\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{amstext} \usepackage{amssymb} \newcommand{\ra}{\rightarrow} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\E}{\mathbb{E}} \newcommand{\para}{\paragraph} \newcommand{\sub}{\paragraph} \newtheorem{definition}{Definition} \newtheorem{formula}[definition]{Formula} \begin{document} \title{Complex Exponents: Supplement to 6.8} \date{} \maketitle \begin{definition} \label{c exps} The exponent $e^{i\theta}$ is defined to be : \[ e^{i\theta}=\cos \theta + i\sin \theta \] \end{definition} Try these examples in your calculator too. \para{Example:} \[ e^{i\pi}=\cos \pi + i\sin \pi = -1 \] \para{Example:} \[ e^{i\frac{\pi}{3}}=\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3}) = \frac{1}{2}+i\frac{\sqrt{3}}{2} \] \para{Trigonometric Form of a Complex Number:} Recall that standard form of a complex number is $a+bi$. Complex numbers can be graphed on the plane by graphing the the point $(a,b)$. We could also specify the angle($\theta$) and the radius($r$) of the ray from the origin to the point $(a,b)$, (see figure~6.45, page~508). How do we go back and forth from these ways of expressing complex numbers? Look at the picture on page~508, (as well as the formulas). \begin{formula} \label{trig to std} Trig form to standard form: \begin{eqnarray*} a & = & r \cos \theta \\ b & = & r \sin \theta \end{eqnarray*} \end{formula} \begin{formula} \label{std to trig} Standard form to Trig form: \[ \begin{array}{cccl} r & = & \sqrt{a^2+b^2} & \mbox{ note that $r$ is always positive so no $\pm$} \\ \tan \theta & = & \frac{b}{a} & \mbox{ you can find $\theta$ by using $\arctan$} \end{array} \] \end{formula} \para{Remark:} $a+bi = (r\cos \theta) + i (r \sin \theta) = r(\cos \theta + i \sin \theta) = r e^{i\theta}$ \begin{definition} \label{trig} The trigonometric form of a complex number $z=a+bi$ is: \[ z=r e^{i\theta} \] where $r$ and $\theta$ are as above in formula~\ref{std to trig}. Important: $\theta$ is in radians. \end{definition} \para{Remark:} The complex numbers with the same $r$ and $\theta$ and $\theta+2\pi$ are the same complex number. Why? (Try drawing the picture) \sub{Example 1:} (See the example~1 in the book) Find $r$ and $\theta$ just like in the book, but write the answer as in definition~\ref{trig}. (Remember: $\theta$ is in radians) \[ z = 4 e^{i\frac{4\pi}{3}} \] \[ z \approx 2\sqrt{10}e^{i\frac{18.4\pi}{180}} \] \sub{Example 2:} (See example~2 in the book.) Here you start with $z=\sqrt{8}e^{i(-\frac{\pi}{3})}$. Use definition~\ref{c exps} to get $z=\sqrt{2} - i \sqrt{6}$. \para{Multiplication and division of complex numbers:} Here is where you will be glad we are using complex exponents. (Compare what I do here with the nasty formulas in the book.) We can multiply in standard form by distributing: \[ (a+bi)(c+di)=ac+adi+bci+bdi^2=ac-bd+i(ad+bc) \] What if we use complex exponents? We use properties of exponents: \begin{eqnarray*} (r_1 e^{i\theta_1}) (r_2 e^{i\theta_2}) & = & r_1 r_2 e^{i\theta_1 + i\theta_2} \\ & = & r_1 r_2 e^{i(\theta_1 + \theta_2)} \end{eqnarray*} Now you can convert this to standard form if you want. \sub{Example 3:} Multiply the complex numbers: \[ z_1 = 2 e^{i \frac{2\pi}{3}} \] \[ z_2 = 8 e^{\frac{11\pi}{6}} \] \begin{eqnarray*} z_1 z_2 & = & (2 e^{i \frac{2\pi}{3}}) (8 e^{i \frac{11\pi}{6}}) \\ & = & 16 e^{i (\frac{2\pi}{3} + \frac{11\pi}{6} )} \\ & = & 16 e^{i (\frac{5\pi}{2})} \\ & = & 16 i \end{eqnarray*} Where the last equality is by using definition~\ref{c exps}. Lets divide these two numbers too: \begin{eqnarray*} \frac{z_1}{z_2} & = & \frac{2 e^{i \frac{2\pi}{3}}}{8 e^{i \frac{11\pi}{6}}} \\ & = & \frac{1}{4} e^{i (\frac{2\pi}{3} - \frac{11\pi}{6} )} \\ & = & \frac{1}{4} e^{i (-\frac{7\pi}{6})} \\ & = & \frac{1}{4} e^{i (\frac{5\pi}{6})} \end{eqnarray*} You could put this is standard form if you wanted using definition~\ref{c exps}. \para{Powers:} These are easy, just use properties of exponents. \sub{Example 5:} Find $(-1+\sqrt{3}i)^{12}$. Think about how you would do this without complex exponents: Either distributing and multiplying out a 12th power, or using the binomial theorem. Either way is a pain. We'll first find the trig form of the number using formula~\ref{std to trig}: \[ -1+\sqrt{3}i=2 e^{i\frac{2\pi}{3}} \] Now raise to the 12th power: \begin{eqnarray*} (-1+\sqrt{3}i)^{12} & = & (2 e^{i\frac{2\pi}{3}})^{12} \\ & = & 2^{12} (e^{i\frac{2\pi}{3}})^{12} \\ & = & 2^{12} e^{(i\frac{2\pi}{3})12} \\ & = & 2^{12} e^{i 8\pi} \\ & = & 2^{12} (1+0) = 2^{12} = 4096 \end{eqnarray*} \para{Roots of a complex number} What is a root of a number? \begin{definition} \label{roots} A complex number $u$ is an nth root of $z$ if $z=u^n$. \end{definition} What does this mean? Here are some examples: \sub{Remark:} We usually say square root instead of 2th root, and cube root instead of 3th root. \sub{Example:} 2 is a square root of 4, so is -2. \sub{Example:} (see Example~5 above) $-1+\sqrt{3}i$ is a 12th root of 4096. \sub{Example:} (see Example~5 above) 2 is a 12th root of 4096. In the above examples, numbers can have more then one root. What happens if you hit $\sqrt[12]{4096}$ on your calculator? You will get 2. Why don't you get $-1+\sqrt{3}i$? Well, its like the square root button. $\sqrt{4}=2$. Square root is a function, so it can only give us one number back, the positive one. The same thing happens when you take a higher root, you only get the positive real number back. (Its a bit more complicated then this, see below.) How to we find all the nth roots of a number? Well we solve the equation $u^n=z$, but we use trig form to make this a bit easier. Look how this is done in the book on pages~513-514. Here is the formula: \begin{formula} \label{formula roots} The complex number $z=r e^{i\theta}$ has exactly $n$ distinct nth roots, and these are: \[ \sqrt[n]{r} e^{i \frac{\theta + 2k\pi}{n}} \] Where $k=0,1,2,\ldots,n-1$. \end{formula} The way to think about this is by taking the $\frac{1}{n}$ power of $z$. Of course this doesn't explain the $+2k\pi$. Where does the $2k\pi$ come from? It comes from the fact that $\cos$ and $\sin$ don't have unique solutions. \para{Remark:} Hitting nth root or $\frac{1}{n}$ on your calculator corresponds to $k=0$ in formula~\ref{formula roots}. \sub{Example 7:} Find the cube roots of $z=-2+2i$. To use formula~\ref{formula roots}, we need to put $z$ in trig form. We do this and get $z=\sqrt{8} e^{\frac{3\pi}{4}}$. Now take the nth root using formula~\ref{formula roots}. (There is a small mistake in the book, it should be $\sqrt[6]{8}(\cos(\frac{}{}) \cdots)$. Note that $\sqrt[6]{8}=\sqrt{2}$) \[ \begin{array}{lllllllll} k=0 & & \mbox{root} & = & \sqrt{2} e^{i \frac{\pi}{4}} & = & 1 + i \\ k=1 & & \mbox{root} & = & \sqrt{2} e^{i \frac{\frac{3\pi}{4} + 2\pi}{3}} & = & \sqrt{2} e^{i \frac{11\pi}{12}} & \approx & -1.3660 + .3660 i \\ k=2 & & \mbox{root} & = & \sqrt{2} e^{i \frac{\frac{3\pi}{4} + 4\pi}{3}} & = & \sqrt{2} e^{i \frac{19\pi}{12}} & \approx & .3660 - 1.3660 i \\ \end{array} \] This gives us the three cube roots. See how we let $k=0,1,2$? (We could have found the exact answer for these using the sum and difference formulas.) \sub{Exercise:} Use formula~\ref{formula roots} to find the 6th roots of 1. See the book page~513 for another way to do this. Make sure your answer matches up, check your answers by raising to the 6th power. \para{Calculator Helps:} TI: $P \triangleright R$ converts standard form to trig form. $R \triangleright P$ converts trig form to standard form. (I don't really understand these, but see the book page 510). HP: (There are several ways to do this.) See the blue shift MTH key (POLAR). This toggles between standard form and trig form. Lets play with $1-2i$. To enter a number in standard form, enter either $(1,2)$ or $[1,2]$. Now toggle the POLAR button. You number $(1,2)$ will change from $(1,2)$ to $(2.2360, \angle 1.10714)$ if you are in radian mode and $(2.2360, \angle 63.4349)$ if you are in degree mode. Notice the $\angle$ in the display. This tells you that you are in trig mode and which entry is the angle. Notice that the upper left corner of your display also changes telling you if you are in standard mode or trig mode. To enter a number in trig mode, you will have to enter the angle sign too. (This is blue shift SPC). Play around with it. \para{Exercise:} Convert $z=1-2i$ to trig form. Graph $z$ on the plane using both standard from and trig form (make sure they match up). Find $z^5$. Graph this too. Find all 5th roots of $z$, (how many are there?). Graph these on the plane too. See how your calculator can help you. \end{document}