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32199c7b2c
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32199c7b2c
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0deb97de3b
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@@ -19,8 +19,8 @@ tags: [ uni, nottingham, mmme1026, maths, complex_numbers ]
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Where:
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Where:
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- $x$ is the real part of $z$ (Re($z$))
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- $x$ is the real part of $z$ (which you may seen written as $\Re(z) = x$ or Re$(z) = x$)
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- $y$ is the imaginary part of $z$(Im($z$))
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- $y$ is the imaginary part of $z$ (which you may seen written as $\Im(z) = y$ or Im$(z) = y$)
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- Two complex numbers are equal if both their real and imaginary parts are equal
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- Two complex numbers are equal if both their real and imaginary parts are equal
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@@ -146,15 +146,23 @@ $$e^{i\theta} = \cos\theta + i\sin\theta$$
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**Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$
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**Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$
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<details>
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<summary>
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### Example 1
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### Example 1
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Write $z = -1 + i$ in exponential form
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Write $z = -1 + i$ in exponential form
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</summary>
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> $\arg z = \frac {3\pi} 4$
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> $\arg z = \frac {3\pi} 4$
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> $|z| = \sqrt 2$
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> $|z| = \sqrt 2$
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>
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>
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> So $z = \sqrt2e^{i\frac{3\pi} 4}$
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> So $z = \sqrt2e^{i\frac{3\pi} 4}$
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</details>
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<details>
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<summary>
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### Example 2
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### Example 2
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The equations for a mechanical vibration problem are found to have the following mathematical
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The equations for a mechanical vibration problem are found to have the following mathematical
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@@ -162,6 +170,8 @@ solution:
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$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$
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$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$
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</summary>
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where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
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where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
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constants.
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constants.
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Although $z(t)$
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Although $z(t)$
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@@ -200,6 +210,8 @@ b. Hence find the constants $b$ and $\varphi$ such that
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y(t) &= \frac 1 a \sin(\omega t - \delta) \\
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y(t) &= \frac 1 a \sin(\omega t - \delta) \\
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> \end{align*}
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> \end{align*}
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</details>
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## Products of Complex Numbers
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## Products of Complex Numbers
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Suppose we have 2 complex numbers:
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Suppose we have 2 complex numbers:
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@@ -222,22 +234,35 @@ z_3 = z_1 z_2 &= (r_1e^{i\theta_1})(r_2e^{i\theta_2}) \\
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Let $z = re^{i\theta}$. Consider $z^n$.
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Let $z = re^{i\theta}$. Consider $z^n$.
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Since $z = r(\cos\theta + i\sin\theta)$,
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\begin{align*}
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\begin{align*}
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\text{Since } z = r(\cos\theta + i\sin\theta) \\
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z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\
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z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\
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\text{But also} \\
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\end{align*}
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But also
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\begin{align*}
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z^n &= (re^{i\theta})^n \\
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z^n &= (re^{i\theta})^n \\
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&= r^n(e^{i\theta})^n \\
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&= r^n(e^{i\theta})^n \\
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&= r^ne^{in\theta} \\
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&= r^ne^{in\theta} \\
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&= r^n(\cos{n\theta} + i\sin{n\theta}) &\text{(2)} \\
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&= r^n(\cos{n\theta} + i\sin{n\theta}) &\text{(2)} \\
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\text{By equating (1) and (2), we find:}\\
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\end{align*}
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By equating (1) and (2), we find de Moivre's theorem:
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\begin{align*}
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r^n(\cos\theta +i\sin\theta)^n &= r^n(\cos{n\theta} + i\sin{n\theta}) \\
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(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta})
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(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta})
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\end{align*}
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\end{align*}
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<details>
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<summary>
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### Example 1
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### Example 1
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Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.
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Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.
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</summary>
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> \begin{align*}
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> \begin{align*}
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r &= |1+i| = \sqrt2 \\
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r &= |1+i| = \sqrt2 \\
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\theta &= \arg{1+i} = \frac \pi 4 \\
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\theta &= \arg{1+i} = \frac \pi 4 \\
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@@ -249,7 +274,10 @@ Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.
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&= 2^7 (1 - i) \\
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&= 2^7 (1 - i) \\
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&= 128 - 128i
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&= 128 - 128i
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> \end{align*}
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> \end{align*}
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</details>
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<details>
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<summary>
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### Example 2
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### Example 2
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Use de Moivre's theorem to show that
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Use de Moivre's theorem to show that
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@@ -260,6 +288,8 @@ Use de Moivre's theorem to show that
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\sin{2\theta} &= 2\sin\theta\cos\theta
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\sin{2\theta} &= 2\sin\theta\cos\theta
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\end{align*}
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\end{align*}
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</summary>
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> Let $n=2$:
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> Let $n=2$:
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> \begin{align*}
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> \begin{align*}
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@@ -268,11 +298,17 @@ Use de Moivre's theorem to show that
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\text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
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\text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
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> \end{align*}
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> \end{align*}
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</details>
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<details>
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<summary>
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### Example 3
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### Example 3
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Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
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Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
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$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.
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$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.
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</summary>
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> \begin{align*}
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> \begin{align*}
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r &= \sqrt{2} \\
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r &= \sqrt{2} \\
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\arg \omega = \theta &= \frac 3 4 \pi \\
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\arg \omega = \theta &= \frac 3 4 \pi \\
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@@ -283,14 +319,20 @@ $w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.
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&= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
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&= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
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> \end{align*}
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> \end{align*}
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</details>
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## Complex Roots of Polynomials
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## Complex Roots of Polynomials
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<details>
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<summary>
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### Example
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### Example
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Which complex numbers $z$ satisfy
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Find which complex numbers $z$ satisfy
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$$z^3 = 8i$$
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$$z^3 = 8i$$
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</summary>
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> 1. Write $8i$ in exponential form,
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> 1. Write $8i$ in exponential form,
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>
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>
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> $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
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> $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
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@@ -333,3 +375,5 @@ $$z^3 = 8i$$
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> Some of these complex roots may be real numbers.
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> Some of these complex roots may be real numbers.
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>
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>
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> 4. There are three solutions
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> 4. There are three solutions
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</details>
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