## 305 -3. Complex numbers.

Mathematicians first approached complex numbers cautiously. Although it was clear that they were useful in solving certain problems at least formally (for example, they are needed to even make sense of the formulas we found in the previous lectures) what was not clear was that they made sense. Perhaps indiscriminate use of them would lead to contradictions.

Gauß solved this problem by realizing that one can define ${\mathbb C}$ and its operations in terms of ${\mathbb R}$ and its operations. As long as we are willing to accept that ${\mathbb R}$ makes sense, then no contradictions will come up from the use of complex numbers.

Definition. The set ${\mathbb C}$ of complex numbers is simply ${\mathbb R}\times{\mathbb R}.$ We define addition and multiplication as follows:

• If $z=(a,b)$ and $w=(c,d),$ then addition is just coordinatewise addition, $z+w=(a+c,b+d).$
• With $z,w$ as above, we set $z\times w=z\cdot w=zw=(ac-bd,ad+bc).$

One easily checks that ${\mathbb R}$ can be identified with the $x$-axis $\{(a,0)\in{\mathbb C}:a\in{\mathbb R}\}:$ One identifies $a\in{\mathbb R}$ with $(a,0)\in{\mathbb C}.$ One easily verifies that $(a+b,0)=(a,0)+(b,0)$ and $(a,0)(b,0)=(ab,0).$ We say that the identification is a homomorphism.

Notice that complex addition is just addition of vectors, which has a clear geometric interpretation. Complex multiplication also has a geometric interpretation, as we will see below, but it is more subtle.

Of course, the point of introducing complex numbers is to make sense of square roots of negative numbers. Define $i=(0,1).$ Then one easily checks that $bi=(0,b)$ (i.e., $(b,0)(0,1)=(0,b)$). Due to these correspondences, one simply writes $a+bi$ instead of $(a,b).$

Theorem. ${\mathbb C}$ with the operations $+,\times$ as defined above is a field. This means that the following properties hold:

1. (Commutativity of addition). For all complex numbers $z,w,$ we have that $z+w=w+z.$
2. (Commutativity of multiplication). Similarly, $zw=wz$ for all complex numbers $z,w.$
3. (Associativity of addition). For all complex numbers $z,w,v,$ we have that $z+(w+v)=(z+w)+v.$
4. (Associativity of multiplication). Similarly, $z(wv)=(zw)v$ for all complex numbers $z,w,v.$
5. (Distributivity). For all complex numbers $z,v,w,$ one has that $(z+v)w=zw+vw.$
6. (Additive identity). There is a complex number ${\hat 0}$ such that $z+{\hat 0}=z$ for all complex numbers $z.$
7. (Multiplicative identity). There is a complex number $\hat 1$ such that $z\hat 1=z$ for all complex numbers $z.$
8. (Additive inverses). With $\hat 0$ as above, for any complex number $z$ there is a complex number $-z$ such that $z+(-z)=\hat 0.$
9. (Multiplicative inverses). For any $z\ne{\hat 0}$ there is a complex number $z^{-1}$ such that $zz^{-1}={\hat 1}.$ Here, ${\hat 0}$ and ${\hat 1}$ are as above.

The proof of the theorem is more or less immediate. For example, one easily checks that we can take ${\hat 0}=(0,0)=0,$ ${\hat 1}=(1,0)=1,$ and $-z=-1\times z.$ (In fact, one checks that these are the only values that work.) Only item 9. requires some care. For this, assume that $z=a+bi.$ Then $z^{-1}=c+di$ if and only if $(ac-bd)+(ad+bc)i=1,$ i.e., if and only if the system

$\displaystyle\left\{\begin{array}{rcl}ac-bd&=&1\\ ad+bc&=&0\end{array}\right.$

in the two unknowns $c,d$ has a solution. It is easy to verify that, indeed, the system has a unique solution given by $c=a/(a^2+b^2)$ and $d=-b/(a^2+b^2).$ In other words, it is indeed the case that whenever $z\ne0$ then there is an inverse $z^{-1}$ given by $\displaystyle z^{-1}=\frac a{a^2+b^2}-\frac b{a^2+b^2}i.$

There is another way of proceeding here.

Definition. The complex conjugate of $z=a+bi$ is the complex number $\bar z=a-bi.$

Note that $z\bar z=a^2+b^2,$ so $z\bar z$ is a nonnegative complex number, and it is always positive, except when $z=0.$

Definition. The norm, or magnitude, or size, or modulus, or absolute value, of the complex number $z$ is $\sqrt{z\bar z}.$

Notice that if $z$ is real, i.e., if $z=(a,0)$ for some $a,$ then its absolute value as a real number coincides with the absolute value as a complex number.

