6.2 Geometrical Representation

Since $\mbox{{\bf C}}=\mbox{{\bf R}}\times\mbox{{\bf R}}$, we can identify complex numbers with points in a plane.

Then $\mbox{{\bf R}}$ is identified with the $x$-axis, and points on the $y$-axis are of the form $iy$ where $y$ is real. I will call the $x$-axis the real axis, and I'll call the $y$-axis the imaginary axis. If $z\in\mbox{{\bf C}}$, then $z^*$ represents the result of reflecting $z$ about the real axis. Also $-z$ represents the result of reflecting $z$ through the origin.

If $z=(a,b)$ and $w=(c,d)$ are two points in $\mbox{{\bf C}}$, and $u=(c,b)$, then $z,u,w$ are the vertices of a right triangle having legs of length $\vert c-a\vert$, and $\vert d-b\vert$. By the Pythagorean theorem, the distance from $w$ to $z$ is $\sqrt{(c-a)^2+(d-b)^2}$. Also,

$$\begin{eqnarray*} \vert w-z\vert&=&\vert(c+id)-(a+ib)\vert \\ &=&\vert(c-a)+i(d... ...{(c-a)^2+(d-b)^2} \\ &=&\mbox{ distance from } w \mbox{ to } z, \end{eqnarray*}$$

and in particular, for $z=0$,

$$\vert w\vert=\mbox{ distance from } w \mbox{ to }0.$$

Claim: If $z,w\in\mbox{{\bf C}}$, then $z+w$ is the fourth vertex of the parallelogram having consecutive vertices $z,0,w$.

To make this look like a geometry proof, I'll denote points by upper case letters, and let $AB$ denote the distance from $A$ to $B$. Let $O=0$, $W=w$, $Z=z$, $S=z+w$. Then

$$ZS=\vert(z+w)-z\vert=\vert w\vert=\vert w-0\vert=OW$$

$$WS=\vert(z+w)-w\vert=\vert z\vert=\vert z-0\vert=OZ.$$

Hence, since the quadrilateral $OWSZ$ has opposite sides equal, it is a parallelogram.

We can now give a geometrical interpretation for the triangle inequality (which motivates its name). In the figure above,

$$\vert z+w\vert\leq \vert z\vert+\vert w\vert$$

says

$$OS\leq OZ+ZS;$$

i.e, the sum of two sides of a triangle is greater than or equal to the third side. This is proposition 20 of book 1 of Euclid [19] `` In any triangle, two sides taken together in any manner are greater than the remaining one." (Euclid did not consider triangles in which all three vertices lie on a line.)

It was the habit of the Epicureans, says Proclus ...to ridicule this theorem as being evident even to an ass, and requiring no proof, and their allegation that the theorem was `` known" $(\gamma \nu \acute{\omega} \rho \iota \mu o\nu)$ even to an ass was based on the fact that, if fodder is placed at one angular point and the ass at another, he does not, in order to get his food, traverse the two sides of the triangle but only the one side separating them [19, vol. I page 287].

6.8   Definition (Circle, disc.) Let $\alpha\in\mbox{{\bf C}}, r\in\mbox{${\mbox{{\bf R}}}^{+}$}$. The circle with center $\alpha$ and radius $r$ is $\mbox{{\bf C}}$

$$\begin{eqnarray*} C(\alpha,r)&=&\{z\in\mbox{{\bf C}}\colon\vert z-\alpha\vert=r\} \\ &=&\mbox{ set of points whose distance from } a\mbox{ is } r. \end{eqnarray*}$$

The open disc with center $\alpha$ and radius $r$ is

$$D(\alpha,r)=\{z\in\mbox{{\bf C}}\colon \vert z-\alpha\vert<r\} \glossary{$D(\alpha,r)$, open disc},$$

and the closed disc with center $\alpha$ and radius $r$ is

$$\bar D(\alpha,r)=\{z\in\mbox{{\bf C}}\colon \vert z-\alpha\vert\leq r\}. \glossary{$\bar D(\alpha,r)$, closed disc}$$

$C(0,1)$ is called the unit circle, and $D(0,1)$ is called the unit disc. A complex number $z$ is in the unit circle if and only if $\vert z\vert=1$.

