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We'll begin our discussion of complex numbers with a simple picture
of the real number line. Zero in the centre, $1,2,3,\dots$ to the
right, $-1,-2,-3,\dots$ to the left. Think of each real number not
as a point on the line, but as the corresponding vector, as the arrow
emanating from $0$ and ending at, say, $2$. Then real number addition
is vector addition; to add two numbers, say, $1$ and $2$, place
the tail of one at the tip of the other. The sum is the vector ending
at the final tip, $1+2=3$. This works for negative numbers too; $2+(-3)=-1$,
$2+(-2)=0$. Multiplication is rescaling; to multiply a vector/number
by $3$, you scale up its length by a factor of $3$, and leave the
direction the same; $2\rightarrow6$, and $-2\rightarrow-6$. To multiply
by a negative number $v$; you still rescale by the length of $v$,
but you also reverse direction, rotating around $180^{\circ}$, just
like the negative vectors are all $180^{\circ}$ away from their positive
counterparts. So $-3\cdot-2=6$.

We can extend the law of addition to all vectors in the plane containing
the real ones. Given two vectors of whatever lengths in whatever directions,
you add them by putting tail to tip and completing the triangle. This
plus this equals that, this plus this equals that.

To multiply a vector by a real number (a real vector), we use the
same rule as before; if the real number is positive, pointing horizontally
to the right, then you just rescale its target by its length, this
times $2$ is that. If the real number is negative, pointing horizontally
to the left, it rescales its target by its length, and then rotates
it by $180^{\circ}$. To multiply by an arbitrary vector in the plane,
say this one, you rescale by its length, which we call its \textbf{magnitude},
and rotate counterclockwise by the angle it forms with the positive
reals, with the horizontal, which we call its \textbf{argument}. So
multiplying by this doubles lengths, and rotates $45^{\circ}$ counterclockwise;

$(\nearrow)\cdot(\uparrow)=(\nwarrow)$

Multiplying by this halves lengths and rotates $300^{\circ}$ counterclockwise
(or $60^{\circ}$ clockwise), taking this to that, and this to that.
Mulitplying by $1$, the unit vector to the right, fixes the length,
and rotates by $0^{\circ}$, ie, leaves everything the same. So $1$
is the multiplicative identity. The unit vector pointing north is
notable, multiplying by it just rotates things $90^{\circ}$. In particular,
$(\uparrow)\cdot(\uparrow)=(\leftarrow)$, which is $-1$. We usually
write $i$ for $\uparrow$, and we've just shown that $i^{2}=-1$.

## Arrow Arithmetic

Complex numbers are vectors in the plane, with addition given by vector
addition, and multiplication given by dilation and rotation. Real
numbers form the horizontal axis, and imaginary numbers form the vertical
axis. Why does it make sense to call these arrows "numbers"?
Because they satisfy all the basic rules of arithmetic:

- Addition is commutative.
    - the two orders form a paralellogram, with the sum on the diagonal, $z+w=w+z$
- Addition is associative.
    - $v+(z+w)=(v+z)+w$ since both compute the straight-line path from $0$ to the end of the $\overset{w}{\rightarrow}\overset{z}{\rightarrow}\overset{v}{\rightarrow}$ trail.
- Multiplication is commutative.
    - Multiplying $(r_{1},\theta_{1})$ by $(r_{2},\theta_{2})$ rescales by $r_{2}$ and rotates by $\theta_{2}$, giving $(r_{1}r_{2},\theta_{1}+\theta_{2})$, which is manifestly symmetric. Put another way, multiplication of complex numbers multiplies magnitudes and adds angles/arguments.
- Multiplication is associative.
    - use formula in polar coordinates
- Multiplication distributes over addition.
    - Take any two complex numbers $v,w$ and add them, making a parallelogram. To find $z(v+w)$, rotate and scale our sum. But that's the same as rotating and scaling each of $v$ and $w$ individually, to $zv$ and $zw$, then adding them together: $zv+zw=z(v+w)$. The distributive law for complex numbers is the fact that a rotated, rescaled paralellogram is still a parallelogram, or, more basically, that dilation and rotation of the plane preserve angles.


