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### Complex Numbers

#### Discussion

Now that we have learned a variety of techniques for handling first order equations, we will move on to higher order equations. Unfortunately, higher order equations can be a lot trickier. Fortunately, we will avoid most of the difficulties by only concentrating on the easy ones, which are also the most common in basic applications in engineering and physics (and when you get to the non-basic applications you should be ready to learn about the harder ones). The easy equations usually reduce down to solving polynomial equations, which is college algebra stuff. We won't be concerned with just the real roots of our polynomial equations though, we will have to consider the complex roots as well. And the complex roots we encounter will then have to be used in exponential functions. So before we begin solving the differential equations, we will take a couple of days to go over the theory of complex variables.We let $i=\sqrt{-1}$. A complex number is a number of the form $a+bi$ where $a$ and $b$ are real numbers. Arithmetic with complex numbers is almost exactly the same as with real numbers except every time you square an $i$ you replace it with $-1$. There is also one new operation called conjugation where you replace $i$ by $-i$.

#### Complex Arithmetic

*Real Part*

$\Re(a + bi) = a$

$\Re(7 - 4i) = 7$

*Imaginary Part*

$\Im(a + bi) = b$

$\Im(3 - 11i) = -11$

*Conjugation*

$\overline{(a + bi)} = a - bi$

$\overline{(-2 + 5i)} = -2 - 5i$

*Addition*

$(a + bi) + (c + di) = (a + c) + (b + d)i$

$(2 + 3i) + (4 - 7i) = (2 + 4) + (3 - 7)i = 6 - 4i$

*Subtraction*

$(a + bi) - (c + di) = (a - c) + (b - d)i$

$(3 + 2i) - (-1 - 4i) = (3 - -1) + (2 - -4)i = 4 + 6i$

*Multiplication*

Here you just multiply out term by term and collect the like
terms.
$$\begin{align}
(a + bi)(c + di) &= ac + adi + bci + bdi^2 \\
&= ac + adi + bci - bd \\
&= (ac - bd) + (ad + bc)i \\
(3 - 4i)(2 + i) &= (6 - -4) + (3 - 8)i = 10 - 5i
\end{align} $$
Division:

This is the one tricky operation. The trick is to multiply the
numerator and the denominator by the conjugate of the denominator.
This will yield a real denominator and then everything is easy.
$$ \eqalign
{\frac{a+bi}{c+di}&=\frac{(a+bi)(\overline{c+di})}{(c+di)(\overline{c+di})}
\cr
&=\frac{(a+bi)(c-di)}{(c+di)(c-di)} \cr
&=\frac{ac+bci-adi+bd}{c^2+cdi-cdi+d^2} \cr
&=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}\cr
&=\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i \cr }
$$
In the last step we use the fact that $c^2+d^2$ is a real number and we
know that the quotient of a sum is the sum of the quotients and that we know
how to divide by real numbers.
The fact that when you multiply a complex number by its conjugate you get a
real product is very important and will show up again.
$$ \eqalign
{\frac{2-7i}{3+i}&=\frac{(2-7i)(3-i)}{(3+i)(3-i)} \cr
&=\frac{(6-7)+(-2-21)i}{(9+1)+(-3+3)i} \cr
&=\frac{-1-23i}{10} \cr
&=-0.1-2.3i \cr}
$$

#### Complex Exponentials

Now that we can do arithmetic with complex numbers, we want to do calculus with them. We start with the exponential function. The trick is to expand the function in a Taylor series about the point $z=0$ and see what happens. In the following, $t$ denotes a real variable. You also need to use the following relations: $ i^2=-1$, $ i^3=-i$, $ i^4=1$, $ i^5=i$, etc. $$ \eqalign {e^{it}&=1+it+\frac{(it)^2}{2!}+\frac{(it)^3}{3!}+\frac{(it)^4}{4!} +\frac{(it)^5}{5!}+\cdots \cr &=1+it+\frac{i^2t^2}{2!}+\frac{i^3t^3}{3!}+\frac{i^4t^4}{4!} +\frac{i^5t^5}{5!}+\cdots \cr &=1+it-\frac{t^2}{2!}-i\frac{t^3}{3!}+\frac{t^4}{4!}+i\frac{t^5}{5!}-\cdots \cr &=\left(1-\frac{t^2}{2!}+\frac{t^4}{4!}-\cdots\right) +i\left(t-\frac{t^3}{3!}+\frac{t^5}{5!}-\cdots\right) \cr &=\cos(t)+i\sin(t) \cr} $$ as we recognize the Taylor series about 0 for cosine and sine. Using this relationship, we can now define the exponential of a general complex number as follows: $$ e^{x+iy}=e^xe^{iy}=e^x(\cos(y)+i\sin(y))$$ We will use this relation often in solving differential equations.

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett