Algebra/Complex Numbers


 * See also: High School Mathematics Extensions, Complex numbers, Complex numbers at Wikipedia

Complex numbers are the extension of the real numbers, i.e., the number line, into a number plane. They allow us to turn the rules of plane geometry into arithmetic. Complex numbers have fundamental importance in describing the laws of the universe at the subatomic level, including the propagation of light and quantum mechanics. They also have practical uses in many fields, including signal processing and electrical engineering.

Introduction
Currently, we are able to solve many different kinds of equations for $$x$$, such as $$x+7=12$$ , or $$3x=4$$ , or $$x+10=0$$. In each of these cases the solution for $$x$$ is a real number: respectively 5, 4/3, and -10.

However, there is no real number x that satisfies the equation $$x^2=-1$$, since the square of any real number is nonnegative.

Conceptually it would be nice to have some kind of number to be the solution of $$x^2=-1$$. This "number" would not be a real number, however, and we refer to such a number as an imaginary number.

Now, is this really the reason?

Well - Definitely not!

This mistake occurs in many teaching books from the attempts to "solve problems by force", as we could explain psychologically. This has nothing to do with reality, and gives the false feeling that mathematicians are "Cranks" full of to much spare time on their hands with nothing to do.

The reason, be surprised, has to do with a problem called Cubic functions.

We then extend the real number system to accommodate this special number. It turns out that there will be two imaginary solutions of the equation $$x^2=-1$$. One of them will be called $$i$$ and, following the normal rules for arithmetic, the other solution is $$-i$$. We may be inclined to say that $$\sqrt{-1}=i$$. That would, however, be incorrect solely because in words this says that "the square root of -1 is $$i$$", but there is no basis for preferring $$i$$ over $$-i$$ (or vice versa) as the square root of -1. Rather, the two square roots have equal standing.

We say that all numbers of the form a +b$$i$$, where $$a$$ and $$b$$ are any real numbers, is the set of complex numbers, and we denote this set $$\Complex$$. The real numbers $$\R$$ may be considered to be the subset of complex numbers $$\Complex=\{a+bi\}$$ for which b = 0. Complex numbers can be added, subtracted, multiplied, and divided (except by 0). We will explore some of the properties of these numbers later.

There are in fact two commonly used definitions of complex numbers, but they are immediately seen to be logically equivalent.

For a negative root like $$\sqrt{-4}$$, we split the number into two parts such that one part is $$\sqrt{-1}$$ like $$\sqrt{-4}=\sqrt{4(-1)}=\sqrt{4}\cdot\sqrt{-1}$$ which leads to $$2i$$

Definition 1
A complex number is an expression of the form x + yi, in which x and y are real numbers and i is a new number, called the imaginary unit, for which expressions the normal rules of calculation apply together with the extra rule: i2=-1.

Definition 2
A complex number is a pair of real numbers (x,y), satisfying the properties:


 * $$(a,b)+(c,d)=(a+c,b+d)$$
 * $$(a,b)\cdot(c,d)=(ac-bd,ad+bc)$$

In both cases a complex number consists of two real numbers x and y. The real number x is called the real part and the real number y the imaginary part of the complex number.

From the properties we deduce that complex numbers of the form (x,0) behave just like the real numbers, so we identify (1,0) with 1 and hence (x,0) with x. Furthermore we see that:


 * $$(0,1)\cdot(0,1)=(-1,0)=-1$$.

It is common use to write i instead of (0,1), so:
 * $$i=(0,1)$$ and $$i^2=-1$$.

Any complex number (x,y) may now be written as x + yi.

Some examples
A complex number is a number that is in the form $$z=a+bi$$, where a and b are real numbers. We say that a is the real part of z and write $${\rm Re}(z)$$, and that b is the imaginary part of z, and write $${\rm Im}(z)$$

A number of the form bi is sometimes called a pure imaginary number, as it has no real part. The pure imaginary numbers are also complex numbers, because bi = 0 + bi. In the same way, all real numbers are also complex numbers, because a = a + 0i. So the set of complex numbers includes real numbers, pure imaginary numbers, and the sums of reals and pure imaginaries.

