Practice questions

Chapter 6: Polar coordinates and complex numbers

Try each problem first. Then check your work with the solution.

1. Polar coordinates.

(a) Find the rectangular coordinates of ${\left(6,\dfrac{5\pi}{6}\right)}$.

(b) Find polar coordinates of ${(-2,-2)}$, with ${r>0}$ and ${0\le\theta<2\pi}$.

(d) Write ${r=3\sec\theta}$ in rectangular form.

Solution:

(a)

$\begin{align*}x&=6\cos\dfrac{5\pi}{6}=6\left(-\dfrac{\sqrt{3}}{2}\right)\\&=-3\sqrt{3}\end{align*}$

$y=6\sin\dfrac{5\pi}{6}=6\cdot\dfrac{1}{2}=3$

The point is ${(-3\sqrt{3},\ 3)}$.

(b)

${r=\sqrt{4+4}=2\sqrt{2}}$. The point is in quadrant III, and ${\tan\theta=1}$, so ${\theta=\pi+\dfrac{\pi}{4}=\dfrac{5\pi}{4}}$.

The point is ${\left(2\sqrt{2},\ \dfrac{5\pi}{4}\right)}$.

(c)

Replace ${x^2+y^2}$ by ${r^2}$ and ${y}$ by ${r\sin\theta}$:

${r^2=6r\sin\theta}$

Divide by ${r}$: ${r=6\sin\theta}$. (This loses nothing: the pole is on the graph anyway, at ${\theta=0}$.)

(d)

${r=\dfrac{3}{\cos\theta}}$, so ${r\cos\theta=3}$. That is ${x=3}$: a vertical line.

2. Graphs of polar equations.

(a) Describe the graph of ${r=4\cos\theta}$.

(b) Sketch the graph of ${r=2-2\sin\theta}$.

(d) How many petals does the rose ${r=4\sin 4\theta}$ have?

(e) Where does the limaçon ${r=1+2\sin\theta}$ pass through the pole, for ${0\le\theta<2\pi}$?

Solution:

(a)

It is a circle through the pole with diameter ${4}$, centered on the polar axis. In rectangular form, ${(x-2)^2+y^2=4}$: center ${(2,0)}$, radius ${2}$.

(b)

This is a cardioid. Key values: at ${\theta=0}$, ${r=2}$. At ${\theta=\dfrac{\pi}{2}}$, ${r=0}$ (the pole). At ${\theta=\pi}$, ${r=2}$. At ${\theta=\dfrac{3\pi}{2}}$, ${r=4}$ (the farthest point, straight down).

The cardioid ${r=2-2\sin\theta}$, with the four key points in orange.

(c)

Here ${n=3}$ is odd, so the rose has ${3}$ petals, each ${5}$ units long.

(d)

Here ${n=4}$ is even, so the rose has ${2\cdot 4=8}$ petals.

(e)

Set ${r=0}$: ${1+2\sin\theta=0}$, so ${\sin\theta=-\dfrac{1}{2}}$. In ${[0,2\pi)}$ that is ${\theta=\dfrac{7\pi}{6}}$ and ${\theta=\dfrac{11\pi}{6}}$.

3. Trigonometric form of complex numbers.

(a) Write ${-2+2i}$ in trigonometric form.

(b) Write ${1-\sqrt{3}\,i}$ in trigonometric form.

$z_2=2\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)$

Find ${z_1z_2}$ and ${\dfrac{z_1}{z_2}}$ in the form ${a+bi}$.

Solution:

(a)

${r=\sqrt{4+4}=2\sqrt{2}}$. The point ${(-2,2)}$ is in quadrant II with reference angle ${\dfrac{\pi}{4}}$, so ${\theta=\dfrac{3\pi}{4}}$:

$\begin{align*}&-2+2i\\&=2\sqrt{2}\left(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}\right)\end{align*}$

(b)

${r=\sqrt{1+3}=2}$. The point ${(1,-\sqrt{3})}$ is in quadrant IV with reference angle ${\dfrac{\pi}{3}}$, so ${\theta=2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}}$:

$\begin{align*}&1-\sqrt{3}\,i\\&=2\left(\cos\dfrac{5\pi}{3}+i\sin\dfrac{5\pi}{3}\right)\end{align*}$

(c)

Product: modulus ${6\cdot 2=12}$, argument $\dfrac{2\pi}{3}+\dfrac{\pi}{6}=\dfrac{5\pi}{6}$:

$\begin{align*}z_1z_2&=12\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i\right)\\&=-6\sqrt{3}+6i\end{align*}$

Quotient: modulus ${\dfrac{6}{2}=3}$, argument $\dfrac{2\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{2}$:

${\dfrac{z_1}{z_2}=3(0+1\cdot i)=3i}$

4. De Moivre's theorem and roots.

(a) Find ${(1-\sqrt{3}\,i)^6}$.

(b) Find ${(\sqrt{3}+i)^4}$ in the form ${a+bi}$.

(d) Find the three cube roots of ${-27}$.

Solution:

(a)

From 3(b), the modulus is ${2}$ and the argument is ${\dfrac{5\pi}{3}}$. So the power has modulus ${2^6=64}$ and argument ${6\cdot\dfrac{5\pi}{3}=10\pi}$, which is five full turns:

$\begin{align*}(1-\sqrt{3}\,i)^6&=64(\cos 10\pi+i\sin 10\pi)\\&=64\end{align*}$

(b)

${\sqrt{3}+i}$ has modulus ${2}$ and argument ${\dfrac{\pi}{6}}$. The fourth power has modulus ${16}$ and argument ${\dfrac{4\pi}{6}=\dfrac{2\pi}{3}}$:

$\begin{align*}(\sqrt{3}+i)^4&=16\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right)\\&=-8+8\sqrt{3}\,i\end{align*}$

(c)

${i=\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}}$

The square roots have modulus ${1}$ and arguments ${\dfrac{\pi/2+2k\pi}{2}}$, that is ${\dfrac{\pi}{4}}$ and ${\dfrac{5\pi}{4}}$:

${\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i}$and$-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}i$

(d)

${-27=27(\cos\pi+i\sin\pi)}$. The cube roots have modulus ${3}$ and arguments ${\dfrac{\pi+2k\pi}{3}}$: ${\dfrac{\pi}{3}}$, ${\pi}$, ${\dfrac{5\pi}{3}}$.

${\dfrac{\pi}{3}}$: ${\dfrac{3}{2}+\dfrac{3\sqrt{3}}{2}i}$
${\pi}$: ${3(-1)=-3}$
${\dfrac{5\pi}{3}}$: ${\dfrac{3}{2}-\dfrac{3\sqrt{3}}{2}i}$

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