Complex Numbers
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These are the solutions to Mathematics questions on Complex Numbers.
The TI-84 Plus CE shall be used for applicable questions.
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Forms of a Complex Number
$
(1.) \underline{\text{ Standard Form or Rectangular Form}} \\[3ex]
Z = x + yi \\[5ex]
(2.) \underline{\text{ Polar Form}} \\[3ex]
Z = r(\cos\theta + i\sin\theta) \\[5ex]
(3.) \underline{\text{ Exponential Form}} \\[3ex]
Z = re^{i\theta} \\[5ex]
(4.)\;\; i = \sqrt{-1} \\[3ex]
(5.)\;\; r = |Z| \\[3ex]
(6.)\;\; r = \sqrt{x^2 + y^2} \\[4ex]
(7.)\;\; \theta = \tan^{-1}\left(\dfrac{y}{x}\right) \\[6ex]
(8.) \underline{\text{ Euler's Formula}} \\[3ex]
e^{i\theta} = \cos\theta + i\sin\theta \\[5ex]
(9.) \underline{\text{ General Formula for n-th roots}} \\[3ex]
Z_k = \sqrt[n]{r} * e^{\dfrac{i(\theta + 2\pi k)}{n}} \\[7ex]
(10.) \underline{\text{ Abraham De Moivre's Theorem}} \\[3ex]
\text{Given the Polar Form}: Z = r(\cos\theta + i\sin\theta) \\[3ex]
\text{The n-th roots are}: Z_k = \sqrt[n]{r}\left[\cos\left(\dfrac{\theta + 2k\pi}{n}\right) +
i\sin\left(\dfrac{\theta + 2k\pi}{n}\right)\right] \\[7ex]
$
where
$Z$ = complex number
$x$ = real part of the complex number
$y$ = imaginary part of the complex number
$i$ = imaginary unit
$yi$ = pure imaginary number
$r$ = modulus or absolute value of the complex number
$\theta$ = argument or phase of the complex number
$\cos\theta$ = real component of the complex number
$\sin\theta$ = imaginary component of the complex number
$e^{i\theta}$ = Euler's formula
$k$ is the index that represents the different possible values of a root of a complex number
$n$ is the degree of the root of the complex number
$k = 0, 1, 2, 3, ..., n - 1$
(1.) Using Abraham De Moivre's theorem, solve the equation $Z^2 - \dfrac{1 - \sqrt{3}i}{1 + \sqrt{-3}} = 0$.
Check your solution.
$ i = \sqrt{-1} \\[3ex] i^2 = -1 \\[5ex] \sqrt{-3} \\[3ex] = \sqrt{3 * -1} \\[3ex] = \sqrt{3} * \sqrt{-1} \\[3ex] = \sqrt{3} * i \\[3ex] = i\sqrt{3} \\[5ex] Z^2 - \dfrac{1 - \sqrt{3}i}{1 + \sqrt{-3}} = 0 \\[5ex] Z^2 - \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} = 0 \\[5ex] Z^2 = \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \\[5ex] \text{Convert the RHS to rectangular form} \\[3ex] \text{Conjugate of the denominator} = 1 - i\sqrt{3} \\[3ex] \implies \\[3ex] \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \\[5ex] = \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} * \dfrac{1 - i\sqrt{3}}{1 - i\sqrt{3}} \\[5ex] = \dfrac{1 - i\sqrt{3} - i\sqrt{3} + 3i^2}{1^2 - (i\sqrt{3})^2...\text{Difference of Two Squares}} \\[5ex] = \dfrac{1 - 2i\sqrt{3} + 3(-1)}{1 - 3i^2} \\[5ex] = \dfrac{1 - 2i\sqrt{3} - 3}{1 - 3(-1)} \\[5ex] = \dfrac{-2 - 2i\sqrt{3}}{4} \\[5ex] = \dfrac{-2}{4} - \dfrac{2i\sqrt{3}}{4} \\[5ex] = -\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \\[5ex] = x + yi \\[3ex] \text{real part} = x = -\dfrac{1}{2} \\[5ex] \text{This is negative: location is the left side of the complex plane} \\[3ex] \text{imaginary part} = y = -\dfrac{\sqrt{3}}{2} \\[5ex] \text{This is negative: location is below the real axis} \\[3ex] \implies \\[3ex] -\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \text{ is in the 3rd Quadrant} \\[5ex] \text{Convert to polar form} \\[3ex] \text{modulus} = r = \sqrt{x^2 + y^2} \\[3ex] r = \sqrt{\left(-\dfrac{1}{2}\right)^2 + \left(-\dfrac{\sqrt{3}}{2}\right)^2} \\[5ex] r = \sqrt{\dfrac{1}{4} + \dfrac{3}{4}} \\[5ex] r = \sqrt{\dfrac{4}{4}} \\[5ex] r = \sqrt{1} \\[3ex] r = 1 \\[5ex] \text{argument} = \theta = \tan^{-1}\left(\dfrac{y}{x}\right) \\[5ex] \theta = \tan^{-1}\left(-\dfrac{\sqrt{3}}{2} \div -\dfrac{1}{2}\right) \\[5ex] \theta = \tan^{-1}\left(-\dfrac{\sqrt{3}}{2} * -\dfrac{2}{1}\right) \\[5ex] \theta = \tan^{-1}(\sqrt{3}) \\[3ex] \tan^{-1}(\sqrt{3}) = 60^\circ ...