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### Geometry of Complex Numbers

#### Discussion

In section 1 we covered the basic arithmetic of complex numbers, which we will use to solve differential equations in later sections. In this section we will give a geometric interpretation of complex numbers. This interpretation isn't*necessary*to solve differential equations, but most students find it helpful in understanding how complex numbers work.

Complex numbers don't lie on a number line but rather on the complex plane. Complex numbers can be written $x+iy$, where $x$ and $y$ are real. We naturally graph the number $x+iy$ at the point $(x,y)$. The $x$--axis is referred to as the Real axis and the $y$--axis as the Imaginary axis. We can also identify points in the plane using polar coordinates rather than rectangular coordinates. The polar coordinates of a complex number are called the modulus and the argument of the complex number, where modulus is the distance from the origin ($r$) and the argument is the angle made with the Real axis by the segment from the origin to the point ($\theta$). The modulus and argument of a complex number therefore have the same definitions as the usual conversion from rectangular to polar coordinates (see Figure 0).

*Modulus*
$$|x+iy|=\sqrt{x^2+y^2} $$
*Argument*
$$
\arg(x+iy)=\left\{\eqalign{
&\arctan(y/x), \cr
&\arctan(y/x)+\pi, \cr
&\arctan(y/x)-\pi, \cr}\qquad
\eqalign{x\ge0\cr x<0, y\ge0 \cr x<0, y<0 \cr } \right.
$$
Recall that $\arctan$ is a $\pi$-periodic function, usually defined to
take values from $-\pi/2$ to $\pi/2$. On the other hand, the argument of a
complex number is a polar angle, which can range from $-\pi$ to $\pi$.
That is why we have the somewhat complicated expression for the argument.
If you are trying to compute an argument using a calculator, you can punch
in $\arctan(y/x)$ but then you are responsible for deciding if you need to
adjust by a factor of $\pm\pi$. On the other hand, many computer languages
now include a `atan2(x,y)`

function that computes the argument
of the complex number $x+iy$ directly.

Once you have gone from rectangular to polar coordinates, the natural thing is to go from polar to rectangular coordinates. Suppose we have a complex number $z$ with modulus $|z|$ and argument $\arg(z)$. Then the real and imaginary parts of $z$ can be determined using the formulas for conversion from polar to rectangular coordinates.

*Real Part:*
$\Re(z)=|z|\cos(\arg(z)) $

*Imaginary Part:*
$\Im(z)=|z|\sin(\arg(z))$

We are now ready for a sneaky trick, that turns out to be very useful. If we write $z=\Re(z)+i\Im(z)$ and use the preceding rules for the real and imaginary parts and do a little bit of algebra and remember what we learned about complex exponentials in the previous section and don't worry about having the word ``and'' too many times in the same sentence we get $$ \eqalign { z&=\Re(z)+i\Im(z) \cr &=|z|\cos(\arg(z))+i|z|\sin(\arg(z)) \cr &=|z|\bigl(\cos(\arg(z))+i\sin(\arg(z))\bigl) \cr &=|z|\bigl(e^{0+i\arg(z)}\bigl) \cr &=|z|e^{i\arg(z)} \cr } $$ This formula is called the polar form of the complex number $z$. In the next section we will see how converting complex numbers from rectangular form $x+iy$ into polar form $re^{i\theta}$ and back again can simplify trig identities. For now we will concentrate on how understanding the polar form helps us understand the geometric meaning of the basic operations of complex arithmetic. Addition and subtraction of complex numbers are just vector addition and subtraction like you studied in Calculus III with this geometric interpretation of complex numbers as vectors in the plane. Multiplication and division are more complicated.

To understand the multiplication of complex numbers geometrically, it is easiest to work in polar coordinates. Suppose we want to multiply $re^{i\theta}$ and $se^{i\phi}$. We compute $$ \eqalign { re^{i\theta}\times se^{i\phi}&=rse^{i\theta}e^{i\phi} \cr &=rse^{i(\theta+\phi)} \cr } $$ So the modulus of the product is the product of the moduli and the argument of the product is the sum of the arguments. In other words, to multiply two complex numbers you multiply the lengths of the corresponding vectors (in polar coordinates) and add the angles (in polar coordinates). Division is just the reverse of course. You divide the moduli and subtract the angles.

Let us check how this interpretation works for $i^2$. The complex number $i$ is represented by the vector $(0,1)$ which has modulus 1 and argument $\pi/2$, that is to say it is a vector of length 1 and it is at a right angle with the Real axis. When we multiply this number by itself, we multiply the length by itself and $1\times1=1$ so the product will also have modulus 1. We add the argument to itself, and twice a right angle is a straight angle, so the vector should point in the negative real direction. And the vector of length 1 in the negative real direction represents the number $-1$, so we do indeed have $i^2=-1$.

