Add examples for de moivre and euler
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@ -42,14 +42,14 @@ $$\bar{z} = z -iy$$
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#### Example
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\begin{align*}
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z_1 &= 5 + i \\
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z_2 &= 1 -i \\
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\\
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\frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
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&= \frac{5 + i + 5i -1}{1 + 1} \\
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&= \frac{4 + 6i}{2} = 2 + 3i
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\end{align*}
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> \begin{align*}
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z_1 &= 5 + i \\
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z_2 &= 1 -i \\
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\\
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\frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
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&= \frac{5 + i + 5i -1}{1 + 1} \\
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&= \frac{4 + 6i}{2} = 2 + 3i
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> \end{align*}
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### Algebra and Conjugation
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@ -146,14 +146,59 @@ $$e^{i\theta} = \cos\theta + i\sin\theta$$
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**Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$
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Example:
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> Write $z = -1 + i$ in exponential form:
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>
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> $\arg z = \frac {3\pi} 4$
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> $|z| = \sqrt 2$
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>
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> So $z = \sqrt2e^{i\frac{3\pi} 4}$
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### Example 1
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Write $z = -1 + i$ in exponential form
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> $\arg z = \frac {3\pi} 4$
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> $|z| = \sqrt 2$
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>
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> So $z = \sqrt2e^{i\frac{3\pi} 4}$
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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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solution:
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$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$
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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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Although $z(t)$
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is complex and cannot directly represent a physical solution, it turns out that the real and
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imaginary parts $x(t)$ and $y(t)$ in $z(t) = x(t) + iy(t)$ can. Polar notation can be used to extract
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this physical information efficiently as follows:
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a. Put the denominator in the form
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$$ae^{i\delta}$$
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where you should give explicit expressions for $a$ and $\delta$ in terms of $\gamma$, $\gamma_0$,
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and $\gamma$.
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> \begin{align*}
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a &= \sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2} \\
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\delta &= \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}
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> \end{align*}
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b. Hence find the constants $b$ and $\varphi$ such that
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$$x(t) = b\cos(\omega t + \varphi)$$
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and write a similar expression for $y(t)$.
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> \begin{align*}
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z &= \frac{e^i\omega t}{ae^{i\delta}} = \frac 1 a e^{i (\omega t - \delta)} \\
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x + iy &= \frac 1 a \cos(\omega t - \delta) + \frac 1 a \sin(\omega t - \delta) \\
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\therefore \Re z &= x = \frac 1 a \cos(\omega t - \delta), \\
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\Im z &= y = \frac 1 a \sin(\omega t - \delta) \\
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\\
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b &= \frac 1 a = \frac 1 {\sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2}} \\
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\varphi &= -\delta = \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}\\
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\\
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y(t) &= \frac 1 a \sin(\omega t - \delta) \\
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> \end{align*}
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## Products of Complex Numbers
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@ -193,17 +238,17 @@ z^n &= (re^{i\theta})^n \\
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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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\begin{align*}
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r &= |1+i| = \sqrt2 \\
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\theta &= \arg{1+i} = \frac \pi 4 \\
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\\
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\text{So } 1 + i &= \sqrt2(\cos{\frac pi 4} + i\sin{\frac \pi 4}) = \sqrt2e^{i\frac \pi 4} \text{ and}\\
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(i+i)^{15} &= (\sqrt2)^{15}\left(\cos{\frac \pi 4} + i\sin{\frac \pi 4}\right)^{15} \\
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&= 2^{\frac 15 2} \left(\cos{\frac{15\pi} 4} + i\sin{\frac{15\pi} 4}\right) \\
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&= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
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&= 2^7 (1 - i) \\
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&= 128 - 128i
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\end{align*}
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> \begin{align*}
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r &= |1+i| = \sqrt2 \\
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\theta &= \arg{1+i} = \frac \pi 4 \\
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\\
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\text{So } 1 + i &= \sqrt2(\cos{\frac pi 4} + i\sin{\frac \pi 4}) = \sqrt2e^{i\frac \pi 4} \text{ and}\\
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(i+i)^{15} &= (\sqrt2)^{15}\left(\cos{\frac \pi 4} + i\sin{\frac \pi 4}\right)^{15} \\
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&= 2^{\frac 15 2} \left(\cos{\frac{15\pi} 4} + i\sin{\frac{15\pi} 4}\right) \\
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&= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
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&= 2^7 (1 - i) \\
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&= 128 - 128i
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> \end{align*}
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### Example 2
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@ -215,13 +260,28 @@ Use de Moivre's theorem to show that
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\sin{2\theta} &= 2\sin\theta\cos\theta
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\end{align*}
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Let $n=2$:
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> Let $n=2$:
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\begin{align*}
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(\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
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\text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{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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> \begin{align*}
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(\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
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\text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{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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### Example 3
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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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> \begin{align*}
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r &= \sqrt{2} \\
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\arg \omega = \theta &= \frac 3 4 \pi \\
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\\
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\omega^n &= r^n(cos{n\theta} + i\sin{n\theta}) \\
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\bar\omega^n &= r^n(cos{n\theta} - i\sin{n\theta}) \\
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\omega^n + \bar\omega^n &= r^n(2\cos{n\theta}) \\
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&= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
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> \end{align*}
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## Complex Roots of Polynomials
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@ -231,45 +291,45 @@ Which complex numbers $z$ satisfy
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$$z^3 = 8i$$
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1. Write $8i$ in exponential form,
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$|8i| = 8$ and $\arg{8i} = \frac \pi 2$
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$\therefore 8i = 8e^{i\frac \pi 2}$
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2. Let the solution be $r = re^{i\theta}$.
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Then $z^3 = r^3e^{3i\theta}$.
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3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
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i. Compare modulus:
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$r^3 = 8 \rightarrow r = 2$
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ii. Compare argument:
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$$3\theta = \frac \pi 2$$
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is a solution but there are others since
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$$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
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so we get a solution whenever
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$$3\theta = \frac \pi 2 + 2n\pi$$
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for any integer `n`
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- $n = 0 \rightarrow z = \sqrt3 + i$
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- $n = 1 \rightarrow z = -\sqrt3 + i$
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- $n = 2 \rightarrow z = -2i$
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- $n = 3 \rightarrow z = \sqrt3 + i$
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- $n = 4 \rightarrow z = -\sqrt3 + i$
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- The solutions start repeating as you can see
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In general, an $n$-th order polynomial has exactly $n$ complex roots.
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Some of these complex roots may be real numbers.
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4. There are three solution
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> 1. Write $8i$ in exponential form,
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>
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> $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
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>
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> $\therefore 8i = 8e^{i\frac \pi 2}$
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>
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>
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> 2. Let the solution be $r = re^{i\theta}$.
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>
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> Then $z^3 = r^3e^{3i\theta}$.
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>
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> 3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
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>
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> i. Compare modulus:
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>
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> $r^3 = 8 \rightarrow r = 2$
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>
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> ii. Compare argument:
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>
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> $$3\theta = \frac \pi 2$$
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>
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> is a solution but there are others since
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>
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> $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
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>
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> so we get a solution whenever
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>
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> $$3\theta = \frac \pi 2 + 2n\pi$$
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>
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> for any integer `n`
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>
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> - $n = 0 \rightarrow z = \sqrt3 + i$
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> - $n = 1 \rightarrow z = -\sqrt3 + i$
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> - $n = 2 \rightarrow z = -2i$
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> - $n = 3 \rightarrow z = \sqrt3 + i$
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> - $n = 4 \rightarrow z = -\sqrt3 + i$
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> - The solutions start repeating as you can see
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>
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> In general, an $n$-th order polynomial has exactly $n$ complex roots.
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> Some of these complex roots may be real numbers.
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>
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> 4. There are three solutions
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