\text{60 is in the 1st Quadrant...discard} \\[3ex] 60^\circ + 180^\circ ...\text{3rd Quadrant Identity} \\[3ex] 60^\circ + 180^\circ = 240^\circ ...\text{240 is in the 3rd Quadrant...keep} \\[3ex] \text{Convert DEG to RAD} \\[3ex] 240^\circ * \dfrac{\pi RAD}{180^\circ} = \dfrac{4\pi}{3}\;RAD \\[5ex] \implies \\[3ex] \theta = \dfrac{4\pi}{3}\;RAD \\[5ex] r(\cos\theta + i\sin\theta)...\text{Polar Form} \\[3ex] 1 * \left[\cos\left(\dfrac{4\pi}{3}\right) + i\sin\left(\dfrac{4\pi}{3}\right)\right] \\[5ex] \cos\left(\dfrac{4\pi}{3}\right) + i\sin\left(\dfrac{4\pi}{3}\right) \\[5ex] $ Let us now solve for Z
We have been working on the RHS
The RHS has been expressed in exponential form
Let us now equate the LHS to the RHS
$ \underline{\text{ Abraham De Moivre's Theorem}} \\[3ex] \text{Given the Polar Form}: Z = r(\cos\theta + i\sin\theta) \\[3ex] \text{The n-th roots are}: Z_k = \sqrt[n]{r}\left[\cos\left(\dfrac{\theta + 2k\pi}{n}\right) + i\sin\left(\dfrac{\theta + 2k\pi}{n}\right)\right] \\[6ex] Z^2 = \cos\left(\dfrac{4\pi}{3}\right) + i\sin\left(\dfrac{4\pi}{3}\right) \\[5ex] \text{From the LHS} \\[3ex] \text{root} = n = 2 \\[3ex] \text{index} = k = 0, 1 \\[3ex] \implies \\[3ex] \text{When } k = 0 \\[3ex] Z_0 = \sqrt[2]{1}\left[\cos\left(\dfrac{\dfrac{4\pi}{3} + 2(0)\pi}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + 2(0)\pi}{2}\right)\right] \\[7ex] Z_0 = 1 * \left[\cos\left(\dfrac{\dfrac{4\pi}{3} + 0}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + 0}{2}\right)\right] \\[7ex] Z_0 = \cos\left(\dfrac{\dfrac{4\pi}{3}}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3}}{2}\right) \\[7ex] Z_0 = \cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right) \\[7ex] \text{When } k = 1 \\[3ex] Z_1 = \sqrt[2]{1}\left[\cos\left(\dfrac{\dfrac{4\pi}{3} + 2(1)\pi}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + 2(1)\pi}{2}\right)\right] \\[7ex] Z_1 = 1 * \left[\cos\left(\dfrac{\dfrac{4\pi}{3} + 2\pi}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + 2\pi}{2}\right)\right] \\[7ex] Z_1 = 1 * \left[\cos\left(\dfrac{\dfrac{4\pi}{3} + \dfrac{6\pi}{3}}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + \dfrac{6\pi}{3}}{2}\right)\right] \\[7ex] Z_1 = \cos\left(\dfrac{\dfrac{10\pi}{3}}{2}\right) + i\sin\left(\dfrac{\dfrac{10\pi}{3}}{2}\right) \\[7ex] Z_1 = \cos\left(\dfrac{5\pi}{3}\right) + i\sin\left(\dfrac{5\pi}{3}\right) \\[5ex] .............................................................................................. \\[3ex] $ Alternatively: We can convert to exponential form and use the General Formula for n-th roots
