IB Mathematics AA HL Flashcards- Polar Forms Euler’s Form

1// The form: ( z = x + iy) is called the cartesian form, or the rectangular form. In this section, we will see a different form that may be more useful in some cases: polar form (also called modulus-argument for).

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POLAR FORM// Using trigonometry, we see the following:

\[
x = r\cos\theta \quad y = r\sin\theta
\]

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So, if \( z = x + yi = r\cos\theta + (r\sin\theta)i \), then:

\[
Z = r(\cos\theta + i\sin\theta) \quad \text{in F.B.}
\]

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where \( r = |z| \) and \( \theta = \arg(z) \).

1// We will often shorten \( z = r(\cos\theta + i\sin\theta) \) to \( z = r\text{cis}\theta \) → in F.B.

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E.G. 1 // Convert \( z = \sqrt{3} – i \) to polar form:

1. \( r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2 \), \( \tan\theta = \frac{-1}{\sqrt{3}} \), \( \theta = 150^\circ \) or \( 330^\circ \)?

\(\therefore z = 2\text{cis}330^\circ\)

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E.G. 2 // Convert \( z = 3(\cos\frac{\pi}{5} + i\sin\frac{5\pi}{6}) \) into cartesian form:

\[
x = r\cos\theta = 3\cos\left(\frac{5\pi}{6}\right) = -\frac{3\sqrt{3}}{2}, \quad y = r\sin\theta = 3\sin\left(\frac{5\pi}{6}\right) = \frac{3}{2}
\]

\(\therefore z = -\frac{3\sqrt{3}}{2} + \frac{3}{2}i\)

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MULTIPLICATION// If we want to multiply \( Z_1 = r_1(\cos(\theta_1) + i\sin(\theta_1)) \) & \( Z_2 = r_2(\cos(\theta_2) + i\sin(\theta_2)) \), you could show, using the compound angle formulae [see 3.10], that we get:

\[
Z_1Z_2 = r_1r_2[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)] = r_1r_2[\text{cis}(\theta_1 + \theta_2)] \quad \text{→ Not in F.B.}
\]

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E.G. 3 // \( Z_1 = 4\text{cis}(\frac{\pi}{3}), Z_2 = 2\text{cis}(\frac{4\pi}{3}) \), find \( Z_1Z_2 \):

\(\therefore Z_1Z_2 = 4 \cdot 2\text{cis}(\frac{\pi}{3} + \frac{4\pi}{3}) = 8\text{cis}(\pi)\)

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DIVISION// Try to show this formula is true is a long process, but using the final formula is very easy—far better than dividing in Cartesian form.

Take \( \frac{Z_1}{Z_2} = \frac{r_1\text{cis}\theta_1}{r_2\text{cis}\theta_2} \):

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Using compound angle and pyth. identity:

\[
\frac{Z_1}{Z_2} = \frac{r_1}{r_2}\text{cis}(\theta_1 – \theta_2) \quad \text{→ Not in F.B.}
\]

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E.G. 4 // Take \( Z_1 = \sqrt{2}\text{cis}45^\circ, Z_2 = 2\text{cis}(-30^\circ) \):

(a) Find \( \frac{Z_1}{Z_2} \):

\[
\frac{Z_1}{Z_2} = \frac{\sqrt{2}}{2}\text{cis}(45^\circ – (-30^\circ)) = \frac{\sqrt{2}}{2}\text{cis}75^\circ
\]

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(b) Find \( \frac{1}{Z_2} \):

\[
\frac{1}{Z_2} = \frac{1}{2}\text{cis}(-(-30^\circ)) = \frac{1}{2}\text{cis}30^\circ
\]

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EULER’S FORM// Again, this requires theory that has not been covered yet (see 5.19), but to sum it up quickly, it requires knowing the Maclaurin series for various expressions. Firstly,

\[
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots
\]

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Then,

\[
\cos x + i\sin x = 1 + \frac{-x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + \dots
\]

and

\[
\cos x + i\sin x = 1 + ix – \frac{x^2}{2!} + \frac{ix^3}{3!} + \frac{-x^4}{4!} + \dots
\]

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Consider

\[
e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \dots = \cos x + i\sin x
\]

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So if \( z = \cos \theta + i \sin \theta \) was polar form, we now have a new form:

\[
Z = r(\text{cis} \theta) = r e^{i \theta} \quad \text{in F.B.}
\]

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1// Because of the simplicity of this formula, and exponent rules, operations with complex numbers become even easier:

E.G. 5 // Express \( z = 5 + 5i \) in exponential form:

\[
|z| = 5\sqrt{2}, \quad \tan \theta = \frac{5}{5} = 1 \quad \Rightarrow \quad \theta = \frac{\pi}{4} \quad \Rightarrow \quad z = 5\sqrt{2} e^{i \frac{\pi}{4}}
\]

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E.G. 6 // Simplify:

\[
\frac{\left(\cos 60^\circ + i \sin 60^\circ\right) \times \left(\cos 30^\circ + i \sin 30^\circ\right)}{\left(\cos 40^\circ + i \sin 40^\circ\right)} = e^{i 60^\circ} \cdot e^{i 30^\circ} \cdot e^{-i 40^\circ} = e^{i 50^\circ} = \text{cis}(50^\circ)
\]

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E.G. 7 // Use Euler’s formula to prove de Moivre’s thm:

\[
(r \text{cis} \theta)^n = r^n (\text{cis}(n \theta))
\]

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Convert \( (r \text{cis} \theta)^n \) into Euler’s form:

\[
(r \text{cis} \theta)^n = (r e^{i \theta})^n
\]

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Expand:

\[
(re^{i \theta})^n = r^n (e^{i \theta \cdot n})
\]

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Convert back:

\[
r^n e^{i (n \theta)} = r^n \text{cis}(n \theta)
\]

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E.G. 8 // Find the value of \( e^{i \pi} + 1 \):

Convert \( e^{i \pi} \) into polar form:

\[
e^{i \pi} = \cos \pi + i \sin \pi
\]

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Now we simply evaluate:

\[
e^{i \pi} + 1 = \cos \pi + i \sin \pi + 1 = -1 + 0i + 1 = 0
\]

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1// In E.G.8, we have just shown that the very famous formula of \( e^{i \pi} = -1 \) is true. People say this is the ‘most beautiful’ formula in mathematics, because it mixes imaginary numbers, a negative, and two unrelated, irrational constants. Having said that, \( \pi \) is actually just the units used in radians for a half-turn.

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