# Complex Exponentiation

**Complex exponentiation** extends the notion of exponents to the complex plane. That is, we would like to consider functions of the form \(e^z\) where \(z = x + iy\) is a complex number.

Why do we care about complex exponentiation? Although they are functions involving the imaginary number \(i = \sqrt{-1}\), complex exponentiation can be a powerful tool for analyzing a variety of applications in the real world. A couple of examples are as follows:

- Circuit analysis

- Motion of an object in an electromagnetic field

#### Contents

## Complex Numbers

This wiki assumes some familiarity with complex numbers \(z = x +iy,\) where \(x\) and \(y\) are real numbers and \(i\) is the imaginary number, \(i = \sqrt{-1}.\)

## The Complex Plane

The complex plane provides a way of visualizing the complex number \(z = x + i y\), where \(x\) and \(y\) are real numbers. It is similar to a Cartesian plane, where the \(x\)-axis represents the real part of a complex number, and the \(y\)-axis represents the imaginary part.

Consider the complex number \(z = x + i y\). As will be shown in the next couple of sections, this can also be represented as \(z = re^{i\theta},\) where \(r = \sqrt{x^2 + y^2}\) and \(\theta\) is the angle between the vector in the complex plane and the \(x\)-axis, as defined in this figure:

The

absolute valueof a complex number \(z = x + iy\) is given by \(\left |z \right | = \sqrt{x^2 + y^2}.\)

The

argument(orphase) of a complex number \(z = x + iy\) is given by \(\theta\) such that \(x = \left |z \right | \cos \theta\) and \(y = \left |z \right | \sin \theta.\)

From this, we can convert to the polar coordinates

- \(x = r \cos \theta\)
- \(y = r \sin \theta\).

Or, for the unit circle, we have

- \(x = \cos \theta\)
- \(y = \sin \theta\).

What is the

absolute valueof the complex number \(z = 6 + 8i\)?

The absolute value of a complex number is defined as \[\left |z \right | = \sqrt{x^2 + y^2}.\] So, \[\left |z \right | = \sqrt{6^2 + 8^2}= \sqrt{100} = 10.\ _\square\]

## Euler's Formula - Derivation

One of the most fundamental equations used in complex theory is Euler's formula, which relates the exponent of an imaginary number, \(e^{i\theta},\) to the two parametric equations we saw above for the unit circle in the complex plane:

- \(x = \cos \theta\)
- \(y = \sin \theta.\)

So, what can we do with \(e^{i\theta}?\) Well, for one we can consider taking the Maclaurin series expansion of it. The Maclaurin series expansion of a function is given by

\[ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n, \]

where \(f^{(n)}(0)\) denotes the \(n^\text{th}\) derivative of \(f(x)\) evaluated at \(x = 0\).

Or, writing out this sum, we get

\[f(\theta) = \sum_{n=0}^{\infty}f^{(n)}(0)\frac{\theta^{n}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}\theta^2+\cdots +\dfrac{f^{(k)}(0)}{k!}\theta^n+\cdots.\]

So, if we let \(f(\theta) = e^{i\theta}\), then taking the first few derivatives gives

\[\begin{align} f(\theta) = e^{i\theta} &\implies f(0) =1\\ f'(\theta) = ie^{i\theta} &\implies f'(0) =i\\ f''(\theta) = -e^{i\theta} &\implies f''(0) =-1\\ f'''(\theta) = -ie^{i\theta} &\implies f'''(0) =-i\\ f''''(\theta) = e^{i\theta} &\implies f''''(0) =1. \end{align}\]

So, a Maclaurin series expansion for \(f(\theta) = e^{i\theta}\) gives us

\[e^{i\theta} = 1 + i\theta - \frac{1}{2!}\theta^2 - i\frac{1}{3!}\theta^3 + \frac{1}{4!}\theta^4 + \cdots. \]

Now, if we group real and imaginary parts together, we have

\[e^{i\theta} = \left(1 - \frac{1}{2!}\theta^2 + \frac{1}{4!}\theta^4 + \cdots\right) + i\left(\theta - \frac{1}{3!}\theta^3 + \cdots\right). \]