### Alternate Forms for Trigonometric Functions

#### Discussion

Quick, what does the graph of $3\cos(2x) + 4\sin(2x)$ look like? This is a difficult question for most people to answer. While you know what the graphs of trig functions look like, adding them usually requires plotting both graphs and then trying to add the heights. It turns out that using some trig identities you can rewrite $3\cos(2x) + 4\sin(2x)$ in the form $A\cos(\omega x-\phi)$ for the right choices of $A$, $\omega$ and $\phi$. This is just a cosine wave with amplitude $A$, circular frequency $\omega$, and phase shift $\phi$, so it is easier to visualize. But most students have difficulty remembering how the trig identities go, even when they are in trig, let alone several years later. Using complex polar coordinates, it is relatively easy to convert sums of cosine and sine functions to this nicer form.The key to rewriting $3\cos(2x) + 4\sin(2x)$ using complex numbers is the following pair of facts: $\Re\left(3e^{2xi}\right)=3\cos(2x)$ and $\Re\left(-4ie^{2xi}\right)=4\sin(2x)$. So $$3\cos(2x)+4\sin(2x)=\Re\left(3e^{2xi}-4ie^{2xi}\right) =\Re\left((3-4i)e^{2xi}\right)$$ We now rewrite $3-4i$ in polar form (using the results from the previous section on complex variables). $$r=|3-4i|=\sqrt{3^2+4^2}=5\qquad\theta=\arctan(-4/3)\approx-.927$$ So $3 - 4i \approx 5e^{-.927i}$. Now we plug this into our representation for $3\cos(2x) + 4\sin(2x)$ and obtain $$ \begin{align} 3\cos(2x)+4\sin(2x)&=\Re\left((3-4i)e^{2xi}\right) \\ &\approx\Re\left(5e^{-.927i}e^{2xi}\right) \\ &=\Re\left(5e^{(2x-.927)i}\right) \\ &=5\cos(2x-.927) \end{align} $$ We are now ready to write a paradigm for this procedure.

#### Paradigm

Write $3\cos(3x)-\sin(3x)$ in the form $A\cos(\omega x+\phi)$
*Step 1:* Write the function as the $\Re$ part of a complex number
times a complex exponential.

The trick here is to make the real part of the complex number the
coefficient of the $\cos$ term and the imaginary part of the complex
number the *negative* of the coefficient of the $\sin$ term.
$\cos(3x)=\Re[e^{i3x}]$ and $\sin(3x)=\Re[-ie^{i3x}]$ so
$3\cos(3x)-\sin(3x)=\Re[3e^{i3x}-(-ie^{i3x})]=\Re[(3+i)e^{i3x}]$

*Step 2:* Write the coefficient of the complex expression in polar
form.

This is just the conversion to polar coordinates. $|3+i|=\sqrt{3^2+1^2}=\sqrt{10}$. $\arg(3+i)=\arctan(1/3)\approx.322$. So $3+i\approx\sqrt{10}e^{.322i}$. (Don't forget in the exercises that you sometimes have to adjust the $\arctan$ by $\pm\pi$ to get the correct $\arg$.)

*Step 3:* Multiply out the coefficient in polar form with the complex
exponential.

$(3+i)e^{i3x}\approx\sqrt{10}e^{.322i}e^{i3x}=\sqrt{10}e^{i(3x+.322)}$

*Step 4:* Take $\Re$ part.

$3\cos(3x)-\sin(3x)=\Re[(3+i)e^{i3x}]\approx\Re[\sqrt{10}e^{i(3x+.322)}] =\sqrt{10}\cos\bigl(3x+.322\bigl)$.

Of course, you can go the other way too. People don't normally worry as much about this because the trig identities used in changing expressions of the form $A\cos(\omega x-\phi)$ to $a\cos(\omega x)+b\sin(\omega x)$ are just the addition of angle formulas and those are more familiar to most people.

EXAMPLE: Write $2\cos(2.7x-1.3)$ in the form $a\cos(2.7x)+b\sin(2.7x)$.

One way is just to reverse the process of the paradigm as follows $$ \eqalign { 2\cos(2.7x-1.3)&=\Re\bigl(2e^{(2.7x-1.3)i}\bigl) \cr &=\Re\bigl(2e^{-1.3i}e^{2.7ix}\bigl) \cr &=\Re\bigl((2\cos(-1.3)+2i\sin(-1.3))(\cos(2.7x)+i\sin(2.7x))\bigl) \cr &\approx\Re\bigl((0.5350 - 1.9271i)(\cos(2.7x)+i\sin(2.7x))\bigl) \cr &=\Re\bigl(0.5350\cos(2.7x) \cr &\qquad\qquad+1.9271\sin(2.7x)+0.5350i\sin(2.7x) - 1.9271i\cos(2.7x)\bigl) \cr &=0.5350\cos(2.7x)+1.9271\sin(2.7x) \cr } $$ Another way is to use the trig identities $$ \eqalign { \cos(\theta\pm\phi)&=\cos(\theta)\cos(\phi)\mp\sin(\theta)\sin(\phi) \cr \sin(\theta\pm\phi)&=\sin(\theta)\cos(\phi)\pm\cos(\theta)\sin(\phi) \cr } $$ which gives the following $$ \eqalign { 2\cos(2.7x-1.3)&=2\cos(2.7x)\cos(1.3)+2\sin(2.7x)\sin(1.3) \cr &\approx 0.5350\cos(2.7x)+1.9271\sin(2.7x) \cr } $$ As you can see, the trig identities work out quicker this time, as long as you keep straight where you have $\pm$ and where you have $\mp$ in the identities.

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©2010, 2014 Andrew G. Bennett