Suppose now that $z\ne0.$ Then, if the notation is to make any sense, we must have $\displaystyle z^{-1}=\frac1z=\frac{\bar z}{z\bar z}=\frac1{|z|^2}\bar z,$ which coincides with the formula found before.

Definition. If $w\ne0$ then $z/w=zw^{-1},$ i.e., $z/w=z\bar w/|w|^2.$

Conjugation is also a homomorphism:

• $\overline{zw}=\bar z\bar w,$ and
• $\overline{z+w}=\bar z +\bar w.$

Also, notice that $\overline{\bar z}=z.$

Definition. The argument of the complex number $z\ne0$ is the oriented angle between the $x$-axis and the vector $z.$

The argument is only defined up to integer multiples of $2\pi,$ i.e., if $\alpha$ is an argument of $z,$ then so are $\alpha+2\pi,$ $\alpha-6\pi,$ etc. Years ago, people would say that the argument is a multivalued function or some such nonsense. Sometimes it is convenient to act as if the argument of 0 is defined. If so, we will simply say that any real $\alpha$ is an argument of $0.$

An easy geometric argument shows the following:

Lemma. For any nonzero complex number $z$, we have $z=|z|(\cos\alpha+i\sin\alpha),$ for $\alpha$ any argument of $z.$ If $z=0,$ the equality holds for any $\alpha.$ The modulus of any number of the form $\cos\alpha+i\sin\alpha$ is 1.

This representation is rather useful, as one easily verifies that

$zw=|z||w|(\cos(\alpha+\beta)+i\sin(\alpha+\beta)),$

where $\alpha$ is any argument of $z$ and $\beta$ is any argument of $w.$ Thus, we have:

Lemma. The modulus of a product is the product of the moduli, and the argument of a product is the sum of the arguments (mod $2\pi$).

This means that complex multiplication corresponds to a dilation followed by a rotation.

A particularly important particular case of the above is the following result:

Lemma. (De Moivre). For any positive integer $n$  and any complex number $z=|z|(\cos\alpha+i\sin\alpha),$ we have $z^n=|z|^n(\cos(n\alpha)+i\sin(n\alpha)).$

Proof. There are several ways of presenting this result. Let’s try an argument by induction. First, the result is true for $n=1$ and there is nothing to show then. If the result holds for $n,$ then $z^{n+1}=z^nz=|z|^n|z|\cos((n\alpha+\alpha)+i\sin(n\alpha+\alpha),$ where we are using the previous lemma. This is clearly equivalent to the case $n+1$ of the statement, and we are done. ${\sf QED}$

We can define $z^{-n}$ as $(z^{-1})^n,$ and it is equally easy to check that De Moivre’s formula also holds for exponents that are negative integers.

The reason why one cares about De Moivre’s formula is that it allows us to find $n$-th roots of any complex number.

Definition. Let $n$ be a positive integer. The complex number $w$ is an $n$-th root of $z$ iff $w^n=z.$

Let’s say that $z=r(cos\alpha+i\sin\alpha)$ and that $w=s(\cos\beta+i\sin\beta)$ where $r=|z|,$ $s=|w|,$ $\alpha$ is an argument of $z,$ and $\beta$ is an argument of $w.$ That $w$ is an $n$-th root of $z$ means that

1. $s^n=r.$ Since $s,r$ are nonnegative real numbers, there is only one possible value of $s,$ namely what one usually denotes $\root n\of r.$
2. Either $z=0,$ in which case $w=0$ as well, or else $n\beta=\alpha,$ where the equality is of course up to an integer multiple of $2\pi.$

Condition 2 means that $\displaystyle \beta=\frac{\alpha+2k\pi}n$ for some integer $k.$ Notice that if $k_1-k_2$ is a multiple of $n,$ then the $\beta$ corresponding to $k_1$ and the $\beta$ corresponding to $k_2$ are the same (modulo $2\pi$). On the other hand, if $k_1-k_2$ is not a multiple of $n,$ then the values of $\beta$ we obtain are different, so the values of $w$ they correspond to are also different.

This means that there are exactly $n$ distinct complex $n$-th roots of any nonzero complex number $z.$

Example. Take $z=1,$ so $\alpha=0,$ and let $n=3.$ the cubic roots of $1$ will be the numbers $\displaystyle \cos\left(\frac{2k\pi}3\right)+i\sin\left(\frac{2k\pi}3\right)$ for $k=0,1,2.$ These are, respectively, the numbers $1,$ $\displaystyle\frac{-1}2+\frac{\sqrt3}2i,$ and $\displaystyle\frac{-1}2-\frac{\sqrt3}2i.$ Of course, these numbers coincide with the numbers we obtain when we solve the cubic equation $x^3-1=0.$