One might immediately recognize that the series in the first set of parentheses \(\big(\)representing the real part of \(e^{i\theta}\big)\) is the expansion for \(\cos x,\) and that the series in the second set of parentheses \(\big(\)representing the imaginary part of \(e^{i\theta}\big)\) is the expansion for \(\sin x.\) Therefore,

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

This famous result is known as the Euler formula after the mathematician Leonhard Euler who discovered it in 1748, which is a very powerful tool when it comes to finding properties associated with complex numbers.

Show that the following identities hold:

\[\begin{align} \cos(x+y) &= \cos x \cos y - \sin x \sin y\\ \sin(x+y) &= \sin x \cos y + \cos x \sin y, \end{align}\]

using Euler's formula.

From Euler's formula, we know that

\[e^{i(x+y)} = \cos(x+y) + i \sin(x+y).\qquad (1)\]

Also, we have

\[\begin{align} e^{i(x+y)} &= e^{ix}e^{iy} \\ &= (\cos x + i\sin x)(\cos y + i\sin y)\\ &= \cos x \cos y - \sin x \sin y + i (\sin x \cos y + \cos x \sin y).\qquad (2) \end{align}\]

Equating the real and imaginary parts of (1) and (2) gives

\[\begin{align} \cos(x+y) &= \cos x \cos y - \sin x \sin y\\ \sin(x+y) &= \sin x \cos y + \cos x \sin y.\ _\square \end{align}\]

Show that \(e^{i\pi} = -1.\)

Using Euler's formula,

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

So,

\[\begin{align} e^{i\pi} &= \cos{\pi} + i \sin{\pi}\\ &= -1 + 0\\ &= -1.\ _\square \end{align}\]

Show that de Moivre's theorem is correct, namely that the following equality holds true:

\[ (\cos{\phi} + i \sin \phi)^n= \cos(n \phi) + i \sin(n \phi). \]

Using Euler's formula,

\[ (\cos{\phi} + i \sin \phi )^n = e^{i n\phi} = \cos(n\phi) + i \sin(n\phi).\ _\square\]

## Complex Exponentiation - Beyond Euler's Formula

We have seen that

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

Now let's consider again the following representation of a complex variable \(z = x + i y\):

If we consider \(r\) and \(\theta\), it is straightforward to see that

- \(r = \sqrt{x^2 + y^2}\)
- \(x = r \cos \theta\)
- \(y = r \sin \theta.\)

So, from Euler's equation, it follows that

\[z = x + i y = r(\cos\theta + i \sin\theta) = re^{i\theta}.\]

This makes multiplying two complex numbers together intuitive and easy to visualize.

Show that multiplying together two complex numbers is the same as adding together their angles and multiplying together their absolute values.

Lets suppose you have two complex numbers:

- \(z_1 = x_1 + i y_1\)
- \(z_2 = x_2 + i y_2.\)
You can convert them to polar coordinates, using the formulae:

- \(r_n = \sqrt{x_n^2 + y_n^2}\)
- \(x_n = r_n\cos\theta_n, \ y_n = r_n\sin\theta_n\)
This gives us the equations

- \(z_1 = r_1e^{i\theta_1}\)
- \(z_2 = r_2e^{i\theta_2}.\)
Multiplying,

\[z_1z_2 = r_1r_2e^{i(\theta_1 + \theta_2)}.\]

So, we can see that multiplying together two complex numbers in the complex plane is as easy as adding their angles together and multiplying their absolute values together. \(_\square\)

If \(z = 3 + 3i\), what is \(z^{10}\)?

