2.1 Com­plex Num­bers

Quan­tum me­chan­ics is full of com­plex num­bers, num­bers in­volv­ing

$${\rm i}=\sqrt{-1}.$$

Note that $\sqrt{-1}$ is not an or­di­nary, real, num­ber, since there is no real num­ber whose square is $\vphantom{0}\raisebox{1.5pt}{$-$}$1; the square of a real num­ber is al­ways pos­i­tive. This sec­tion sum­ma­rizes the most im­por­tant prop­er­ties of com­plex num­bers.

First, any com­plex num­ber, call it $c$, can by de­f­i­n­i­tion al­ways be writ­ten in the form

$$c = c_r+{\rm i}c_i$$ (2.1)

where both $c_r$ and $c_i$ are or­di­nary real num­bers, not in­volv­ing $\sqrt{-1}$. The num­ber $c_r$ is called the real part of $c$ and $c_i$ the imag­i­nary part.

You can think of the real and imag­i­nary parts of a com­plex num­ber as the com­po­nents of a two-di­men­sion­al vec­tor:

$$\begin{picture}(200,70)(-100,-20) \thinlines \put(-50,0... ...tor(2,1){70}} \put(0,23){\makebox(0,0)[tl]{$c$}} \end{picture}$$

The length of that vec­tor is called the “mag­ni­tude,” or “ab­solute value” $\vert c\vert$ of the com­plex num­ber. It equals

$$\vert c\vert = \sqrt{c_r^2+c_i^2}.$$

Com­plex num­bers can be ma­nip­u­lated pretty much in the same way as or­di­nary num­bers can. A re­la­tion to re­mem­ber is:

$$\frac{1}{{\rm i}} = -{\rm i}$$ (2.2)

which can be ver­i­fied by mul­ti­ply­ing the top and bot­tom of the frac­tion by ${\rm i}$ and not­ing that by de­f­i­n­i­tion ${\rm i}^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$1 in the bot­tom.

The com­plex con­ju­gate of a com­plex num­ber $c$, de­noted by $c^*$, is found by re­plac­ing ${\rm i}$ every­where by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm i}$. In par­tic­u­lar, if $c$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_r+{{\rm i}}c_i$, where $c_r$ and $c_i$ are real num­bers, the com­plex con­ju­gate is

$$c^* = c_r - {\rm i}c_i$$ (2.3)

The fol­low­ing pic­ture shows that graph­i­cally, you get the com­plex con­ju­gate of a com­plex num­ber by flip­ping it over around the hor­i­zon­tal axis:

$$\begin{picture}(200,100)(-100,-50) \thinlines \put(-50,... ...2,-1){70}} \put(0,-23){\makebox(0,0)[bl]{$c^*$}} \end{picture}$$

You can get the mag­ni­tude of a com­plex num­ber $c$ by mul­ti­ply­ing $c$ with its com­plex con­ju­gate $c^*$ and tak­ing a square root:

$$\vert c\vert = \sqrt{c^* c}$$ (2.4)

If $c$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_r+{\rm i}{c}_i$, where $c_r$ and $c_i$ are real num­bers, mul­ti­ply­ing out $c^*c$ shows the mag­ni­tude of $c$ to be

$$\vert c\vert = \sqrt{c_r^2+c_i^2}$$

which is in­deed the same as be­fore.

From the above graph of the vec­tor rep­re­sent­ing a com­plex num­ber $c$, the real part is $c_r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert c\vert\cos\alpha$ where $\alpha$ is the an­gle that the vec­tor makes with the hor­i­zon­tal axis, and the imag­i­nary part is $c_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert c\vert\sin\alpha$. So you can write any com­plex num­ber in the form

$$c = \vert c\vert\left(\cos\alpha+{\rm i}\sin\alpha\right)$$

The crit­i­cally im­por­tant Euler for­mula says that:

$$\cos\alpha + {\rm i}\sin\alpha = e^{{\rm i}\alpha} %$$ (2.5)

So, any com­plex num­ber can be writ­ten in po­lar form as

$$c = \vert c\vert e^{{\rm i}\alpha} %$$ (2.6)

where both the mag­ni­tude $\vert c\vert$ and the phase an­gle (or ar­gu­ment) $\alpha$ are real num­bers.

Any com­plex num­ber of mag­ni­tude one can there­fore be writ­ten as $e^{{\rm i}\alpha}$. Note that the only two real num­bers of mag­ni­tude one, 1 and $\vphantom{0}\raisebox{1.5pt}{$-$}$1, are in­cluded for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, re­spec­tively $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$. The num­ber ${\rm i}$ is ob­tained for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$$\raisebox{.5pt}{$/$}$​2 and $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm i}$ for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$\pi$$\raisebox{.5pt}{$/$}$​2.

(See de­riva­tion {D.7} if you want to know where the Euler for­mula comes from.)

Key Points

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

Com­plex num­bers in­clude the square root of mi­nus one, ${\rm i}$, as a valid num­ber.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

All com­plex num­bers can be writ­ten as a real part plus ${\rm i}$ times an imag­i­nary part, where both parts are nor­mal real num­bers.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The com­plex con­ju­gate of a com­plex num­ber is ob­tained by re­plac­ing ${\rm i}$ every­where by $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm i}$.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The mag­ni­tude of a com­plex num­ber is ob­tained by mul­ti­ply­ing the num­ber by its com­plex con­ju­gate and then tak­ing a square root.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The Euler for­mula re­lates ex­po­nen­tials to sines and cosines.

2.1 Re­view Ques­tions

1.

Mul­ti­ply out $(2+3{\rm i})^2$ and then find its real and imag­i­nary part.

So­lu­tion math­c­plx-a

2.

Show more di­rectly that 1$\raisebox{.5pt}{$/$}$​${\rm i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm i}$.

So­lu­tion math­c­plx-b

3.

Mul­ti­ply out $(2+3{\rm i})(2-3{\rm i})$ and then find its real and imag­i­nary part.

So­lu­tion math­c­plx-c

4.

Find the mag­ni­tude or ab­solute value of $2+3{\rm i}$.

So­lu­tion math­c­plx-d

5.

Ver­ify that $(2-3{\rm i})^2$ is still the com­plex con­ju­gate of $(2+3{\rm i})^2$ if both are mul­ti­plied out.

So­lu­tion math­c­plx-e

6.

Ver­ify that $e^{-2{\rm i}}$ is still the com­plex con­ju­gate of $e^{2{\rm i}}$ af­ter both are rewrit­ten us­ing the Euler for­mula.

So­lu­tion math­c­plx-f

7.

Ver­ify that $\left(e^{{\rm i}\alpha}+e^{-{\rm i}\alpha}\right)$$\raisebox{.5pt}{$/$}$​2 $\vphantom0\raisebox{1.5pt}{$=$}$ $\cos\alpha$.

So­lu­tion math­c­plx-g

8.

Ver­ify that $\left(e^{{\rm i}\alpha}-e^{-{\rm i}\alpha}\right)$$\raisebox{.5pt}{$/$}$​$2{\rm i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sin\alpha$.

So­lu­tion math­c­plx-h