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2.3 Complex Numbers

Definition of Complex Numbers

Imaginary Components

Complex numbers expand the real numbers by adding the imaginary square root of -1

i= 1,i \stackrel{\triangle}{=}\ \sqrt{-1},

thus 9=3i,2=2i\sqrt{-9}=3i, \sqrt{-2}=\sqrt{2}i, etc.

Complex: Real plus Imaginary

Speaking more broadly, a complex number has both real and imaginary components. A complex number zz can be expressed as

z=x+yiz = x+yi

with xx being the real components and yy the imaginary. Of course, we can treat any purely real or imaginary number as a complex number by setting yy or xx to zero, respectively.

Complex Conjugate

A key property of complex number is the complex conjugate zz^{*} defined as

z=(a+bi)=abiz^{*} = (a+bi)^{*} = a-bi

i.e. the complex conjugate leaves the real components unchanged and flips the sign of the imaginary components. From this it immediately follows that any purely real number is its own complex conjugate.

Magnitude of Complex Numbers

The magnitude of a complex number, denoted as z|z|, is given as

z=a+bi=(a+bi)(a+bi)=(a+bi)(abi)=a2abi+abi+b2=a2+b2\begin{split} |z| &= |a+bi|\\ &= \sqrt{(a+bi)(a+bi)^{*}}\\ &= \sqrt{(a+bi)(a-bi)}\\ &= \sqrt{a^2-abi+abi+b^2}\\ &= \sqrt{a^2+b^2} \end{split}

where we have used the fact that b2i2=b2-b^2i^2=b^2.

Eq. (4) suggests that complex numbers may be viewed as vectors in the real plane with length z|z|.

Conversion between polar and Cartesian forms of complex numbers.

Figure 1:Conversion between polar and Cartesian forms of complex numbers.

You may recognize Figure 1 as an example of an isomorphism between C\mathbb{C} and R2\mathbb{R^2}. If you’re not familiar with the term isomorphism, you can consider it a fancy way of saying “mapping.”

Polar Form

Figure 1 suggests the polar form of complex with magnitude r=zr=|z| and angle θ=arctan(yx)\theta=\arctan{(\frac{y}{x})}, allowing the substitution

z=rcos(θ)+irsin(θ).z = r \cos{(\theta)} + ir\sin{(\theta)}.

While Eq. (5) may be suggestive as a heuristic, we will see in the next section that it actually touches upon a deep concept in mathematics.

Euler’s Formula: The “Jewel” of Mathematics

Euler’s formula, called “our jewel” by Richard Feynman, establishes a profound connection between trigonometry and complex numbers.

Euler’s Formula:

eiθ=cos(θ)+isin(θ)e^{i\theta} = \cos(\theta) + i \sin(\theta)

Or, multiplying through by rr for magnitudes different than 1:

reiθ=rcos(θ)+irsin(θ)r e^{i\theta} = r \cos(\theta) + i r \sin(\theta)

Euler’s Identity

When θ=π\theta=\pi, we have:

eiπ=1+i0e^{i\pi} = -1 + i \cdot 0

or

eiπ+1=0e^{i\pi} + 1 = 0

This is known as Euler’s identity, which relates five fundamental mathematical constants: ee, ii, π\pi, 1, and 0.

Proof of Euler’s Formula

Let f(θ)=cos(θ)+isin(θ)eiθf(\theta) = \frac{\cos(\theta) + i \sin(\theta)}{e^{i\theta}}

df(θ)dθ=ddθcos(θ)+isin(θ)eiθ=ddθeiθ(cos(θ)+isin(θ))=ieiθ(cos(θ)+isin(θ))+eiθ(sin(θ)+icos(θ))=eiθ(icos(θ)+sin(θ)sin(θ)+icos(θ))=eiθ(0)=0\begin{split} \frac{d f(\theta)}{d\theta} &= \frac{d}{d\theta} \frac{\cos(\theta) + i \sin(\theta)}{e^{i\theta}}\\ &= \frac{d}{d\theta} e^{-i \theta}(\cos(\theta) + i \sin(\theta))\\ &= -i e^{-i \theta}(\cos(\theta) + i \sin(\theta)) + e^{-i \theta}(-\sin(\theta) + i \cos(\theta))\\ &= e^{-i \theta}(-i\cos(\theta) + \sin(\theta) - \sin(\theta) + i \cos(\theta))\\ &= e^{-i \theta}(0)\\ &= 0 \end{split}

Since dfdθ=0\frac{df}{d\theta} = 0, f(θ)f(\theta) is a constant, and we need only establish its value at a single point.

