Euler’s Formula Without Calculus, pt. 2 – Complex Exponents

ng. (inside joke, see part one.)

We’re trying to find a reasonable definition for

x^{a+b\imath}

with x, a, b real numbers. Continue with the previous strategy – demand the new definition satisfy the normal properties of exponentiation, and see what this leaves you with. The properties I’ll use are:

x^a*x^b = x^{a+b}
x^0 = 1 \, \, \, \, for x \neq 0
\left(x^a\right)^b = x^{ab}

Begin with

x^{a+b\imath} = x^a*x^{\imath b}

We can already do x^a, so we’ll just focus on

x^{\imath b} = \left(e^{ln(x)}\right)^{\imath b} = e^{\imath ln(x)b}

I did that funny thing with changing the base to e just because I want to focus just on complex exponents, and not have to worry about what the base is. If I understand how complex exponents work for one base, I understand them all. I chose e as the base because I know where we’re going with this, and it works out well in the end, but you can imagine that e is a nice base to work in for exponentiation because, after all, it is the natural logarithm.

Let’s define a new constant c so that

e^{\imath \ln(x) b} = e^{\imath c}

It’s true that we expanded the domain of numbers before to allow the complex numbers to extend the real numbers. That let us solve the problem

x = (-1)^{1/2},

but if we expand the complex numbers to allow a new type of number beyond that, just so we can answer

x = e^{\imath c},

where will the process end? (In fact, mathematicians and physicists sometimes expand the complex numbers to things like quaternions and spinors, both of which have some applications in physics, especially mechanics, which I don’t particularly understand.)

Let’s assume that e^{\imath c} is a complex number of some sort. Then in general

e^{\imath c} = a(c) + \imath b(c),

with a(c), b(c) real-valued functions of c. We’ll occasionally denote them by just a, b.

The rule that states

x^0 = 1

implies

e^0 = a(0) + \imath b(0) = 1

So

a(0) = 1

b(0) = 0.

Introduce a new complex exponential and multiply it by the one we already have.

e^{\imath c}*e^{\imath d} = e^{\imath (c+d)} = a(c+d) + \imath b(c+d)

But we should be able to multiply them the FOIL way, too.

e^{\imath c}*e^{\imath d} = \left( a(c) + \imath b(c) \right) \left( a(d) + \imath b(d) \right) = a(c)a(d) - b(c)b(d) + \imath \left( a(c)b(d) + a(d)b(c) \right)

Equating the real parts and imaginary parts of the two equations (which is clearly legal given the ordered pair interpretation of complex numbers) gives

a(c+d) = a(c)a(d) - b(c)b(d)

b(c+d) = a(c)b(d) + a(d)b(c)

Those are just the angle addition formulas for the sine and cosine function. Perhaps we should define

e^{\imath c} = \cos(c) + \imath \sin(c)?

One check on whether this is a good definition is to see if the initial values work out, since we previously identified

a(0) = 1 = \cos(0)

b(0) = 0 = \sin(0).

That’s all good. The other property we should check is whether

\left(e^{\imath c}\right)^d = (\cos(c) + \imath \sin(c))^d = e^{cd} = \cos(cd) + \imath \sin(cd)

That is just De Moivre’s formula, which can be (and originally was) proven without any calculus. Just use induction or the binomial theorem or something.

I suppose there’s a gap in checking this property, because the binomial theorem can only prove De Moivre’s formula for integer d. But we could probably get around this by thinking of d as a rational, using repeated multiplication in the numerator and De Moivre’s theorem together, and then taking everything to the power of the denominator to show that it all works out. I’m getting tired and won’t lose sleep over it.

So we have a nice guess, which is that (as it is normally written)

e^{\imath \theta} = \cos(\theta) + \imath \sin(\theta).

It’s called Euler’s Formula. It’s great because the trig identities are built into it (that might not be the only reason it’s great, but it’s a decent reason for students to like it). For example,

e^{\imath \theta}*e^{- \imath \theta} = 1 = (\cos(\theta) + \imath \sin(\theta))*(\cos(-\theta) + \imath \sin(-\theta)) = \cos^2(\theta) + \sin^2(\theta)

which is the Pythagorean theorem in trigonometry. I exploited that the cosine function is even and the sine function is odd, and simplified the algebra in one step.

We also have

sin(\theta) = \frac{e^{\imath \theta} - e^{-\imath \theta}}{2 \imath}

cos(\theta) = \frac{e^{\imath \theta} + e^{-\imath \theta}}{2},

which you can verify by simply plugging in the Euler’s formula. That can make calculus with trig functions much easier.

The question remains whether this is the only viable answer. There might, after all, be other definitions of a(c), b(c) that also satisfy all the conditions we put on them.