$ re^{i\theta}...\text{Exponential Form} \\[3ex] 1 * e^{i * \dfrac{4\pi}{3}} \\[5ex] e^{\dfrac{4i\pi}{3}} \\[5ex] \underline{\text{ General Formula for n-th roots}} \\[3ex] Z_k = \sqrt[n]{r} * e^{\dfrac{i(\theta + 2\pi k)}{n}} \\[5ex] \text{When } k = 0 \\[3ex] Z_0 = \sqrt[2]{1} * e^{\dfrac{i\left(\dfrac{4\pi}{3} + 2\pi * 0\right)}{2}} \\[5ex] Z_0 = 1 * e^{\dfrac{i\left(\dfrac{4\pi}{3} + 0\right)}{2}} \\[5ex] Z_0 = e^{\dfrac{i\left(\dfrac{4\pi}{3}\right)}{2}} \\[5ex] Z_0 = e^{i\left(\dfrac{2\pi}{3}\right)} \\[6ex] \text{When } k = 1 \\[3ex] Z_1 = \sqrt[2]{1} * e^{\dfrac{i\left(\dfrac{4\pi}{3} + 2\pi * 1\right)}{2}} \\[5ex] Z_1 = 1 * e^{\dfrac{i\left(\dfrac{4\pi}{3} + 2\pi\right)}{2}} \\[5ex] Z_1 = 1 * e^{\dfrac{i\left(\dfrac{4\pi}{3} + \dfrac{6\pi}{3}\right)}{2}} \\[5ex] Z_1 = e^{\dfrac{i\left(\dfrac{10\pi}{3}\right)}{2}} \\[5ex] Z_1 = e^{i\left(\dfrac{5\pi}{3}\right)} \\[6ex] .............................................................................................. \\[3ex] \text{Unit Circle: Trigonometric Functions of Special Angles} \\[3ex] \cos \dfrac{2\pi}{3} = \cos 120^\circ = -\dfrac{1}{2} \\[5ex] \sin \dfrac{2\pi}{3} = \sin 120^\circ = \dfrac{\sqrt{3}}{2} \\[5ex] \cos \dfrac{5\pi}{3} = \cos 300^\circ = \dfrac{1}{2} \\[5ex] \sin \dfrac{5\pi}{3} = \sin 300^\circ = -\dfrac{\sqrt{3}}{2} \\[6ex] \text{Convert to polar form using Euler's formula; then back to rectangular form} \\[3ex] e^{i\theta} = \cos\theta + i\sin\theta \\[5ex] Z_0 = e^{i\left(\dfrac{2\pi}{3}\right)} \\[5ex] Z_0 = \cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3} \\[5ex] Z_0 = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} \\[6ex] Z_1 = e^{i\left(\dfrac{5\pi}{3}\right)} \\[5ex] Z_1 = \cos\dfrac{5\pi}{3} + i\sin\dfrac{5\pi}{3} \\[5ex] Z_1 = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} \\[5ex] $ Check
$ Z_0 = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} \\[5ex] Z_0 = \dfrac{-1 + i\sqrt{3}}{2} \\[5ex] $
$ Z_1 = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} \\[5ex] Z_1 = \dfrac{1 - i\sqrt{3}}{2} \\[5ex] $
Check your solution.
$ i = \sqrt{-1} \\[3ex] i^2 = -1 \\[5ex] \sqrt{-3} \\[3ex] = \sqrt{3 * -1} \\[3ex] = \sqrt{3} * \sqrt{-1} \\[3ex] = \sqrt{3} * i \\[3ex] = i\sqrt{3} \\[5ex] Z^2 - \dfrac{1 - \sqrt{3}i}{1 + \sqrt{-3}} = 0 \\[5ex] Z^2 - \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} = 0 \\[5ex] Z^2 = \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \\[5ex] \text{Convert the RHS to rectangular form} \\[3ex] \text{Conjugate of the denominator} = 1 - i\sqrt{3} \\[3ex] \implies \\[3ex] \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \\[5ex] = \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} * \dfrac{1 - i\sqrt{3}}{1 - i\sqrt{3}} \\[5ex] = \dfrac{1 - i\sqrt{3} - i\sqrt{3} + 3i^2}{1^2 - (i\sqrt{3})^2...\text{Difference of Two Squares}} \\[5ex] = \dfrac{1 - 2i\sqrt{3} + 3(-1)}{1 - 3i^2} \\[5ex] = \dfrac{1 - 2i\sqrt{3} - 3}{1 - 3(-1)} \\[5ex] = \dfrac{-2 - 2i\sqrt{3}}{4} \\[5ex] = \dfrac{-2}{4} - \dfrac{2i\sqrt{3}}{4} \\[5ex] = -\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \\[5ex] = x + yi \\[3ex] \text{real part} = x = -\dfrac{1}{2} \\[5ex] \text{This is negative: location is the left side of the complex plane} \\[3ex] \text{imaginary part} = y = -\dfrac{\sqrt{3}}{2} \\[5ex] \text{This is negative: location is below the real axis} \\[3ex] \implies \\[3ex] -\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \text{ is in the 3rd Quadrant} \\[5ex] \text{Convert to polar form} \\[3ex] \text{modulus} = r = \sqrt{x^2 + y^2} \\[3ex] r = \sqrt{\left(-\dfrac{1}{2}\right)^2 + \left(-\dfrac{\sqrt{3}}{2}\right)^2} \\[5ex] r = \sqrt{\dfrac{1}{4} + \dfrac{3}{4}} \\[5ex] r = \sqrt{\dfrac{4}{4}} \\[5ex] r = \sqrt{1} \\[3ex] r = 1 \\[5ex] \text{argument} = \theta = \tan^{-1}\left(\dfrac{y}{x}\right) \\[5ex] \theta = \tan^{-1}\left(-\dfrac{\sqrt{3}}{2} \div -\dfrac{1}{2}\right) \\[5ex] \theta = \tan^{-1}\left(-\dfrac{\sqrt{3}}{2} * -\dfrac{2}{1}\right) \\[5ex] \theta = \tan^{-1}(\sqrt{3}) \\[3ex] \tan^{-1}(\sqrt{3}) = 60^\circ ...