Here are some examples
 * 1 + 4i: a complex number, real part 1, imaginary part 4
 * 2 - 2i: a complex number, real part 2, imaginary part -2.
 * -4i: a complex number, real part 0, imaginary part -4
 * 2: a complex number (also a real number), real part 2, imaginary part 0.

Notice that the number 2 is a complex number and a real number. This fact is clearer if we write 2 = 2 + 0i.

Any complex number may be written in three main forms, which we will explore later. The form x + yi is known as the Cartesian form.

Complex numbers and matrices
Complex numbers can be identified with a certain set of matrices. If we think of the 2×2 identity matrix as the number 1, and we think of $$i$$ which we introduced above as the matrix
 * $$\begin{pmatrix}

0 & -1 \\ 1 & 0 \end{pmatrix}$$,

then the complex number $$a+bi$$ then has the form
 * $$\begin{pmatrix}

a & -b \\ b & a \end{pmatrix}$$.

Properties
Complex numbers obey most of the properties of real numbers. Take two complex numbers, $$z_1=a+bi\,$$ and $$z_2=c+di\,$$.

Addition
How can we add these two complex numbers?

We don't even have to think about $$i$$ as being "special" in any way, just treat it as any other symbol and proceed by the standard rules of algebra, grouping along the way.

We obtain:
 * $$z_1+z_2=(a+bi)+(c+di)=(a+c)+(b+d)i\,$$

If one uses the matrix analogy above, regular matrix addition works to add complex numbers in the same way. Verify for yourself that this is true.

Subtraction
Subtraction proceeds just as before.
 * $$z_1-z_2=(a+bi)-(c+di)=(a-c)+(b-d)i\,$$

Multiplication
By the normal rules, taking into account that $$i^2=-1\,$$, we find:
 * $$z_1z_2=(a+bi)(c+di)=ac+adi+bci+bdi^2=(ac-bd)+(bc+ad)i\,$$

If one uses the matrix analogy above, regular matrix multiplication works to multiply complex numbers in the same way. Verify for yourself that this is true.

Conjugates
The conjugate of a complex number $$z$$, written $$\bar{z}$$, is the same number with the sign of the imaginary part changed: the conjugate of a + bi = a - bi (and vice versa).

Let us examine what happens when we have a complex number z = a + bi, what is the product of z and its conjugate?


 * (a+bi)(a-bi) = a2 + abi - abi - b2i2 = a2  +  b2

Notice the imaginary parts cancel out, so the product is a real number. This will aid us greatly in the division of a complex number, as we will see.

Notice also that this is the sum of two squares, analogous to the difference of two squares.

Matrix transposition behaves as conjugation if one uses the matrix analogy.

Division
How do we compute the quotient
 * $$\frac{a+bi}{c+di}$$

of two complex numbers? It is not difficult. Let the quotient be:


 * $$\frac{a+bi}{c+di}=r+si$$,

then cross multiplication gives:
 * $$\,a+bi=(r+si)(c+di)=rc-sd+i(rd+sc)$$,

hence
 * $$\,a=rc-sd$$,

and
 * $$\,b=rd+sc$$.

So we have to solve two linear equations. Solution:


 * $$r=\frac{ac+bd}{c^2+d^2}$$,

and
 * $$s=\frac{bc-ad}{c^2+d^2}$$.

Note that this is complete nonsense unless $$c^2+d^2 \ne 0$$. In fact, it is easy to see that the pair of linear equations have a solution exactly when this is true, i.e., when c and d are not both zero. Or in other words, when $$c+di \ne 0$$ as a complex number. Thus we can divide by a complex number only when it is non-zero.

Luckily there is a little trick to speed up this computation. We multiply the denominator with a well chosen number as to make it real. We realize the denominator by multiplying with the conjugate of the denominator; a complex number times its conjugate is a real number:


 * $$\,(c+di)(c-di)=c^2-(di)^2=c^2+d^2$$.

hence:
 * $$\frac{a+bi}{c+di} = \frac{a+bi}{c+di}\cdot \frac{c-di}{c-di} = \frac{ac-adi+bci+bd}{c^2+d^2} = \frac{(ac+bd)+(bc-ad)i}{c^2+d^2}=\frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i$$.