I never realized it before writing this post, but there are other such functions. For example, consider

a'(\theta) = \cos(2\theta),

b'(\theta)= \sin(2\theta).

Those functions will have all the right properties – initial values, adding exponents and multiply exponents-to-exponents. It has some mildly unaesthetic deficiencies. For example,

e^{\imath \theta} = \cos( \theta) + \imath \sin( \theta) \Rightarrow \textrm{cosh}(\theta) = \cos(\imath \theta).

while

e^{\imath \theta} = \cos(2 \theta) + \imath \sin(2 \theta) \Rightarrow \textrm{cosh}(\theta) = \cos(\imath \theta/2).

Which is certainly strange, but not illegal. So I can’t claim to have nailed down Euler’s formula completely without calculus, but we can narrow it down to a class of functions, of which Euler’s formula is obviously the most natural (you can actually prove it must be

a(\theta) = \cos(\alpha \theta)

for some \alpha by considering \theta = 0 and \theta = \theta_1, showing with the angle addition formulas that this completely fixes a(\theta), and finally showing that it corresponds to some choice of \alpha in the desired form. I won’t bore you with the details.)

With that in hand, you can take the polar representation of points in a plane to write an arbitrary complex number as

a + b\imath = Re^{\imath \theta}.

R = \sqrt{a^2 + b^2}.

\tan(\theta) = \frac{b}{a}.

Of course many values of \theta will fit the above prescription, so we’ll restrict to

0 \leq \theta < 2\pi.

With that,

(a + \imath b)^{c + \imath d} = (Re^{\imath \theta})^{c + \imath d} = R^ce^{-\theta d}*e^{\imath \ln(R)d + \theta c}

which shows that the complex numbers are closed under exponentiation (unlike the reals). For example

\imath^\imath = \left(e^{\imath \pi/2}\right)^{\imath} = e^{-\pi/2}.

Finally, be careful with zero.

a^\imath = e^{\ln(a)\imath}

but \ln(0) is undefined. In fact

lim_{a \to 0} a^\imath

is similarly undefined, because it never settles down, but just races faster and faster around the unit circle.

8 Responses to “Euler’s Formula Without Calculus, pt. 2 – Complex Exponents”

  1. Euler’s Formula Without Calculus, pt. 1 - Exponentiation « Arcsecond Says:

    […] As Simple As Improbable, But No Less Unconfusing « Propagation of Error Euler’s Formula Without Calculus, pt. 2 – […]

  2. Tommi Says:

    I’m not yet familiar with complex numbers, but reading these two posts gives good baseline and some good ways to remember things. Good work.

  3. meichenl Says:

    Glad to hear it!

  4. Nik Says:

    I was doing quite nicely until the last line. What do you mean by “races faster and faster around the unit circle.”

  5. meichenl Says:

    Hi Nikita,

    It’s a bit subtle, so I’ll try to clarify that point at the top of the next post.

  6. Nik Says:

    Thanks Mark.

  7. David Lewis Says:

    Very nice work.

    I would like to see the details of proving that

    a(\theta) = \cos(\alpha \theta)

    for some alpha. Is there a reference? In fact, I cannot find this treatment anywhere on the web. Is there one, or is it completely original with you?

  8. Colin Robinson Says:

    I found your article very thought provoking. Deriving Euler’s equation without calculus, or with as little calculus as possible, is certainly an interesting exercise.

    Re what you said about the relation between cos and cosh functions. You describe the statement “cosh θ = cos i θ / 2 ” as “strange but not illegal”. (I think actually the implied statement is “cosh θ = cos 2 i θ”, but that is by the way.) Whether such a statement is “illegal” or not depends on how we define the cos and cosh functions, and on how we define circular and hyperbolic angles which are the arguments of those functions.

    If circular angle is defined as twice the area of the section of the unit circle, and if hyperbolic angle is defined as twice the area of the section of the unit hyperbola, then for both sin and sinh, as the angle becomes small, sin θ approximates to θ, that is, sin θ / θ approaches a limit of 1, and sinh θ / θ does the same, as can be seen intuitively by drawing the unit circle and unit hyperbola on a sheet of paper. Extending that result into the realm of complex numbers, sin (i θ) becomes approximately equal to i θ when θ becomes small. All of which is consistent with the statement “sinh θ = – i sin i θ”. Deriving cos from sin and cosh from sinh by means of the formulas (sin θ)^2 + (cos θ)^2 = 1, and (cosh θ)^2 – (sinh θ)^2 = 1, the formula relating cosh to cos then has to be cosh θ = cosh i θ. Not “cosh θ = cos i θ / 2 ” or “cosh θ = cos 2 i θ” or any other such variation.

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