\text{60 is in the 1st Quadrant...discard} \\[3ex] 60^\circ + 180^\circ ...\text{3rd Quadrant Identity} \\[3ex] 60^\circ + 180^\circ = 240^\circ ...\text{240 is in the 3rd Quadrant...keep} \\[3ex] \text{Convert DEG to RAD} \\[3ex] 240^\circ * \dfrac{\pi RAD}{180^\circ} = \dfrac{4\pi}{3}\;RAD \\[5ex] \implies \\[3ex] \theta = \dfrac{4\pi}{3}\;RAD \\[5ex] r(\cos\theta + i\sin\theta)...\text{Polar Form} \\[3ex] 1 * \left[\cos\left(\dfrac{4\pi}{3}\right) + i\sin\left(\dfrac{4\pi}{3}\right)\right] \\[5ex] \cos\left(\dfrac{4\pi}{3}\right) + i\sin\left(\dfrac{4\pi}{3}\right) \\[5ex] $ Let us now solve for Z
We have been working on the RHS
The RHS has been expressed in exponential form
Let us now equate the LHS to the RHS
$ \underline{\text{ Abraham De Moivre's Theorem}} \\[3ex] \text{Given the Polar Form}: Z = r(\cos\theta + i\sin\theta) \\[3ex] \text{The n-th roots are}: Z_k = \sqrt[n]{r}\left[\cos\left(\dfrac{\theta + 2k\pi}{n}\right) + i\sin\left(\dfrac{\theta + 2k\pi}{n}\right)\right] \\[6ex] Z^2 = \cos\left(\dfrac{4\pi}{3}\right) + i\sin\left(\dfrac{4\pi}{3}\right) \\[5ex] \text{From the LHS} \\[3ex] \text{root} = n = 2 \\[3ex] \text{index} = k = 0, 1 \\[3ex] \implies \\[3ex] \text{When } k = 0 \\[3ex] Z_0 = \sqrt[2]{1}\left[\cos\left(\dfrac{\dfrac{4\pi}{3} + 2(0)\pi}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + 2(0)\pi}{2}\right)\right] \\[7ex] Z_0 = 1 * \left[\cos\left(\dfrac{\dfrac{4\pi}{3} + 0}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + 0}{2}\right)\right] \\[7ex] Z_0 = \cos\left(\dfrac{\dfrac{4\pi}{3}}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3}}{2}\right) \\[7ex] Z_0 = \cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right) \\[7ex] \text{When } k = 1 \\[3ex] Z_1 = \sqrt[2]{1}\left[\cos\left(\dfrac{\dfrac{4\pi}{3} + 2(1)\pi}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + 2(1)\pi}{2}\right)\right] \\[7ex] Z_1 = 1 * \left[\cos\left(\dfrac{\dfrac{4\pi}{3} + 2\pi}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + 2\pi}{2}\right)\right] \\[7ex] Z_1 = 1 * \left[\cos\left(\dfrac{\dfrac{4\pi}{3} + \dfrac{6\pi}{3}}{2}\right) + i\sin\left(\dfrac{\dfrac{4\pi}{3} + \dfrac{6\pi}{3}}{2}\right)\right] \\[7ex] Z_1 = \cos\left(\dfrac{\dfrac{10\pi}{3}}{2}\right) + i\sin\left(\dfrac{\dfrac{10\pi}{3}}{2}\right) \\[7ex] Z_1 = \cos\left(\dfrac{5\pi}{3}\right) + i\sin\left(\dfrac{5\pi}{3}\right) \\[5ex] .............................................................................................. \\[3ex] $ Alternatively: We can convert to exponential form and use the General Formula for n-th roots
