{36} \\[5ex] = -\dfrac{12}{36} + \dfrac{2i\sqrt{13}}{36} \\[5ex] = -\dfrac{1}{3} + \dfrac{i\sqrt{13}}{18} \\[5ex] Real\:\: part = -\dfrac{1}{3} \\[5ex] Imaginary\:\: part = \dfrac{\sqrt{13}}{18} $

(15.) Simplify $[x - (3 + 7i)][x - (3 - 7i)]$

We can do this question in at least two ways.
Choose whatever way you prefer.

$ \underline{First\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] [x - (3 + 7i)][x - (3 - 7i)] \\[3ex] = (x - 3 - 7i)(x - 3 + 7i) \\[3ex] x(x) = x^2 \\[3ex] x(-3) = -3x \\[3ex] x(7i) = 7xi \\[3ex] -3(x) = -3x \\[3ex] -3(-3) = 9 \\[3ex] -3(7i) = -21i \\[3ex] -7i(x) = -7xi \\[3ex] -7i(-3) = 21i \\[3ex] -7i(7i) = -49i^2 = -49(-1) = 49 \\[3ex] = x^2 - 3x + 7xi - 3x + 9 - 21i -7xi + 21i + 49 \\[3ex] = x^2 - 6x + 58 \\[5ex] \underline{Second\:\: Approach:\:\: Difference \:\:of\:\: Two\:\: Squares} \\[3ex] [x - (3 + 7i)][x - (3 - 7i)] \\[3ex] = (x - 3 - 7i)(x - 3 + 7i) \\[3ex] = [(x - 3) - 7i][(x - 3) + 7i] \\[3ex] = (x - 3)^2 - (7i)^2 \\[3ex] = [(x - 3)(x - 3)] - (7^2 * i^2) \\[3ex] = [x^2 - 3x - 3x + 9] - (49 * -1) \\[3ex] = x^2 - 6x + 9 - (-49) \\[3ex] = x^2 - 6x + 9 + 49 \\[3ex] = x^2 - 6x + 58 $

(18.)

$ i^{22} - 1 \\[3ex] = i^{20} * i^2 - 1 \\[3ex] = ((i^4)^5 * -1) - 1 \\[3ex] = (1^5 * -1) - 1 \\[3ex] = (1 * -1) - 1 \\[3ex] = -1 - 1 \\[3ex] = -2 $