Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

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Solved Examples - Multiply Complex Numbers

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 any wrong answer.

Multiply the complex numbers as indicated.
Express your answer in standard form as applicable.
Write the real part as applicable.
Write the imaginary part as applicable.

(1.)


Sum is the answer you get after addition.

$ (3 - 4i) + (5 + 3i) \\[3ex] 3 - 4i + 5 + 3i \\[3ex] 3 + 5 - 4i + 3i \\[3ex] 8 - i \\[3ex] Real\:\: part = 8 \\[3ex] Imaginary\:\: part = -1 $
(2.)


$ (-3 + 7i) - (-7i - 12) \\[3ex] = -3 + 7i + 7i + 12 \\[3ex] = -3 + 12 + 7i + 7i \\[3ex] = 9 + 14i \\[3ex] Real\:\: part = 9 \\[3ex] Imaginary\:\: part = 14 $
(3.) 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 $
(4.)


$ -\sqrt {-53} \\[3ex] = -1 * \sqrt{-1} * \sqrt{53} \\[3ex] = - 1* i * \sqrt{53} \\[3ex] = -i\sqrt{53} $
(5.) $\left(\dfrac{1}{6} + \dfrac{i\sqrt{13}}{6}\right)^2$


$ \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} $
(6.)


$ \sqrt{\dfrac{-49}{75}} \\[5ex] = \sqrt {\dfrac{-1 * 49}{75}} \\[5ex] = \sqrt{-1} * \sqrt{\dfrac{49}{75}} \\[5ex] = i * \dfrac{\sqrt{49}}{\sqrt{75}} \\[5ex] = i * \dfrac{7}{\sqrt{25 * 3}} \\[5ex] = i * \dfrac{7}{\sqrt{25} * \sqrt{3}} \\[5ex] = i * \dfrac{7}{5 * \sqrt{3}} \\[5ex] = \dfrac{7i}{5\sqrt{3}} \\[5ex] = \dfrac{7i}{5\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] = \dfrac{7i\sqrt{3}}{5\sqrt{3}*\sqrt{3}} \\[5ex] = \dfrac{7i\sqrt{3}}{5(3)} \\[5ex] = \dfrac{7i\sqrt{3}}{15} \\[5ex] $
(7.) $[x - (3 + 7i)][x - (3 - 7i)]$


We can do this question in two ways.

Choose whatever way you prefer.

$ \underline{First\:\: Method:}\:\: Normal\:\: Expansion \\[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\:\: Method:}\:\: 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 $
(8.)


$ -2\left|3 - \dfrac{p}{3}\right| + 1 = -5 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -5 - 1 \\[5ex] -2\left|3 - \dfrac{p}{3}\right| = -6 \\[5ex] \left|3 - \dfrac{p}{3}\right| = \dfrac{-6}{-2} \\[5ex] \left|3 - \dfrac{p}{3}\right| = 3 \\[5ex] This\:\: means\:\: that \\[5ex] 3 - \dfrac{p}{3} = 3 \:\:OR\:\: -(3 - \dfrac{p}{3} = 3 \\[5ex] 3 - \dfrac{p}{3} = 3 \\[5ex] 3 - 3 = \dfrac{p}{3} \\[5ex] 0 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 0 \\[5ex] p = 0(3) \\[3ex] p = 0 \\[3ex] OR \\[3ex] -(3 - \dfrac{p}{3} = 3 \\[5ex] Divide\:\: both\:\: sides\:\: by\:\: -1 \\[3ex] 3 - \dfrac{p}{3} = -3 \\[5ex] 3 + 3 = \dfrac{p}{3} \\[5ex] 6 = \dfrac{p}{3} \\[5ex] \dfrac{p}{3} = 6 \\[5ex] p = 6(3) \\[3ex] p = 18 \\[3ex] $ Check
Check for both values.
$ \underline{LHS} \\[3ex] -2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 0 \\[3ex] -2\left|3 - \dfrac{0}{3}\right| + 1 \\[5ex] -2|3 - 0| + 1 \\[5ex] -2|3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex] $ $p = 0$ is a solution




$ -2\left|3 - \dfrac{p}{3}\right| + 1 \\[5ex] p = 18 \\[3ex] -2\left|3 - \dfrac{18}{3}\right| + 1 \\[5ex] -2|3 - 6| + 1 \\[5ex] -2|-3| + 1 \\[3ex] -2(3) + 1 \\[3ex] -6 + 1 \\[3ex] -5 \\[3ex] $ $p = 18$ is a solution
$ \underline{RHS} \\[3ex] -5 $