$ re^{i\theta}...\text{Exponential Form} \\[3ex] 1 * e^{i * \dfrac{4\pi}{3}} \\[5ex] e^{\dfrac{4i\pi}{3}} \\[5ex] \underline{\text{ General Formula for n-th roots}} \\[3ex] Z_k = \sqrt[n]{r} * e^{\dfrac{i(\theta + 2\pi k)}{n}} \\[5ex] \text{When } k = 0 \\[3ex] Z_0 = \sqrt[2]{1} * e^{\dfrac{i\left(\dfrac{4\pi}{3} + 2\pi * 0\right)}{2}} \\[5ex] Z_0 = 1 * e^{\dfrac{i\left(\dfrac{4\pi}{3} + 0\right)}{2}} \\[5ex] Z_0 = e^{\dfrac{i\left(\dfrac{4\pi}{3}\right)}{2}} \\[5ex] Z_0 = e^{i\left(\dfrac{2\pi}{3}\right)} \\[6ex] \text{When } k = 1 \\[3ex] Z_1 = \sqrt[2]{1} * e^{\dfrac{i\left(\dfrac{4\pi}{3} + 2\pi * 1\right)}{2}} \\[5ex] Z_1 = 1 * e^{\dfrac{i\left(\dfrac{4\pi}{3} + 2\pi\right)}{2}} \\[5ex] Z_1 = 1 * e^{\dfrac{i\left(\dfrac{4\pi}{3} + \dfrac{6\pi}{3}\right)}{2}} \\[5ex] Z_1 = e^{\dfrac{i\left(\dfrac{10\pi}{3}\right)}{2}} \\[5ex] Z_1 = e^{i\left(\dfrac{5\pi}{3}\right)} \\[6ex] .............................................................................................. \\[3ex] \text{Unit Circle: Trigonometric Functions of Special Angles} \\[3ex] \cos \dfrac{2\pi}{3} = \cos 120^\circ = -\dfrac{1}{2} \\[5ex] \sin \dfrac{2\pi}{3} = \sin 120^\circ = \dfrac{\sqrt{3}}{2} \\[5ex] \cos \dfrac{5\pi}{3} = \cos 300^\circ = \dfrac{1}{2} \\[5ex] \sin \dfrac{5\pi}{3} = \sin 300^\circ = -\dfrac{\sqrt{3}}{2} \\[6ex] \text{Convert to polar form using Euler's formula; then back to rectangular form} \\[3ex] e^{i\theta} = \cos\theta + i\sin\theta \\[5ex] Z_0 = e^{i\left(\dfrac{2\pi}{3}\right)} \\[5ex] Z_0 = \cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3} \\[5ex] Z_0 = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} \\[6ex] Z_1 = e^{i\left(\dfrac{5\pi}{3}\right)} \\[5ex] Z_1 = \cos\dfrac{5\pi}{3} + i\sin\dfrac{5\pi}{3} \\[5ex] Z_1 = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} \\[5ex] $ Check
| LHS | RHS |
|---|---|
|
$
Z_0 = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} \\[5ex]
Z_0^2 \\[3ex]
= \left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)^2 \\[5ex]
= \left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)\left(-\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}\right)
\\[5ex]
= \dfrac{1}{4} - \dfrac{i\sqrt{3}}{4} - \dfrac{i\sqrt{3}}{4} + \dfrac{3i^2}{4} \\[5ex]
= \dfrac{1}{4} - \dfrac{2i\sqrt{3}}{4} - \dfrac{3}{4} \\[5ex]
= \dfrac{-2}{4} - \dfrac{2i\sqrt{3}}{4} \\[5ex]
= -\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2}
$
$ Z_1 = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} \\[5ex] Z_1^2 \\[3ex] = \left(\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right)^2 \\[5ex] = \left(\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}\right) \\[5ex] = \dfrac{1}{4} - \dfrac{i\sqrt{3}}{4} - \dfrac{i\sqrt{3}}{4} + \dfrac{3i^2}{4} \\[5ex] = \dfrac{1}{4} - \dfrac{2i\sqrt{3}}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{-2}{4} - \dfrac{2i\sqrt{3}}{4} \\[5ex] = -\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} \\[3ex] $ We get the same result from squaring either root $ Z_0^2 = Z_1^2 = Z^2 = -\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} $ |
|
| $ Z^2 - \dfrac{1 - \sqrt{3}i}{1 + \sqrt{-3}} \\[5ex] = -\dfrac{1}{2} - \dfrac{i\sqrt{3}}{2} - \dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}} \\[5ex] = \dfrac{-(1 + i\sqrt{3}) - i\sqrt{3}(1 + i\sqrt{3}) - 2(1 - i\sqrt{3})}{2(1 + i\sqrt{3})} \\[5ex] = \dfrac{-1 - i\sqrt{3} - i\sqrt{3} - 3i^2 - 2 + 2i\sqrt{3}}{2(1 + i\sqrt{3})} \\[5ex] = \dfrac{-1 - 2i\sqrt{3} + 3 - 2 + 2i\sqrt{3}}{2(1 + i\sqrt{3})} \\[5ex] = \dfrac{0}{2(1 + i\sqrt{3})} \\[5ex] = 0 $ | 0 |
$ Z_0 = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} \\[5ex] Z_0 = \dfrac{-1 + i\sqrt{3}}{2} \\[5ex] $
$ Z_1 = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} \\[5ex] Z_1 = \dfrac{1 - i\sqrt{3}}{2} \\[5ex] $
(2.) Solve the equation $3(x + 1)^4 - 12 = 60$
Check your solution with a calculator.