## Coordinates

We can't be constantly drawing arrows in the plane; to do anything
quantitative, we need to introduce coordinates. Rectangular coordinates
are easy, each vector has an $x$ and $y$-coordinate, given by its
projections onto the real and imaginary axis. If we use column vector
notation, we have
$z = \left(\begin{array}{c}
a\\
b\end{array}\right)
 = a\left(\begin{array}{c}
1\\
0\end{array}\right)+b\left(\begin{array}{c}
0\\
1\end{array}\right)
 = a\cdot1+b\cdot i
 = a+bi$
Put another way, we are using $1$ and $i$ as basis vectors. For
example,  
$\left(\begin{array}{c}
3\\
2\end{array}\right)=3(\rightarrow)+2(\uparrow)=3+2i$  
and  
$\overset{2}{\nwarrow}=\left(\begin{array}{c}
-\sqrt{2}\\
\sqrt{2}\end{array}\right)=-\sqrt{2}(\rightarrow)+\sqrt{2}(\uparrow)=-\sqrt{2}+\sqrt{2}i$


Polar coordinates represent each complex number by its magnitude and argument, its radius and angle (in radians). So 
$i=\uparrow=(1,\pi/2)$ and $\overset{2}{\nwarrow}=(2,3\pi/4)$.

The familiar angle addition formula from trigonometry now follows
from the distributive law applied to unit vectors.  
    $\cos(\theta+\phi)+i\sin(\theta+\phi)=(\cos\theta+i\sin\theta)(\cos\phi+i\sin\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi+i(\cos\theta\sin\phi+\sin\theta\cos\phi)$


In the workshop session, you'll use a similar trick to compute triple
and quadruple angle formulas, for $\sin$ and $\cos$, deriving De
Moivre's theorem.

## Functions

The most basic transformations of the complex plane are given by multipling
by some fixed complex number $\rho=r(\cos\theta+i\sin\theta)=a+bi$,
which, as we saw before, is a dilation by $r$ plus a rotation by
$\theta$. The distributive law says precisely that this is a linear
transformation of the plane, viewed as a two-dimensional vector space.
And linear maps are given in rectangular coordinates by matrices:
$\left(\begin{array}{c}
x\\
y\end{array}\right)\mapsto\left(\begin{array}{cc}
a & c\\
b & d\end{array}\right)\left(\begin{array}{c}
x\\
y\end{array}\right)=\left(\begin{array}{c}
ax+cy\\
bx+dy\end{array}\right).$
What is the matrix corresponding to multiplication by $\rho$? Well,
the first column is the image of $\left(\begin{array}{c}
1\\
0\end{array}\right)$, ie, of $1$, which just $v=\left(\begin{array}{c}
a\\
b\end{array}\right)$ itself, and the second column is the image of $i=\left(\begin{array}{c}
0\\
1\end{array}\right)$, which is $\rho$ rotated by $90^{\circ}$, which has coordinates
$\left(\begin{array}{c}
-b\\
a\end{array}\right)$. So complex number $a+bi$ is identified with the matrix $
\left(\begin{array}{cc}
a & -b\\
b & a\end{array}\right).$
As a sanity check, note that $i$ corresponds to $\left(\begin{array}{cc}
0 & -1\\
1 & 0\end{array}\right)$, which squares to $\left(\begin{array}{cc}
-1 & 0\\
0 & -1\end{array}\right)\sim-1$.

Another natural transformation of the complex plane is given by squaring,
sending $z\mapsto z^{2}$. This squares the length of each
vector, and doubles its angle. PICTURE. What does $z\mapsto z^{n}$
look like? PICTURE.

Another key example is the exponential map. Recall the power series
for $e^{z}$:

$e^{z}=1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\frac{z^{5}}{5!}+\cdots$

We can raise complex numbers to powers, divide by the real denominators,
and add them up just fine, so we can exponentiate complex values of
$z$. We know what happens to real values, what happens to pure imaginary
ones? Let $y\in\mathbb{R}$. Then  
$\begin{array}{}ee^{iy} & = & 1+iy+\frac{(iy)^{2}}{2!}+\frac{(iy)^{3}}{3!}+\frac{(iy)^{4}}{4!}+\frac{(iy)^{5}}{5!}+\cdots\\
 & = & 1+iy-\frac{y^{2}}{2!}-i\frac{y^{3}}{3!}+\frac{y^{4}}{4!}+i\frac{y^{5}}{5!}+\cdots\\
 & = & (1-\frac{y^{2}}{2!}+\frac{y^{4}}{4!}+\cdots)+i(y-\frac{y^{3}}{3!}+\frac{y^{5}}{5!}-\cdots)\\
 & = & \cos y+i\sin y\end{array}$


Substituting $y=\pi$, we recover Euler's famous identity,
$e^{i\pi}=-1.$
 Given a real argument $y$, the $e^{iy}$ gives the unit vector with
that argument. Given an arbitrary complex number $z=x+iy$, it's exponnt
$e^{z}$ is the complex number with magnitude $e^{x}$ and angle $y$
radians.

What does $z\mapsto e^{z}$ look like? Periodic in the $i$
direction with period $2\pi$. Takes a horizontal strip and wraps
it around, forming an annulus. PICTURE.

Let's take another look at these functions. What do you notice?

Images of grid lines still intersect orthogonally! In fact, all angles
are preserved: these two curves go to those two curves, and the angle
between them (actually, the angles between their tangent vectors at
the intersection point) stays the same.


## Conformality, Holomorphicity

A map $\mathbb{C}\rightarrow\mathbb{C}$ is **conformal** if
it preserves oriented angles. We (meaning me *and* you) will
show that polynomials, exponentials, etc are conformal (almost everywhere).

Consider a smooth map $f$ from the plane to itself; it takes a smooth
curve $\gamma$ through $z$ to a smooth curve $f\circ\gamma$ through
$f(z)$. What happens to the tangent of $\gamma$ at $z$? Given by
the derivative $df(z)$, a linear map taking vectors based at $z$
to vectors based at $f(z)$. If we use rectangular coordinates  
    $z\mapsto f(z)$  
    $x+iy \mapsto u(x,y)+iv(x,y)$  
    $\left(\begin{array}{c}
x\\
y\end{array}\right)\mapsto\left(\begin{array}{c}
u(x,y)\\
v(x,y)\end{array}\right)$  
then the derivative is  
    $df=\left(\begin{array}{cc}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right).$  
If $f$ is conformal, then this matrix had better take the (orthogonal)
standard basis to orthogonal vectors; the little square becomes a
little rectangle. Since the diagonal of the square bisects the right
angle, and $df$ is linear, its image must bisect too, that is, the
little square must become a little *square*. Numerically, this
means that $\left(\begin{array}{c}
\frac{\partial u}{\partial y}\\
\frac{\partial u}{\partial y}\end{array}\right)$ is $\left(\begin{array}{c}
\frac{\partial u}{\partial x}\\
\frac{\partial v}{\partial x}\end{array}\right)$ rotated by $\pi/2$. If we write \[
\left(\begin{array}{c}
a\\
b\end{array}\right)=\left(\begin{array}{c}
\frac{\partial u}{\partial x}\\
\frac{\partial v}{\partial x}\end{array}\right)$
, then $
\left(\begin{array}{c}
\frac{\partial u}{\partial y}\\
\frac{\partial u}{\partial y}\end{array}\right)=\left(\begin{array}{c}
-b\\
a\end{array}\right).$
Put another way, the linear transformation $df(z)$ has the form 
$\left(\begin{array}{cc}
a & -b\\
b & a\end{array}\right),$
ie, it looks just like multiplication by the complex number $a+bi$.
The function $f$ is conformal if its derivative acts like a nonzero
complex number. Analytically, this condition is given by the following
differential equations, called the **Cauchy-Riemann equations**:  
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\mbox{ and }\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$  
A complex function $f=u+iv$ is said to be **holomorphic** if
$f$ satisfies the CR. We've shown that conformal $\Longrightarrow$
holomorphic. Holomorphic functions are slightly more general, as the
Jacobian can vanish; $df$ can be the complex number $0$. This is
what happens at the origin for $f=z^{n}$. We can now check that $e^{z}$
is conformal, and that $z^{n}$ is holomorphic, so conformal away
from $0$.  
**Example:** $z^{2}$ is holomorphic. $z^{2}=(x+iy)^{2}=x^{2}-y^{2}+i(2xy).$ So
$u=x^{2}-y^{2}$ and $v=2xy$.


This class is, essentially, the study of holomorphic functions, which
includes all angle-preserving transformations of the plane. The lesson
here is that complex analysis is geometry; complex numbers are dilations
and rotations, and holomorphic functions (defined in terms of differential
equations) are (almost everywhere) angle-preserving (if they're not
constant).

Why study complex numbers, holomorphic functions, and conformal maps?
Because of the beautiful and deep structure to which they lead.

Mathematically,

- fundamental theorem of algebra
- difficult integrals ($\int_{-\infty}^{\infty}\frac{1}{(x^{2}+1)^{2}}dx=\frac{\pi}{2}$)
- theory of surfaces
- hyperbolic geometry
Applicationally,
- conformal maps preserve maxwell's equations (in 2D) and incompressible,
irrotational fluid flow.
- fourier transform, and its generalization, the laplace transform for
understanding dynamical signals and systems.

We'll do all that and more, in the next six weeks. Hold on, dig in,
and enjoy!