Note that in the multiplication and division of complex numbers, we usually work out the whole problem instead of just memorizing the equation of the answer. The reader will note that we use here the familiar trick from algebra of multiplying by the number 1 in a particularly convenient form: $$\frac{c-di}{c-di}$$. (We leave it to the reader to verify that any non-zero complex number divided by itself is in fact equal to 1.)

Exponents and Roots
Because $$x^{y}=e^{ln{x}\times y}=1+\frac{ln{ x}\times y}{1!}+\frac{(ln{ x}\times y)^{2}}{2!}+\frac{(ln{ x}\times y)^{3}}{3!}+...$$, real numbers can be raised to imaginary numbers. Imaginary and complex numbers cannot be raised to imaginary or complex numbers, as imaginary and complex numbers have no natural logarithms. Example: $$e^{i}=0.5403023058681397+0.8414709848078965i$$ Because $$\sqrt[y]{x}=x^{\frac{1}{y}}$$, imaginary numbers can be root degrees.

Problem set
Given the above rules, answer the following questions.

Note: Use sqrt(x) for $$\sqrt{x}$$  { (7 + 2i) + (11 - 6i) = { 18_19 } + { -4_19 } i
 * type="{}"}

{ (8 - 3i) - (6i) = { 8_19 } + { -9_19 } i
 * type="{}"}

{ (9 + 4i)(3 - 16i) = { 91_19 } + { -132_19 } i
 * type="{}"}

{ 3i $$\times$$ 9i = { -27_19 } + { 0_19 } i
 * type="{}"}

{ $$\frac{i}{2+i}=$${ 1/5|0.2_19 } + { 2/5|0.4_19 } i
 * type="{}"}

{ $$\frac{11 + 3i}{\sqrt{3} - 4i}=$${ [11sqrt(3) - 12]/19 (i)|[11sqrt(3)-12]/19 (i) _19 } + { [44 + 3sqrt(3)]/19 (i)|[44+3sqrt(3)]/19 (i) _19 } i
 * type="{}"}

{ $${(x + yi)}^{-1}=$${ x/(x^2 + y^2) (i)|x/(x^2+y^2) (i) _19 } + { -y(x^2 + y^2) (i)|-y(x^2+y^2) (i) _19 } i
 * type="{}"}

The Argand Plane
We can represent complex numbers geometrically as well. Every complex number can be represented in the form z=x+iy (so x=Re(z) and y=Im(z)). Then we can represent z in the xy-plane by the point (x,y). Notice that this is a one-to-one relationship: for each complex number, we have one corresponding point in the plane, and for each point in the plane there corresponds one complex number. When we use the xy-plane in this way to represent complex numbers, we call the plane the "Argand plane". We will refer to the "y" axis as the imaginary axis, and the "x" axis as the real axis.

Notice that a purely imaginary number is represented in the Argand plane by a point on the imaginary axis. A purely real number is represented by a point on the real axis.

Here is an example of the Argand plane.
 * [[Image:Argand2.png]]

There are two complex numbers drawn in the plane; viz., 1 + i, -2 - i. Their sum is plotted on the graph, -1. The red and blue lines demonstrate how geometrically, a parallelogram can be constructed, and their apex forms their sum.

Modulus and argument

 * [[Image:Modulus and argument.png]]

On this diagram we can see the number 3 + 4i. The red line is the distance away from the origin (the number 0 + 0i). The gray line represents the distances away from the respective axes. We can see that the red line makes an angle &theta; from the real axis.

It is clear that almost all complex numbers have this distance away from the origin and that almost all complex numbers make an angle away from the real axis. We give these two qualities special names; the distance away from the origin is known as the modulus of the complex number, and the angle &theta; is known as the argument of the complex number.

We write the modulus of a complex number z by |z|, and the argument of the complex number as arg z.

We can calculate the modulus and argument by basic trigonometry.

Calculating the modulus
In the above example, we have the number 3 + 4i. We can form a triangle in the Argand plane with base 3 and height 4. By Pythagoras we can find the length of the hypotenuse by $$\sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$$.