6.9   Warning. The word ``circle'' is sometimes used to mean ``disc'', although the word ``disc'' is never used to mean ``circle''. When you see the word ``circle'' used in a mathematical statement, you should determine from the context which of the two words is meant. For example, in the statement ``the area of the unit circle is $\pi$'', the word ``circle'' means ``disc'', since the unit circle is, in fact, a zero-area set. In these notes the word ``circle'' always means ``circle'' except in warning 5.2.

6.10   Theorem. The product of two numbers in the unit circle is in the unit circle.

Proof: Let $\alpha,\beta\in C(0,1)$; i.e., $\vert\alpha\vert=\vert\beta\vert=1$. Then $\vert\alpha\beta\vert=\vert\alpha\vert\vert\beta\vert=1\cdot 1=1$, so $\alpha\beta\in C(0,1)$. $\mid\!\mid\!\mid$

We can also give a geometrical interpretation to the product of two complex numbers. Let $\alpha=A$ and $\beta=B$ be complex numbers and let $C=\alpha\beta$. Let $O=0$ and let $I=1$.

Then $\triangle OIA$ is similar to $\triangle OBC$. The proof consists in showing that

$${{OI}\over {OB}}={{IA}\over {BC}}={{OA}\over {OC}}.$$ (6.11)

6.12   Exercise. Prove the equalities listed in (6.11). Assume $\alpha\notin \{0,1\}$ and $\beta\neq 0$.

From the similarity of $\triangle OIA$ and $\triangle OBC$, we have $\angle IOA=\angle BOC$. In particular, if we take $\alpha=a\in\mbox{${\mbox{{\bf R}}}^{+}$}$, we get the picture

where

angle($1$-$0$-$a$)$=$ angle ($\beta$-$0$-$a\beta$),

which indicates that $a\beta$ lies on the line through $0$ that passes through $\beta$. Also

$$\vert a\beta\vert=\vert a\vert\;\vert\beta\vert=a\vert\beta\vert$$

so the length of $a\beta$ is obtained by multiplying the length of $\beta$ by $a$.

The figure below shows the powers of a complex number a.

Powers of a: $I=1$, $A=$ a, $B=$a$^2$, $C =$ a$^3$, $D =$ a$^4$.

In each case the four triangles $\triangle IOA$, $\triangle AOB$, $\triangle BOC$, and $\triangle COD$ are all similar. In the third figure, where a is in the unit circle, the triangles $\triangle IOA, \triangle AOB,\triangle BOC$ and $\triangle COD$ are in fact congruent.

If $\alpha,\beta$ are points on the unit circle, then

angle($\beta$-$0$-$\alpha\beta$) $=$ angle($1$-$0$-$\alpha$),

which indicates that $\alpha\beta$ is the point in the unit circle such that

angle($1$-$0$-$\alpha\beta$) $=$ angle($1$-$0$-$\alpha$) $+$ angle($1$-$0$-$\beta$).

From trigonometry, you know that the point on the unit circle making angle $\theta$ with the segment $OI$ is $(\cos\theta,\sin\theta)=\cos\theta+i\sin\theta$.

The previous geometrical argument suggests that

$$(\cos\theta+i\sin\theta)(\cos\phi+i\sin\phi)=\left(\cos(\theta+\phi) +i\sin(\theta+\phi)\right).$$ (6.13)

6.14   Exercise. Using standard trigonometric identities, prove (6.13), and show that $(\cos\theta + i\sin\theta)^{-1} = \cos\theta - i\sin\theta \mbox{ for all }\theta\in\mbox{{\bf R}}$.

6.15   Exercise. Let $\theta\in\mbox{{\bf R}}$. Let $n\in\mbox{{\bf N}}$. Prove that

$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta).$$ (6.16)

Then show that formula (6.16) is in fact valid for all $n\in\mbox{{\bf Z}}$. (Formula (6.16) is called De Moivre's Formula.)