$ 3(x + 1)^4 - 12 = 60 \\[3ex] 3(x + 1)^4 = 60 + 12 \\[3ex] 3(x + 1)^4 = 72 \\[3ex] (x + 1)^4 = \dfrac{72}{3} \\[5ex] (x + 1)^4 = 24 \\[3ex] x + 1 = \pm \sqrt[4]{24} \\[3ex] $ Student: Mr. C, I'm curious why this question is posted under Complex Numbers
My thinking is that the question should be posted under "Expressions and Equations"
Besides, how are you sure that the roots of the equation are complex roots?
What if they are rational roots?
Teacher: Good question.
We shal find out the kind of roots.
It is okay to post the topic under "Expressions and Equations". However, to be on the safe side, it is better to post it here.
The set of real numbers is a proper subset of the set of complex numbers.
In other words, all real numbers are complex numbers.
If the equation has real roots only, or complex roots only, or both real and complex roots, that is fine.
But what if the equation has only complex roots and it is posted under "Expressions and Equations" meant only for real numbers?
$ x + 1 = \pm \sqrt[4]{24} \\[3ex] x = -1 \pm \sqrt[4]{24} \\[3ex] \text{Focus on } \sqrt[4]{24} \\[3ex] 24 = 24 + 0i \\[3ex] = x + yi \\[3ex] \text{real part} = x = 24 \\[3ex] \text{This is positive: location is the side side of the complex plane} \\[3ex] \text{imaginary part} = y = 0 \\[3ex] \text{This is located at the origin} \\[3ex] \implies \\[3ex] 24 + 0i \text{ is in the 1st Quadrant} \\[5ex] \text{Convert to polar form} \\[3ex] \text{modulus} = r = \sqrt{x^2 + y^2} \\[3ex] r = \sqrt{24^2 + 0^2} \\[5ex] r = \sqrt{24^2} \\[5ex] r = 24 \\[5ex] \text{argument} = \theta = \tan^{-1}(0) \\[3ex] \theta = 0\;RAD \\[3ex] r(\cos\theta + i\sin\theta)...\text{Polar Form} \\[3ex] 24 * (\cos 0 + i\sin 0) \\[3ex] 24 * (1 + i(0)) \\[3ex] 24(1 + 0i) \\[5ex] \underline{\text{ Abraham De Moivre's Theorem}} \\[3ex] \text{Given the Polar Form}: Z = r(\cos\theta + i\sin\theta) \\[3ex] \text{The n-th roots are}: Z_k = \sqrt[n]{r}\left[\cos\left(\dfrac{\theta + 2k\pi}{n}\right) + i\sin\left(\dfrac{\theta + 2k\pi}{n}\right)\right] \\[6ex] For\;\; \sqrt[4]{24} \\[3ex] \text{root} = n = 4 \\[3ex] \text{index} = k = 0, 1, 2, 3 \\[3ex] \implies \\[3ex] \text{When } k = 0 \\[3ex] Z_0 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 2(0)\pi}{4}\right) + i\sin\left(\dfrac{0 + 2(0)\pi}{4}\right)\right] \\[5ex] Z_0 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 0}{4}\right) + i\sin\left(\dfrac{0 + 0}{4}\right)\right] \\[5ex] Z_0 = \sqrt[4]{24}(\cos 0 + i\sin 0) \\[3ex] Z_0 = \sqrt[4]{24}(1 + i * 0) \\[3ex] Z_0 = \sqrt[4]{24}(1 + 0) \\[3ex] Z_0 = \sqrt[4]{24}(1) \\[3ex] Z_0 = \sqrt[4]{24} \\[5ex] \text{When } k = 1 \\[3ex] Z_1 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 2(1)\pi}{4}\right) + i\sin\left(\dfrac{0 + 2(1)\pi}{4}\right)\right] \\[5ex] Z_1 = \sqrt[4]{24}\left[\cos\left(\dfrac{2\pi}{4}\right) + i\sin\left(\dfrac{2\pi}{4}\right)\right] \\[5ex] Z_1 = \sqrt[4]{24}\left[\cos\left(\dfrac{\pi}{2}\right) + i\sin\left(\dfrac{\pi}{2}\right)\right] \\[5ex] Z_1 = \sqrt[4]{24}(0 + i * 1) \\[3ex] Z_1 = \sqrt[4]{24}(0 + i) \\[3ex] Z_1 = \sqrt[4]{24}(i) \\[3ex] Z_1 = i\sqrt[4]{24} \\[5ex] \text{When } k = 2 \\[3ex] Z_2 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 2(2)\pi}{4}\right) + i\sin\left(\dfrac{0 + 2(2)\pi}{4}\right)\right] \\[5ex] Z_2 = \sqrt[4]{24}\left[\cos\left(\dfrac{4\pi}{4}\right) + i\sin\left(\dfrac{4\pi}{4}\right)\right] \\[5ex] Z_2 = \sqrt[4]{24}(\cos \pi + i\sin \pi) \\[3ex] Z_2 = \sqrt[4]{24}(-1 + i * 0) \\[3ex] Z_2 = \sqrt[4]{24}(-1 + 0) \\[3ex] Z_2 = \sqrt[4]{24}(-1) \\[3ex] Z_2 = -\sqrt[4]{24} \\[5ex] \text{When } k = 3 \\[3ex] Z_3 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 2(3)\pi}{4}\right) + i\sin\left(\dfrac{0 + 2(3)\pi}{4}\right)\right] \\[5ex] Z_3 = \sqrt[4]{24}\left[\cos\left(\dfrac{6\pi}{4}\right) + i\sin\left(\dfrac{6\pi}{4}\right)\right] \\[5ex] Z_3 = \sqrt[4]{24}\left[\cos\left(\dfrac{3\pi}{2}\right) + i\sin\left(\dfrac{3\pi}{2}\right)\right] \\[5ex] Z_3 = \sqrt[4]{24}(0 + i * -1) \\[3ex] Z_3 = \sqrt[4]{24}(0 - i) \\[3ex] Z_3 = \sqrt[4]{24}(-i) \\[3ex] Z_3 = -i\sqrt[4]{24} \\[5ex] But: \\[3ex] x = -1 \pm \sqrt[4]{24} \\[3ex] x = -1 \pm Z_0 \\[2ex] x = -1 \pm \sqrt[4]{24} \\[5ex] x = -1 \pm Z_1 \\[3ex] x = -1 \pm i\sqrt[4]{24} \\[5ex] x = -1 \pm Z_2 \\[3ex] x = -1 \pm -\sqrt[4]{24} \\[5ex] x = -1 \pm Z_3 \\[3ex] x = -1 \pm -i\sqrt[4]{24} \\[5ex] \text{roots are: } \\[3ex] x = -1 + \sqrt[4]{24} \\[3ex] x = -1 - \sqrt[4]{24} \\[3ex] x = -1 + i\sqrt[4]{24} \\[3ex] x = -1 - i\sqrt[4]{24} $
Check your solution with a calculator.
$ 3(x + 1)^4 - 12 = 60 \\[3ex] 3(x + 1)^4 = 60 + 12 \\[3ex] 3(x + 1)^4 = 72 \\[3ex] (x + 1)^4 = \dfrac{72}{3} \\[5ex] (x + 1)^4 = 24 \\[3ex] x + 1 = \pm \sqrt[4]{24} \\[3ex] $ Student: Mr. C, I'm curious why this question is posted under Complex Numbers
My thinking is that the question should be posted under "Expressions and Equations"
Besides, how are you sure that the roots of the equation are complex roots?
What if they are rational roots?
Teacher: Good question.
We shal find out the kind of roots.
It is okay to post the topic under "Expressions and Equations". However, to be on the safe side, it is better to post it here.
The set of real numbers is a proper subset of the set of complex numbers.
In other words, all real numbers are complex numbers.
If the equation has real roots only, or complex roots only, or both real and complex roots, that is fine.
But what if the equation has only complex roots and it is posted under "Expressions and Equations" meant only for real numbers?