This could be done in two ways:

- multiplying out \((3+3i)(3+3i) \cdots (3+3i)\) the long way;
- first converting to polar coordinates.
The first way is a bit tedious, and prone to error. However, converting to polar coordinates can make the task much simpler. In this case,

\[z = 3 + 3i.\]

Its polar coordinates give us

- \(r = 3\sqrt2\)
- \(\theta = \dfrac{\pi}{4}.\)
So, \(z^{10}\) is equivalent to raising the radius to the power of \(10\) and multiplying the angle in the exponent by \(10\), or

\[\begin{align} z &= re^{i\theta}\\\\ z^{10} &= r^{10}e^{i(10\theta)}\\ &= \big(3\sqrt2\big)^{10}e^{i\big(\frac{10\pi}{4}\big)}\\ &= 3^{10}2^{5}e^{2\pi i}e^{i\frac{\pi}{2}}\\ &= 3^{10}2^{5}i.\ _\square \end{align}\]

Note: From Euler's formula, \(e^{2\pi i} = 1\) and \(e^{i\frac{\pi}{2}} = i\).

## Raising a Complex Number to a Complex Number

Consider the general problem of raising a complex number to the power of a complex number. That is, if \(z\) and \(w\) are both complex numbers, what is \(z^w\)? In order to solve \(z^w\), where \(z\) and \(w\) are both complex numbers, we rewrite \(z^w\) as follows:

\[z^w = e^{w \ln z}.\]

So, how can we evaluate \(\ln z?\) Consider the following example:

Show that

\[\ln z = \ln |z| + i\big(2n\pi + \arg(z) \big).\]

We have

\[\begin{align} e^{\ln z}

&= e^{\ln |z| + i\big(2n\pi + \arg(z)\big) }\\ &= |z| e^{i2n\pi} e^{i \arg(z)}\\ &= |z| e^{ i \arg(z)}\\ &= |z| \Big(\cos\big(\arg(z)\big) + i\sin\big(\arg(z)\big)\Big)\\ &= \text{Re}(z) + i\, \text{Im}(z)\\ &= z. \end{align}\]Therefore, \(\ln z = \ln |z| + i\big(2n\pi + \arg(z) \big).\ _\square\)

Using the results from this example,

- \(z^w = e^{w\ln z}\)
- \(z^w = e^{ w\cdot \left(\ln |z| + i\big(2n\pi + arg(z) \big) \right)}\)
- \(z^w = e^ {w\left(\ln |z| + i \arg(z) \right)} e^{ iw(2n\pi)}.\)

The second exponential here, \(e^{ iw(2n\pi)},\) is of course multi-valued, and therefore, in general, if \(z\) and \(w\) are complex, then \(z^w\) is multi-valued, in fact it has an infinite number of possible values. We call \(n = 0\) the **principle** value.

## Complex Roots - Common Pitfalls

For real numbers it is straightforward to find the root of a number.

A couple of examples:

- If \(x^3 = 27,\) then \(x=3\). (one value)
- If \(x^2 = 25,\) then \(x = \pm 5\). (two values)

In general, for even \(n\) a number will have two roots, and for odd \(n\) it will have one root.

However, finding the roots of **complex** numbers is, well (pardon the expression!), much more complex!

In fact, in general, if we have \(z^n = w\), where \(w\) is a given complex number, it will have \(n\) distinct \(n^\text{th}\) roots.

Let's consider an example:

How many complex cube roots are there for the number \(1:\)

\[z^3 = 1?\]

The obvious answer, of course, is \(1\). But are there any others?

Recall that a complex number can be written in polar form: \(z = re^{i\theta}.\) In this case \(r = 1\), so

\[z = e^{i\theta}\implies z^3 = e^{3i\theta}.\]

In general, \(1 = e^{2n\pi i},\) so

\[\begin{align} e^{2n\pi i}&= e^{3i\theta}\\ 2n \pi &= 3 \theta\\ \theta &= \dfrac{2n pi}{3}. \end{align}\]

In the complex plane, we can let \(0 \leq \theta < 2\pi.\) So, we have three roots, corresponding to, \(n = 0, 1\) and \(2:\)

\[\theta_0 = 0,\ \theta_1 = \dfrac{2\pi}{3},\ \theta_2 = \dfrac{4\pi}{3}.\]

