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 - Divide 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.

Divide the complex numbers as indicated.
Express your quotient in standard form.
Write the real part.
Write the imaginary part.

(1.) $\dfrac{7}{5 - 12i}$


Conjugate of $5 - 12i = 5 + 12i$

$ \dfrac{7}{5 - 12i} \\[5ex] \rightarrow \dfrac{7}{5 - 12i} * \dfrac{5 + 12i}{5 + 12i} \\[5ex] \rightarrow \dfrac{7(5 + 12i)}{(5 - 12i)(5 + 12i)} \\[5ex] 7(5 + 12i) = 35 + 84i \\[3ex] (5 - 12i)(5 + 12i) = (5)^2 - (12i)^2...Difference\:\: of\:\: two\:\: squares \\[3ex] = 25 - (12^2 * i^2) \\[3ex] = 25 - (144 * -1) \\[3ex] = 25 - (-144) \\[3ex] = 25 + 144 = 169 \\[3ex] \rightarrow \dfrac{35 + 84i}{169} \\[5ex] \rightarrow \dfrac{35}{169} + \dfrac{84}{169}i \\[5ex] Real\:\: part = \dfrac{35}{169} \\[5ex] Imaginary\:\: part = \dfrac{84}{169} $
(2.) $\dfrac{5}{2 + 5i}$


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} $
(3.) $\dfrac{4 + i}{- 3 - 2i}$


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} $
(4.) $\dfrac{33 + 10i}{4 - 5i}$


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 $
(5.)


$ \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.)


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 $