At θ=0\theta = 0:

f(0)=1+0i1=1f(0) = \frac{1 + 0i}{1} = 1

Therefore f(θ)=1f(\theta) = 1 for all θ\theta, or:

eiθ=cos(θ)+isin(θ)e^{i\theta} = \cos(\theta) + i \sin(\theta)

Applications of Euler’s Formula

Solving Problems Involving ii

Euler’s formula can help solve otherwise extremely difficult problems involving ii. To see this, note that given that eiθ=cos(θ)+isin(θ)e^{i\theta} = \cos(\theta) + i \sin(\theta), at θ=π2\theta=\frac{\pi}{2} we have

eiπ2=cos(π2)+isin(π2)=0+1i=i\begin{split} e^{\frac{i\pi}{2}} &= \cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})\\ &=0+1i\\ &=i \end{split}

Solving iii^i

Eq. (16) allows us to solve problems that at first may seem intractable. For example, let us calculate iii^i:

ii=(eiπ2)i=ei2π2=eπ2.\begin{split} i^i &= (e^{\frac{i\pi}{2}})^i\\ &= e^{\frac{i^2\pi}{2}}\\ &= e^{-\frac{\pi}{2}}. \end{split}

Somewhat surprisingly, it turns out that iii^i is a real number slightly greater than 15\frac{1}{5}.

Sine and Cosine Addition Formulae

Use of Euler’s formula can simplify trigonometric derivations by transforming geometric problems into algebraic ones. A pair of trigonometric identities that we will make use of later in the book are the sine and cosine addition formulae, given as

sin(α+β)=sin(α)cos(β)+cos(α)sin(β)\sin{(\alpha + \beta)} = \sin{(\alpha)}\cos{(\beta)} + \cos{(\alpha)}\sin{(\beta)}

and

cos(α+β)=cos(α)cos(β)sin(α)sin(β).\cos{(\alpha + \beta)} = \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)}.

While these can be proven purely geometrically, Euler’s formula gives us an arguably far simpler algebraic method to prove them instead. Using (z)\Re(z) to denote the real component of zz and (z)\Im(z) to denote the imaginary component, we have

cos(α+β)=(ei(α+β))\cos{(\alpha + \beta)} = \Re(e^{i(\alpha + \beta)})

and

sin(α+β)=(ei(α+β)).\sin{(\alpha + \beta)} = \Im(e^{i(\alpha + \beta)}).

Note that

ei(α+β)=eiαeiβ=(cos(α)+isin(α))(cos(β)+isin(β))=cos(α)cos(β)sin(α)sin(β)+isin(α)cos(β)+icos(α)sin(β)=[cos(α)cos(β)sin(α)sin(β)]+i[sin(α)cos(β)+cos(α)sin(β)].\begin{split} e^{i(\alpha + \beta)} &= e^{i\alpha} e^{i\beta} \\ &= (\cos{(\alpha)} + i\sin{(\alpha)})(\cos{(\beta)} + i\sin{(\beta)})\\ &= \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)} \\ &\quad + i\sin{(\alpha)}\cos{(\beta)} + i\cos{(\alpha)}\sin{(\beta)}\\ &= [\cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)}]\\ &\quad + i[\sin{(\alpha)}\cos{(\beta)} + \cos{(\alpha)}\sin{(\beta)}]. \end{split}

Combining Eq. (22) with Eq. (23) we arrive at the sine addition formula

sin(α+β)=sin(α)cos(β)+cos(α)sin(β).\sin{(\alpha + \beta)} = \sin{(\alpha)}\cos{(\beta)} + \cos{(\alpha)}\sin{(\beta)}.

Similarly, combining Eq. (21) with Eq. (23) gives us the cosine addition formula

cos(α+β)=cos(α)cos(β)sin(α)sin(β).\cos{(\alpha + \beta)} = \cos{(\alpha)}\cos{(\beta)} - \sin{(\alpha)}\sin{(\beta)}.