And thus, the length of the hypotenuse is thus the modulus of the complex number, and it is 5 for 3+4i.

Generalization
If z = x + yi, |z| is clearly $$\sqrt{x^2+y^2}$$.

Equivalently, $$|z|=\sqrt{\mathrm{Re}(z)^2+\mathrm{Im}(z)^2}$$.

Calculating the argument
We have the same triangle as we had in calculating the modulus. Remember from trigonometry that tan &theta; is the ratio of the height over the base. So, for 3 + 4i, we have tan &theta; = 4/3, and thus &theta; = arctan 4/3 = 0.9...

With complex numbers, we always take two things:
 * the argument must be in radians
 * the argument lies in the interval [-&pi;,&pi;], and we always adjust the angle so it does.

Note that arg 0 is undefined.

Generalization
If z = x + yi, arg z is clearly arctan (y/x), or, equivalently, arg z = arctan (Im(z)/Re(z)).

The polar form
We are now able to calculate the modulus and argument of a complex number, where these two numbers are able to uniquely describe every number in the Argand plane.

Using these two characteristics of complex numbers, we are now able to formulate a new way of writing these numbers.

Note that in the above diagram, we obtain a triangle that describes the complex number 3 + 4i. Clearly, we can do this for all complex numbers in the Argand plane (except for 0).

To simplify our work, let us look at numbers in the circle of unit length equidistant from 0. From trigonometry, we can parameterize all the points on a circle in the Cartesian plane by (cos &theta;, sin &theta;). In complex number notation we can say that all numbers on this unit circle are in the form cos &theta;+i sin &theta;.

This works well on the unit circle, but how does this generalize to describing all numbers on the plane? We simply make the circle larger or smaller to encompass the number; this is done by multiplying by the modulus.

So then, we obtain the polar form r(cos &theta; + i sin &theta;) = z, where r is the modulus.

Euler's formula
A very significant result in the area of complex numbers is Euler's formula. It basically asserts that
 * rei&theta; = r (cos &theta; + i sin &theta;)

This statement can be verified through a rearrangement of the Taylor Series of the cosine and sine functions.

Note that conjugate complex numbers have an opposing argument. 2e2i and 2e-2i are conjugate pairs.

Using Taylor series
Here is a proof of Euler's formula using Taylor Series expansions as well as basic facts about the powers of i:



i^0 = 1, \ i^1 = i, \ i^2 = -1, \ i^3 = -i, \ i^4 = 1, \ i^5 = i, \ i^6 = -1, \ i^7 = -i, \ i^8 = 1, \dots $$

The functions ex, cos(x) and sin(x) (assuming x is real) can be written as:


 * $$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$


 * $$ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots

$$


 * $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots

$$

and for complex z we define each of these function by the above series, replacing x with iz. This is possible because the radius of convergence of each series is infinite. We then find that


 * $$e^{iz} = 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + \frac{(iz)^6}{6!} + \frac{(iz)^7}{7!} + \frac{(iz)^8}{8!} + \cdots$$


 * $$= 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - \frac{z^6}{6!} - \frac{iz^7}{7!} + \frac{z^8}{8!} + \cdots$$


 * $$= \left( 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \frac{z^8}{8!} + \cdots \right) + i\left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots \right) $$


 * $$= \cos (z) + i\sin (z) \,$$

The rearrangement of terms is justified because each series is absolutely convergent. Taking z = x to be a real number, gives the original identity as Euler discovered it.$$\square$$

Using calculus
Define the complex number z such that
 * $$z=\cos x + i\sin x \,$$

Differentiating z with respect to x:
 * $$\frac{dz}{dx}=-\sin x + i\cos x$$

Using the fact that i2 = -1:
 * $$\frac{dz}{dx}=i^2\sin x + i\cos x=i(\cos x + i\sin x)=zi$$

Separating variables and integrating both sides:
 * $$\int\frac{1}{z}\,dz=\int i\,dx$$
 * $$\ln z=xi + C\,$$

where C is the constant of integration. To finish the proof we have to argue that it is zero. This is easily done by substituting x = 0.
 * $$\ln z = C\,$$

