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\begin{document}

\title{Complex Exponents: Supplement to 7.3}
\date{}
\author{}
\maketitle

\begin{definition} \label{c exps}
The exponential $e^{i\theta}$ is defined for $\theta$ in radians as:
\[
e^{i\theta}=\cos \theta + i\sin \theta
\]
\end{definition}

Here are some examples.  Make sure you try them in your calculator too (in radian mode!)
\begin{example}
\[
e^{i\pi}=\cos \pi + i\sin \pi = -1
\]
\end{example}

\begin{example}
\[
e^{i\frac{\pi}{3}}=\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)
	= \frac{1}{2}+i\frac{\sqrt{3}}{2}
\]
\end{example}













\para{Trigonometric Form of a Complex Number:}
Recall that standard form of a complex number is $a+bi$ where $a$ and $b$ are real numbers.
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 page~524 of text).

How do we go back and forth from these ways of expressing complex numbers?
Look at the picture on page~524, (as well as the formulas).
The is just a right triangle, something we are used to working with.

\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}

\begin{remark}
$a+bi = (r\cos \theta) + i (r \sin \theta) = r(\cos \theta + i \sin \theta)
= r e^{i\theta}$
\end{remark}

\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}

\begin{remark}
The complex numbers with the same $r$ and $\theta$ and $\theta+2\pi$
are the same complex number.  Why? (Try drawing the picture)
\end{remark}

\begin{example}
Find $r$ and $\theta$ just like in the book for $1+i$ and $\sqrt{3}-i$, 
but write the answer as in definition~\ref{trig}.
(Remember: $\theta$ is in radians)

First, for $1+i$, $r=\sqrt{1^2+1^2} = \sqrt{2}$.
And, $\theta$ is in quadrant I, and $\tan \theta = 1$, so $\theta = \pi/4$:
\[
1+i = \sqrt{2}e^{i\pi/4}
\]

Now for $\sqrt{3}-i$.
$r = \sqrt{3+1^2} = \sqrt{4} = 2$.
And, $\theta$ is in quadrant IV and $\tan \theta = -1/\sqrt{3}$, or $\theta = -\pi/6$.
\[
\sqrt{3}-i = 2e^{-i\pi/6}
\]
\end{example}

\begin{example}
Here we go from trig form to standard form.  See Example~4 of the book and Definition~\ref{c exps}.
Notice that $120^\circ = 2\pi/3$.
\begin{eqnarray*}
2e^{i2\pi/3} & = & 2\left( \cos(2\pi/3) + i \sin(2\pi/3) \right) \\
	& = & 2 \left( -\frac{1}{2} + i \frac{\sqrt{3}}{2} \right) 
	=  -1 + i \sqrt{3} \\
\sqrt{8}e^{i7\pi/4} & = & \sqrt{8} \left( \cos(7\pi/4) + i \sin(7\pi/4) \right) \\
	& = & \sqrt{8} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right)
	= \frac{\sqrt{8}}{\sqrt{2}} - i \frac{\sqrt{8}}{\sqrt{2}} \\
	& = & 2-2i 
\end{eqnarray*}
\end{example}








\para{Multiplication and division of complex numbers:}
Here is where you will be glad we are using complex exponents.
(Compare what we 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.  Remember that if the bases are the same then the exponents add.
\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.

\begin{example}
Multiply the complex numbers: $2 e^{i \frac{2\pi}{3}}$ and $8 e^{i\frac{11\pi}{6}}$
\begin{eqnarray*}
\left(2 e^{i \frac{2\pi}{3}}\right) \left(8 e^{i\frac{11\pi}{6}}\right) 
	& = & 16 e^{i (\frac{2\pi}{3} + \frac{11\pi}{6} )} \\
	& = & 16 e^{i (\frac{5\pi}{2})} \\
	& = & 16 i 
\end{eqnarray*}
(Why is $e^{i\pi/2} = i$?)

Lets divide these two numbers too:
\begin{eqnarray*}
\frac{2 e^{i 2\pi/3}}{8 e^{i 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}.
\end{example}













\para{Powers:} Again, we just use properties of exponents.

\begin{example} \label{example mult}
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*}
\end{example}














\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:

\begin{remark}
We usually say square root instead of 2nd root,
and cube root instead of 3th root.
\end{remark}

\begin{example}
$2$ and $-2$ are both square roots of 4.
\end{example}

\begin{example}
(see Example~\ref{example mult} above)
$-1+\sqrt{3}i$ is a 12th root of 4096.
\end{example}

\begin{example}
(see Example~\ref{example mult} above)
2 is a 12th root of 4096.
\end{example}

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 actually a bit more complicated then this, but we'll see how it works in a minute.)

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.
The formula is below, but lets see what happens for square roots.
If you don't understand why this works, don't worry too much.
Make sure you can use Formula~\ref{formula roots}.

