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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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@ -222,15 +222,24 @@ 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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