But z is just equal to:
 * $$\ z = \cos x + i\sin x = \cos 0 + i \sin 0 = 1\,$$

thus
 * $$\ln 1 = C\,$$


 * $$\ C = 0\,$$

So now we just exponentiate
 * $$\ \ln z = xi\,$$


 * $$\ e^{\ln z} = e^{xi}\,$$


 * $$\ z = e^{xi}\,$$


 * $$\ e^{xi} = \cos x + i\sin x\,\square$$

Corollaries
A number of significant results follow as corollaries to Euler's result.

de Moivre's theorem
De Moivre's theorem is useful in calculating powers of complex numbers. It states that
 * (r (cos &theta; + i sin &theta;))n = rn ( cos n&theta; + i sin n&theta;)

This follows clearly (from the laws of exponents) if we rewrite the theorem in the form
 * (rei&theta;)n=rnein&theta;

Equivalent trigonometric forms
From the cosine/sine form of the complex number, we can rewrite the cosine and sine function in terms of exponentials.
 * cos &theta; = 1/2 (ei&theta; + e-i&theta;)
 * sin &theta; = 1/(2i) (ei&theta; - e-i&theta;)

Relation of e, $$\pi$$, i, 1, and 0
By substituting &pi; into the formula, we obtain the following result:

$$e^{\pi i}+1=0$$

The actual mathematical relevance of this equation is actually very little. It is known more for its relation of the many branches of mathematics: e comes from calculus, &pi; from geometry, i comes from algebra, 1 is the multiplicative identity, and 0 is the additive identity. It is also known for its simple mathematical aesthetic.

Forming trigonometric identities
The aforementioned equivalent trigonometric forms, combined with the Binomial theorem, allow us to create some trigonometric identities that would be difficult to form in any other way. These identities can be used to simplify integral problems.

Cosine/sine powers
How can we simplify, say, (cos x)5?

Let us look at a simple example for motivation. First, rewrite as
 * (cos x)5 = (1/2 (eix+e-ix))5 =

1/25 (eix+e-ix)5

By the Binomial theorem,
 * (a+b)5=a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5.

Replacing a with eix/2 and b with e-ix/2, we obtain
 * 10e-ix/32+10eix/32+5e3ix/32+5e-3ix/32+e5ix/32+e-5ix/32=
 * (5/16)(e-ix+eix)+(5/32)(e3ix+e-3ix)+(1/32)(e5ix+e-5ix)=
 * (5/16)(2 cos x)+(5/32)(2 cos 3x)+(1/32)(2 cos 5x)=
 * (5/8) cos x+(5/16) cos 3x+(1/16) cos 5x=(cos x)5

Procedure
We can summarize from the above example the general procedure: To simplify an expression in the form:
 * cos(x)k
 * sin(x)k

the procedure is:
 * write cos(x) or sin(x) in the exponential form, all to the power of k
 * expand using the Binomial theorem
 * collect conjugate pairs
 * write back from the exponential form into the trigonometric form

Cosine/sine multiples
We can also form identities in the form:
 * cos(kx)
 * sin(kx).

Let's look at another example to see how it's done.

Example
Let's expand sin(3x).

Recall de Moivre's theorem stating
 * (cos(x)+i sin(x))3 = cos(3x)+i sin (3x)

We will use this fact to expand out the left side. For ease of manipulation, it may be easier to let c = cos(x), and s=sin(x). Then use the Binomial theorem again to expand out:
 * (c+is)3=c3 + 3i c2s - 3cs2 - i s3

Collect real and imaginary parts
 * (c+is)3=c3 - 3cs2 + i(3c2s - s3)

Now, this is of course equal to cos(3x)+i sin (3x). So, substituting back cos(x) for c and similarly for sin(x), we can equate real and imaginary parts, and we get
 * sin(3x)=3 cos(x)2sin(x) - sin(x)3

and we get
 * cos(3x)=cos(x)3 - 3 cos(x)sin(x)2

for free.

NB: In the cosine expansion, one could write sin(x)2 as 1-cos(x)2 to obtain a formula consisting completely of cosines. (analogously one could write the sine expansion using only sines)