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 ztimes w=zcdot w=zw=(ac-bd,ad+bc).

One easily checks that {mathbb R} can be identified with the x-axis {(a,0)in{mathbb C}:ain{mathbb R}}: One identifies ain{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 zhat 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 zne{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=-1times 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

displaystyleleft{begin{array}{rcl}ac-bd&=&1\ ad+bc&=&0end{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 zne0 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 zbar z=a^2+b^2, so zbar 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{zbar 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 zne0. Then, if the notation is to make any sense, we must have displaystyle z^{-1}=frac1z=frac{bar z}{zbar z}=frac1{|z|^2}bar z, which coincides with the formula found before.

Definition. If wne0 then z/w=zw^{-1}, i.e., z/w=zbar w/|w|^2.

Conjugation is also a homomorphism:

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

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

Definition. The argument of the complex number zne0 is the oriented angle between the x-axis and the vector z. 

The argument is only defined up to integer multiples of 2pi, i.e., if alpha is an argument of z, then so are alpha+2pi, alpha-6pi, 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|(cosalpha+isinalpha), for alpha any argument of z. If z=0, the equality holds for any alpha. The modulus of any number of the form cosalpha+isinalpha is 1.

This representation is rather useful, as one easily verifies that

zw=|z||w|(cos(alpha+beta)+isin(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 2pi). 

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|(cosalpha+isinalpha), we have z^n=|z|^n(cos(nalpha)+isin(nalpha)).

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((nalpha+alpha)+isin(nalpha+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(cosalpha+isinalpha) and that w=s(cosbeta+isinbeta) 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 nof r.
  2. Either z=0, in which case w=0 as well, or else nbeta=alpha, where the equality is of course up to an integer multiple of 2pi. 

Condition 2 means that displaystyle beta=frac{alpha+2kpi}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 2pi). 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 cosleft(frac{2kpi}3right)+isinleft(frac{2kpi}3right) for k=0,1,2. These are, respectively, the numbers 1, displaystylefrac{-1}2+frac{sqrt3}2i, and displaystylefrac{-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.

Advertisements

One Response to 305 -3. Complex numbers.

  1. […] 1–9 of the Theorem from last lecture hold with elements of in the place of complex numbers, in the place of and in the place […]

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: