Appendix B - Complex Numbers

Complex numbers arise naturally from an attempt to solve the equation

a^{2}+1=0.\,\!

It's easy enough to write such an equation down, but how would you solve it? The answer is

a=\pm {\sqrt {-1}}=\pm i.\,\!

We let the symbol $i\,\!$ represent ${\sqrt {-1}}\,\!$ , so that $i^{2}=-1\,\!$ . Then any number of the form


	$z=x+iy,\,\!$	(B.1)

where $x\,\!$ and $y\,\!$ are real is called a complex number. Let's take some other complex number to be $\eta =c+id\,\!$ where $c\,\!$ and $d\,\!$ are real. Then the two complex numbers are equal,

\eta =z.\,\!

This means that

x+iy=c+id,\,\!

is true if and only if

x=c,\;\;{\mbox{ and }}\;\;y=d.\,\!

We refer to $x\,\!$ as the real part of the complex number $z\,\!$ and $y\,\!$ as the imaginary part. Sometimes these are written as Re( $z\,\!$ ) and Im( $z\,\!$ ) respectively.

We may restate the equivalence condition as $z=\eta \,\!$ if and only if the real part of $z\,\!$ is equal to the real part of $\eta \,\!$ and the imaginary part of $z\,\!$ is equal to the imaginary part of $\eta \,\!$ .

To add two complex numbers, add the real parts and the imaginary parts separately:

z+\eta =(x+iy)+(c+id)=(x+c)+i(y+d).\,\!

Complex numbers are multiplied like any other binomial expression (recall foil: first two, outside two, inside two, last two):

z\eta =(x+iy)(c+id)=xc-yd+i(yc+xd),\,\!

where we have used $i^{2}=-1\,\!$ .

The complex conjugate of the complex number $z\,\!$ is denoted $z^{*}\,\!$ and is given by


	$z^{*}=x-iy.\,\!$	(B.2)

One reason for defining this is that a number times its own complex conjugate is real,

zz^{*}=(x+iy)(x-iy)=x^{2}+y^{2}+i(xy-yx)=x^{2}+y^{2}.\,\!

Taking the complex conjugate of the number means just changing the sign in front of the $i\,\!$ . The complex conjugate of the complex conjugate is the original complex number, and

z^{*}z=(x-iy)(x+iy)=x^{2}+y^{2}+i(yx-xy)=x^{2}+y^{2}=zz^{*}.\,\!

We also call this the modulus squared . The modulus is


	$\|z\|={\sqrt {(z^{*}z)}}={\sqrt {x^{2}+y^{2}}}.\,\!$	(B.3)

Note that the complex conjugate of a product is the product of the complex conjugates:

(z\eta )^{*}=z^{*}\eta ^{*}.\,\!

It is often useful to look at a graph for a complex number, consisting of an x-axis for the real part and a y-axis for the complex part. This is shown in Fig. B.1. In this way, we can think of $z\,\!$ as a two-dimensional vector with the magnitude (length) being equivalent to the modulus of the complex number, $|z|={\sqrt {x^{2}+y^{2}}}\,\!$ .

Figure B.1: A complex number in Cartesian coordinates.

Another useful way to represent this is with polar coordinates. We can do this by writing

z=|z|(\cos \theta +i\sin \theta ).\,\!

Clearly if we wanted to rewrite this in terms of $x\,\!$ and $y\,\!$ , we find that $x=|z|\cos \theta \,\!$ and $y=|z|\sin \theta \,\!$ .

A very famous identity exists between the exponential function and sine and cosine,


	$e^{i\theta }=\cos \theta +i\sin \theta ,\,\!$	(B.4)

where $e\,\!$ is the exponential function, which is a number that is approximately 2.71828. Using this equation, we could also write

z=|z|e^{i\theta }.\,\!

This is really just a short-hand notation for us and most of the time, when we calculate something, we will use the $\sin \theta ,\cos \theta \,\!$ functions. But is often the case that people will write this as


	$z=re^{i\theta },\,\!$	(B.5)

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = \sqrt{x^2+y^2}\,\!} as is usual for polar coordinates. So everything is just like polar coordinates with the exception of the inclusion of the factor $i\,\!$ . (See Fig. B.2.)

Figure B.2: A polar coordinate representation of a complex number.

Exercises

Let $z_{1}=5+2i\,\!$ and $z_{2}=3+i\,\!$ . For 2-6, write the answer in terms of the real and complex components, i.e., in the form $x+iy\,\!$ , where $x\,\!$ and $y\,\!$ are real numbers.