Polar Form

The polar form of a complex number $ x+iy$ is $ r(\cos(\theta) + i\sin(\theta))$ where $ (r,\theta)$ are any choice of polar coordinates that represent the point $ (x,y)$ in rectangular coordinates. Recall that you can find the polar form of a point using that

$\displaystyle r=\sqrt{x^2+y^2}$    and $\displaystyle \quad\theta=\tan^{-1}(y/x).

NOTE: The ``existence'' of complex numbers wasn't generally accepted until people got used to a geometric interpretation of them.

Example 4.3.2   Find the polar form of $ 1+i$.
Solution. We have $ r=\sqrt{2}$, so

$\displaystyle 1+i = \sqrt{2}\left( \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right)
= \sqrt{2}\left( \cos(\pi/4) + i \sin(\pi/4) \right).

Example 4.3.3   Find the polar form of $ \sqrt{3} - i$.
Solution. We have $ r=\sqrt{3 + 1} = 2$, so

$\displaystyle \sqrt{3} - i = 2 \left( \frac{\sqrt{3}}{2} + i\frac{-1}{2} \right)
= 2 \left( \cos(-\pi/6) +i \sin(-\pi/6)\right)

[[A picture is useful here.]]

Finding the polar form of a complex number is exactly the same problem as finding polar coordinates of a point in rectangular coordinates. The only hard part is figuring out what $ \theta$ is.

If we write complex numbers in rectangular form, their sum is easy to compute:

$\displaystyle (a+bi) + (c+di) = (a+c) + (b+d)i

The beauty of polar coordinates is that if we write two complex numbers in polar form, then their product is very easy to compute:

$\displaystyle r_1 (\cos(\theta_1) + i \sin(\theta_1))
\cdot r_2 (\cos(\theta_2...
(r_1 r_2) (\cos(\theta_1+\theta_2) + i \sin(\theta_1 + \theta_2)).

The magnitudes multiply and the angles add. The above formula is true because of the double angle identities for $ \sin$ and $ \cos$ (and it is how I remember those formulas!).

$\displaystyle (\cos(\theta_1)$ $\displaystyle + i \sin(\theta_1)) \cdot (\cos(\theta_2) + i \sin(\theta_2))$    
  $\displaystyle = (\cos(\theta_1)\cos(\theta_2) - \sin(\theta_1)\sin(\theta_2)) + i (\sin(\theta_1)\cos(\theta_2) + \cos(\theta_1)\sin(\theta_2)).$    

For example, the power of a singular complex number in polar form is easy to compute; just power the $ r$ and multiply the angle.

Theorem 4.3.4 (De Moivre's)   For any integer $ n$ we have

$\displaystyle (r (\cos(\theta) + i\sin(\theta)))^n
= r^n (\cos(n \theta) + i\sin(n \theta)).

Example 4.3.5   Compute $ (1+i)^{2006}$.
Solution. We have

$\displaystyle (1+i)^{2006}$ $\displaystyle = (\sqrt{2}\left( \cos(\pi/4) + i \sin(\pi/4) \right))^{2006}$    
  $\displaystyle = \sqrt{2}^{2006} \left( \cos(2006 \pi/4) + i \sin(2006 \pi/4) \right))$    
  $\displaystyle = 2^{1003} \left( \cos(3\pi/2) + i \sin(3\pi/2) \right))$    
  $\displaystyle = -2^{1003} i$    

To get $ \cos(2006 \pi/4) = \cos(3\pi/2)$ we use that $ 2006/4 = 501.5$, so by periodicity of cosine, we have

$\displaystyle \cos(2006 \pi/4) = \cos((501.5)\pi - 250(2\pi)) = \cos(1.5\pi) = \cos(3\pi/2).

EXAM 1: Wednesday 7:00-7:50pm in Pepper Canyon 109 (!)
Today: Supplement 1 (get online; also homework online)
Wednesday: Review
Bulletin board, online chat, directory, etc. - see main course website.
Review day - I will prepare no LECTURE; instead I will answer questions.
Your job is to have your most urgent questions ready to go!
Office hours moved: NOT Tue 11-1 (since nobody ever comes then and I'll be at a conference); instead I'll be in my office to answer questions WED 1:30-4pm, and after class on WED too.
Office: AP&M 5111

Quick review:

Given a point $ (x,y)$ in the plane, we can also view it as $ x+iy$ or in polar form as $ r(\cos(\theta) + i\sin(\theta))$. Polar form is great since it's good for multiplication, powering, and for extracting roots:

$\displaystyle r_1(\cos(\theta_1) + i \sin(\theta_1))
r_2(\cos(\theta_2) + i \si...
= (r_1 r_2) (\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)).

(If you divide, you subtract the angle.) The point is that the polar form works better with multiplication than the rectangular form.

Theorem 4.3.6 (De Moivre's)   For any integer $ n$ we have

$\displaystyle (r (\cos(\theta) + i\sin(\theta)))^n
= r^n (\cos(n \theta) + i\sin(n \theta)).

Since we know how to raise a complex number in polar form to the $ n$th power, we can find all numbers with a given power, hence find the $ n$th roots of a complex number.

Proposition 4.3.7 ($ n$th roots)   A complex number $ z=r(\cos(\theta) + i\sin(\theta))$ has $ n$ distinct $ n$th roots:

$\displaystyle r^{1/n}\left(\cos\left(\frac{\theta+2\pi k}{n}\right)
+ i\sin\left(\frac{\theta+2\pi k}{n}\right) \right),

for $ k=0,1,\ldots, n-1$. Here $ r^{1/n}$ is the real positive $ n$-th root of $ r$.

As a double-check, note that by De Moivre, each number listed in the proposition has $ n$th power equal to $ z$.

An application of De Moivre is to computing $ \sin(n\theta)$ and $ \cos(n\theta)$ in terms of $ \sin(\theta)$ and $ \cos(\theta)$. For example,

$\displaystyle \cos(3 \theta) + i\sin(3 \theta)$ $\displaystyle = (\cos(\theta) + i\sin(\theta))^3$    
  $\displaystyle =(\cos(\theta)^3 - 3\cos{}(\theta)\sin(\theta)^2) + i(3\cos(\theta)^2\sin(\theta) - \sin(\theta)^3)$    

Equate real and imaginary parts to get formulas for $ \cos(3\theta)$ and $ \sin(3\theta)$. In the next section we will discuss going in the other direction, i.e., writing powers of $ \sin$ and $ \cos$ in terms of $ \sin$ and cosine.

Example 4.3.8   Find the cube roots of $ 2$.
Solution. Write $ 2$ in polar form as

$\displaystyle 2 = 2 (\cos(0) + i \sin(0)).

Then the three cube roots of $ 2$ are

$\displaystyle 2^{1/3} (\cos(2\pi k/3) + i\sin(2\pi k/3)),

for $ k=0,1,2$. I.e.,

$\displaystyle 2^{1/3}, \quad 2^{1/3}(-1/2 + i \sqrt{3}/2),
\quad 2^{1/3}(-1/2 - i \sqrt{3}/2).

William Stein 2006-03-15