$ x + 1 = \pm \sqrt[4]{24} \\[3ex] x = -1 \pm \sqrt[4]{24} \\[3ex] \text{Focus on } \sqrt[4]{24} \\[3ex] 24 = 24 + 0i \\[3ex] = x + yi \\[3ex] \text{real part} = x = 24 \\[3ex] \text{This is positive: location is the side side of the complex plane} \\[3ex] \text{imaginary part} = y = 0 \\[3ex] \text{This is located at the origin} \\[3ex] \implies \\[3ex] 24 + 0i \text{ is in the 1st Quadrant} \\[5ex] \text{Convert to polar form} \\[3ex] \text{modulus} = r = \sqrt{x^2 + y^2} \\[3ex] r = \sqrt{24^2 + 0^2} \\[5ex] r = \sqrt{24^2} \\[5ex] r = 24 \\[5ex] \text{argument} = \theta = \tan^{-1}(0) \\[3ex] \theta = 0\;RAD \\[3ex] r(\cos\theta + i\sin\theta)...\text{Polar Form} \\[3ex] 24 * (\cos 0 + i\sin 0) \\[3ex] 24 * (1 + i(0)) \\[3ex] 24(1 + 0i) \\[5ex] \underline{\text{ Abraham De Moivre's Theorem}} \\[3ex] \text{Given the Polar Form}: Z = r(\cos\theta + i\sin\theta) \\[3ex] \text{The n-th roots are}: Z_k = \sqrt[n]{r}\left[\cos\left(\dfrac{\theta + 2k\pi}{n}\right) + i\sin\left(\dfrac{\theta + 2k\pi}{n}\right)\right] \\[6ex] For\;\; \sqrt[4]{24} \\[3ex] \text{root} = n = 4 \\[3ex] \text{index} = k = 0, 1, 2, 3 \\[3ex] \implies \\[3ex] \text{When } k = 0 \\[3ex] Z_0 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 2(0)\pi}{4}\right) + i\sin\left(\dfrac{0 + 2(0)\pi}{4}\right)\right] \\[5ex] Z_0 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 0}{4}\right) + i\sin\left(\dfrac{0 + 0}{4}\right)\right] \\[5ex] Z_0 = \sqrt[4]{24}(\cos 0 + i\sin 0) \\[3ex] Z_0 = \sqrt[4]{24}(1 + i * 0) \\[3ex] Z_0 = \sqrt[4]{24}(1 + 0) \\[3ex] Z_0 = \sqrt[4]{24}(1) \\[3ex] Z_0 = \sqrt[4]{24} \\[5ex] \text{When } k = 1 \\[3ex] Z_1 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 2(1)\pi}{4}\right) + i\sin\left(\dfrac{0 + 2(1)\pi}{4}\right)\right] \\[5ex] Z_1 = \sqrt[4]{24}\left[\cos\left(\dfrac{2\pi}{4}\right) + i\sin\left(\dfrac{2\pi}{4}\right)\right] \\[5ex] Z_1 = \sqrt[4]{24}\left[\cos\left(\dfrac{\pi}{2}\right) + i\sin\left(\dfrac{\pi}{2}\right)\right] \\[5ex] Z_1 = \sqrt[4]{24}(0 + i * 1) \\[3ex] Z_1 = \sqrt[4]{24}(0 + i) \\[3ex] Z_1 = \sqrt[4]{24}(i) \\[3ex] Z_1 = i\sqrt[4]{24} \\[5ex] \text{When } k = 2 \\[3ex] Z_2 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 2(2)\pi}{4}\right) + i\sin\left(\dfrac{0 + 2(2)\pi}{4}\right)\right] \\[5ex] Z_2 = \sqrt[4]{24}\left[\cos\left(\dfrac{4\pi}{4}\right) + i\sin\left(\dfrac{4\pi}{4}\right)\right] \\[5ex] Z_2 = \sqrt[4]{24}(\cos \pi + i\sin \pi) \\[3ex] Z_2 = \sqrt[4]{24}(-1 + i * 0) \\[3ex] Z_2 = \sqrt[4]{24}(-1 + 0) \\[3ex] Z_2 = \sqrt[4]{24}(-1) \\[3ex] Z_2 = -\sqrt[4]{24} \\[5ex] \text{When } k = 3 \\[3ex] Z_3 = \sqrt[4]{24}\left[\cos\left(\dfrac{0 + 2(3)\pi}{4}\right) + i\sin\left(\dfrac{0 + 2(3)\pi}{4}\right)\right] \\[5ex] Z_3 = \sqrt[4]{24}\left[\cos\left(\dfrac{6\pi}{4}\right) + i\sin\left(\dfrac{6\pi}{4}\right)\right] \\[5ex] Z_3 = \sqrt[4]{24}\left[\cos\left(\dfrac{3\pi}{2}\right) + i\sin\left(\dfrac{3\pi}{2}\right)\right] \\[5ex] Z_3 = \sqrt[4]{24}(0 + i * -1) \\[3ex] Z_3 = \sqrt[4]{24}(0 - i) \\[3ex] Z_3 = \sqrt[4]{24}(-i) \\[3ex] Z_3 = -i\sqrt[4]{24} \\[5ex] But: \\[3ex] x = -1 \pm \sqrt[4]{24} \\[3ex] x = -1 \pm Z_0 \\[2ex] x = -1 \pm \sqrt[4]{24} \\[5ex] x = -1 \pm Z_1 \\[3ex] x = -1 \pm i\sqrt[4]{24} \\[5ex] x = -1 \pm Z_2 \\[3ex] x = -1 \pm -\sqrt[4]{24} \\[5ex] x = -1 \pm Z_3 \\[3ex] x = -1 \pm -i\sqrt[4]{24} \\[5ex] \text{roots are: } \\[3ex] x = -1 + \sqrt[4]{24} \\[3ex] x = -1 - \sqrt[4]{24} \\[3ex] x = -1 + i\sqrt[4]{24} \\[3ex] x = -1 - i\sqrt[4]{24} $
