Differential Equations - Packet 01

Complex numbers

Arithmetic

A complex number $z\in ℂ$ may be written in ‘Cartesian form’ with distinct real and imaginary components, $z=a+bi$ with $a,b\in ℝ$. For addition (as $z_1+z_2$) and multiplication by a real number (as $\lambda z$) one operates on $a$ and $b$ just as the components of a vector. Complex numbers are multiplied by complex numbers using the formula $i^2=-1$. Thus if $z_1=a_1+b_{1}i$ and $z_2=a_2+b_{2}i$, then we have:

$$z_1\cdot z_2=(a_1{a}_2-b_1{b}_2)+(a_1{b}_2+a_2{b}_1)i.$$

Complex arithmetic Mathlet

Exponents

Complex numbers can be used as exponents in virtue of Euler’s Formula:

$$e^{i\theta }=\cos \theta +i\sin \theta .$$

Deriving Euler’s Formula using power series

Recall the power series formula for $e^x$:

$$e^x=1+\frac{1}{1!}x+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+\dots =\sum _{i=0}^{\infty }\frac{1}{i!}{x}^i.$$

Plug in $x=i\theta$, and recall the power series formulas for $\cos (\theta )$ and $\sin (\theta )$:

$$\begin{array}{cc}{e}^{i\theta }& =1+(i\theta )+\frac{1}{2!}{(i\theta )}^2+\frac{1}{3!}{(i\theta )}^3+\cdots +\\ & =(1-\frac{1}{2!}{\theta }^2+\frac{1}{4!}{\theta }^4-\dots )+(\theta -\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5-\dots )i\\ & =\cos \theta +\sin \theta i.\end{array}$$

Other complex exponents can be calculated from Euler’s Formula by using exponent rules:

$$\begin{array}{cc}{e}^{a+bi}& =e^a{e}^{bi}=e^a(\cos (b)+i\sin (b))\\ & =e^a\cos (b)+e^a\sin (b)\cdot i\end{array}$$

Euler Formula Mathlet

Polar format

A complex number can be written in ‘polar form’ as $z=r{e}^{i\theta }$. This translates to ‘Cartesian form’ using the exponential rule:

$$z=r{e}^{i\theta }=r\cos (\theta )+r\sin (\theta )i.$$

These forms allow one to graph complex numbers in the plane ${ℝ}^2$ using the polar coordinates $(r,\theta )$ or the corresponding Cartesian coordinates $(x,y)=(a,b)$.

The radius $r$ is called the modulus and the angle $\theta$ is called the argument of the complex number $r{e}^{i\theta }$. The modulus of $z=a+bi$ is the number $\sqrt{a^2+b^2}$. It is sometimes written $|z|$.

If a complex number is graphed in the coordinate plane, and this number is multiplied by another complex number $z=r{e}^{i\theta }$, then the product is given by scaling $z$ by $r$, and rotating $z$ by $\theta$ (counterclockwise).

Complex exponentials Mathlet

Conjugate

Complex numbers come in natural pairs. Given $z=a+bi$, there is a conjugate number $\bar{z}=a-bi$. Conjugates correspond to reflection in the real axis.

It is easy to verify a relationship between conjugation and multiplication:

$$z\bar{z}=|z{|}^2.$$

Given $z=a+bi$, we sometimes write $\mathrm{Re}(z)=a$ and $\mathrm{Im}(z)=b$ for the real part and imaginary part. These can be extracted using conjugation:

$$\mathrm{Re}(z)=\frac{z+\bar{z}}{2},\mathrm{Im}(z)=\frac{z-\bar{z}}{2i}.$$

Conjugation preserves algebraic operations:

$$\begin{array}{cc}\overline{z+w}& =\overline{z}+\overline{w}\\ \overline{(zw)}& =(\overline{z})(\overline{w}).\end{array}$$

Division

Division can be taken by complex numbers. The quotient can be calculated using conjugate multiplication. Supposing $z=a+bi$, then:

$$\frac{1}{z}=\frac{\bar{z}}{z\bar{z}}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i.$$

Roots

Whole number roots of complex numbers are found using the exponential form:

$${(r{e}^{i\theta })}^{1/n}=\sqrt[n]{r}\cdot e^{\frac{\theta }{n}\cdot i}.$$

In polar form, this means the root is taken on the modulus directly, and the angle is divided by the root number.

This procedure gives the analogue of $\sqrt{4}=2$. But of course, ${(-2)}^2=4$ as well. For any given $z\ne 0$, there are exactly $n$ distinct complex numbers, $w_1,\dots ,w_n$, such that $w_i^n=z$. These numbers are given by adding $(2\pi i)\frac{k}{n}$ to the angle, for various $k$:

$$\begin{array}{cc}{w}_1& =\sqrt[n]{r}\cdot e^{\frac{\theta }{n}i}\\ w_2& =\sqrt[n]{r}\cdot e^{\frac{\theta }{n}i+(2\pi i)\frac{1}{n}}\\ w_3& =\sqrt[n]{r}\cdot e^{\frac{\theta }{n}i+(2\pi i)\frac{2}{n}}\\ w_4& =\sqrt[n]{r}\cdot e^{\frac{\theta }{n}i+(2\pi i)\frac{3}{n}}\\ \dots & \\ w_n& =\sqrt[n]{r}\cdot e^{\frac{\theta }{n}i+(2\pi i)\frac{n-1}{n}}.\end{array}$$

Complex roots Mathlet