Notes
Polar coordinates
- $x = r\cos\theta$, $y = r\sin\theta$, $r^2 = x^2 + y^2$, $\displaystyle \tan\theta = \frac{y}{x}$. A point has many polar names, and $(-r, \theta) = (r, \theta + \pi)$.
- Basic graphs: $r = k$ is a circle; $r = a\cos\theta$ is a circle through the origin; $r = a + a\cos\theta$ is a cardioid; $r = a\cos n\theta$ is a rose with $n$ petals (odd $n$) or $2n$ petals (even $n$).
Complex numbers
- $i^2 = -1$. The modulus of $a + bi$ is $\displaystyle \sqrt{a^2 + b^2}$. Divide using the conjugate.
- Polar form: $z = r\,\text{cis}\,\theta$. Multiply by multiplying moduli and adding angles.
- De Moivre: $(r\,\text{cis}\,\theta)^n = r^n\,\text{cis}\,n\theta$. The $n$ roots of a number are equally spaced on a circle.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle x = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3}$ and $\displaystyle y = 6 \cdot \frac{1}{2} = 3$.
$\displaystyle r = \sqrt{2}$ and the point is in quadrant III.
$r^2 = 4r\cos\theta$ gives $x^2 + y^2 = 4x$, or $(x - 2)^2 + y^2 = 4$.
$6 - 2i + 3i - i^2 = 6 + i + 1 = 7 + i$.
$(1 - i)^2 = -2i$, and $(-2i)^2 = -4$.
Moduli multiply to 6 and angles add to $\displaystyle \frac{\pi}{2}$.
Free response
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Free-response question
Let $z = 1 + \sqrt{3}i$.
(a) Write $z$ in polar form.
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$\displaystyle |z| = \sqrt{1 + 3} = 2$ and $\displaystyle \tan\theta = \sqrt{3}$ in quadrant I, so $\displaystyle \theta = \frac{\pi}{3}$. Thus $\displaystyle z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$.
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(b) Use De Moivre's Theorem to compute $z^6$.
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$\displaystyle z^6 = 2^6\,\text{cis}\,\left(6 \cdot \frac{\pi}{3}\right) = 64\,\text{cis}\,2\pi = 64$.
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(c) Find the two square roots of $z$ in polar form and in rectangular form for the principal root.
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The roots are $\displaystyle \sqrt{2}\,\text{cis}\,\frac{\pi}{6}$ and $\displaystyle \sqrt{2}\,\text{cis}\,\frac{7\pi}{6}$. The principal root is $\displaystyle \sqrt{2}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i$.
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