\begin{example} \label{square root}
Lets find the square roots of $-1+i\sqrt{3}$ (see Example~\ref{example mult}).
This means we need to find all the complex numbers $z$ so that $z^2 = -1+\sqrt{3}$.
We first put $z$ and $-1+\sqrt{3}$ into complex form: $-1+\sqrt{3} = 2e^{i2\pi/3}$ and $z=re^{i\theta}$.
We now set up the equations representing $z^2 = -1+\sqrt{3}$ (in trig form):
\[
r^2e^{2i\theta} = 2e^{i2\pi/3}
\]
Since $r$ is just a real number, and must be positive, $r=\sqrt{2}$ (this is the normal postive square root that we are used to).
But, what values of $\theta$ will make sure that $e^{2i\theta} = e^{i2\pi/3}$.
The obvious choice is $\theta = \pi/3$, but there are more solutions!
Lets write it out as in Definition~\ref{c exps}:
\begin{eqnarray*}
e^{i2\pi/3} & = & \cos(2\pi/3) + i \sin (2\pi/3) 
	= -\frac{1}{2} + i \frac{\sqrt{3}}{2} \\
e^{2i\theta} & = & \cos (2\theta) + i \sin(2\theta)
\end{eqnarray*}
What we see is that we really have some trig equations to solve:
\begin{eqnarray}
\cos(2\theta) = -\frac{1}{2} \label{eqn 1} \\
\sin(2\theta) = \frac{\sqrt{3}}{2} \label{eqn 2}
\end{eqnarray}
We will only solve this for solutions between $0$ and $2\pi$.
Solving (\ref{eqn 1}) gives:
\[
2\theta = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}
\]
or 
\begin{equation} \label{solve cos}
\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}
\end{equation}
Now, solving (\ref{eqn 2}):
\[
2\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}
\]
or
\begin{equation} \label{solve sin}
\theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}
\end{equation}
The solution we are looking for must satisfy both Equation~(\ref{solve cos}) and Equation~(\ref{solve sin}).
So, that means we must have
\[
\theta = \frac{\pi}{3}, \frac{4\pi}{3}
\]
In other words, the square roots of $-1+i\sqrt{3}$ are given by $r=\sqrt{2}$ and the values for $\theta$ above:
\begin{eqnarray*}
\sqrt{2}e^{i\pi/3} & = & \sqrt{2}( \cos(\pi/3) + i \sin(\pi/3)) \\
	& = & \sqrt{2}\left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) \\
	& = &  \frac{\sqrt{2}}{2} + i \frac{\sqrt{6}}{2} \\
\sqrt{2}e^{i4\pi/3} & = & \sqrt{2}( \cos(4\pi/3) + i \sin(4\pi/3)) \\
	& = & \sqrt{2}\left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) \\
	& = &  -\frac{\sqrt{2}}{2} - i \frac{\sqrt{6}}{2} \\
\end{eqnarray*}
\end{example}

Its really not this hard in practice.
We will usually just use the formula in the book (page~531):
%
\begin{formula} \label{formula roots}
The complex number $z=r e^{i\theta}$ has exactly $n$ distinct nth roots, 
and these are:
\[
r^{1/n} e^{i(\theta + 2k\pi)/n}
\]
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 (like we saw in Example~\ref{square root}).

\begin{remark}
Hitting nth root or $\frac{1}{n}$ power on your calculator
corresponds to $k=0$ in formula~\ref{formula roots}.
\end{remark}

\begin{example}
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 $z=\sqrt{8} e^{i3\pi/4}$.
Now take the nth root using formula~\ref{formula roots}.
We first take care of $(\sqrt{8})^{1/3}$: (notice that $8=2^3$)
\begin{eqnarray*}
(\sqrt{8})^{1/3} & = & \left( 8^{1/2} \right)^{1/3} 
	= 8^{1/6} \\
	& = & (2^3)^{1/6} 
	= 2^{3/6} = 2^{1/2}\\
	& = & \sqrt{2}
\end{eqnarray*}
\[
\begin{array}{lllllllll}
k=0 & & \mbox{root:} & & \sqrt{2} e^{i \frac{\pi}{4}} 
	& = & 1 + i \\
k=1 & & \mbox{root:} & & \sqrt{2} e^{i \frac{\frac{3\pi}{4} + 2\pi}{3}}
	& = & \sqrt{2} e^{i \frac{11\pi}{12}} 
	& \approx & -1.3660 + .3660 i \\
k=2 & & \mbox{root:} & & \sqrt{2} e^{i \frac{\frac{3\pi}{4} + 4\pi}{3}} 
	& = & \sqrt{2} e^{i \frac{19\pi}{12}} 
	& \approx & .3660 - 1.3660 i 
\end{array}
\]
This gives us the three cube roots.
See how we let $k=0,1,2$?
Can you take these numbers and test them out on your calculator?  Here is an example of one:
\[
(.3660 - 1.3660i)^3 = -(1.99979) + i(1.99994)
\]
Looks like its correct.
\end{example}

\begin{exer}
Use formula~\ref{formula roots} to find the 6th roots of 1.
See the book page~531 for another way to do this.  
Make sure your answer matches up, 
check your answers by raising to the 6th power.
\end{exer}














\para{Calculator Helps:}  
On my TI85, I can go to CPLX (2nd 9) to get some functions.
I can enter the number $1+2i$ as $(1,2)$.
Try this:
\[
(1,2)^5 = (41,-38)
\]
which means $(1+2i)^5 = 41-38i$.
We can also put the complex number $1+2i$ into trig form by punching $(1,2)$ and then hitting $\triangleright Pol$ (found in the CPLX menu):
\[
(1,2)\triangleright Pol = (2.23 \angle 63.43) 
\]
which means that $r=2.24$ and $\theta = 63.43^\circ$ (I was in degree mode).
We can do this in radian mode too:
\[
(1,2)\triangleright Pol = (2.23 \angle 1.107) 
\]
Notice that the value for $r$ didn't change.

What do you think the button $\triangleright Rec$ does?
Try it!

\begin{exer}
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{exer}


\para{Homework:}  You are welcome to do this using the book techniques or using complex exponentials.
	I think that complex exponentials is quite a bit easier and more understandable.
	
These you should do first, but not hand in:
\begin{itemize}
\item    Section 7.3: 15, 17, 25 (write this in exponential notation too), 33, 40, 49, 57
\end{itemize}

Hand these in:
\begin{itemize}
\item   Section 7.3: 14, 39, 43, 61
\end{itemize}


\end{document}