So, the three roots are

\[\begin{align} z_0 &= e^{0} = 1\\ z_1 &= e^{i\frac{2\pi}{3}} = -\dfrac{1}{2} + \dfrac{\sqrt3}{2}i\\ z_2 &= e^{i\frac{4\pi}{3}} = -\dfrac{1}{2} - \dfrac{\sqrt3}{2}i. \end{align}\]

In the complex plane, notice all these lie on the unit circle:

They are called the three

roots of unity. \(_\square\)

We have seen that the three roots of unity for \(z^3 = 1\) are simply points on the unit circle in the complex plane at evenly distributed points, starting at \(z = 1\).

Similarly for \(z^n = 1\), we will have \(n\) evenly distributed points on the unit circle at angles \(\frac{2m\pi}{n}\), where \(m\) goes from \(0\) to \(n-1\).

For example for \(n = 5\) we will have this:

More generally, if we have \(z^n = r\), then we will have \(n\) evenly distributed points on a circle of radius \(\sqrt[n]{r}\), with argument \(\frac{2m\pi}{n}\), where \(m\) goes from \(0\) to \(n-1\).

The multi-valued nature of complex roots can lead to some apparent paradoxes and erroneous results.

For example, one might argue (incorrectly) that

\[-1 = i\cdot i = \sqrt{-1} \sqrt{-1} = \sqrt {(-1)\cdot (-1)} = 1.\]

The issue here, of course, is that you have multiple values for both \(\sqrt{-1}\) and \(\sqrt{1},\) so the above is strongly dependent on which branch you choose.

Another common pitfall is finding "the solution" to an exponential and assuming that it is the unique solution.

For example consider the following example:

Find all possible values of \(i^i\).

We know the famous eqution:

\[e^{i\pi} = -1.\]

One might conclude the following:

\[\begin{align} e^{i \pi \times i} &= -1^{i} \\ e^{-1 \times \pi} &= -1^{i} \\ e^{\frac{-\pi}{2}} &= -1^{\frac{i}{2}} \\ e^{-\frac{\pi}{2}} &= \left(-1^{\frac{1}{2}}\right)^{i} \\ \Rightarrow i^i &= e^{-\frac{\pi}{2}}. \end{align} \]

However, even though each of the steps appears to be valid, this is only a partial answer, only one of the many possible values for \(i^i \)!

A more complete approach might be to recognize that \(i\) can take on all the following values:

\[i = e^{i \big(\frac{\pi}{2} + 2n\pi\big)}. \]

So, if we raise it to the \(i^\text{th}\) power, we get

\[i^i = e^{- \big(\frac{\pi}{2} + 2n\pi\big)}. \]

This gives a much more complete picture of all the possible values of \(i^i\). \(_\square\)

So, we can see, through these couple of examples, how important it is to keep track of all possible complex roots in order to arrive at a complete (and correct) solution to a problem involving complex exponentials.

## AC Circuit Application

As mentioned earlier, complex exponentiation--in particular, assuming the results of the form \(e^{i\omega t}\) and its resulting properties--can be used to solve real world problems, even though they involve imaginary numbers.

Consider, for example, the following electrical circuit:

We would like to understand why a sine wave produced by the voltage source in the circuit on the left would produce the waveforms on the right.

In order to analyze this circuit, we first look at the differential equations of the above circuit. In this section, we will assume some understanding of the fundamental behavior of an RLC circuit.

In the above circuit, we have

\[V(t) = V_0e^{i\omega t},\]

where \(\omega\) is the frequency of the input voltage which will be sinusoidal.

**Note**: Electrical engineers will often use \(j\) to denote the imaginary number \(i = \sqrt{-1}\), so as to not confuse it with the variable \(I\) used for current. However, for the sake of continuity, we will continue to use \(i\).

Now, we consider the behavior of each element.

The voltage across the resistor will be

\[V_R = IR,\]

where \(I\) is the current \((\)in amperes, \(\text{A})\) and \(R\) is the resistance \((\)in ohms, \(\Omega).\)

The voltage across the inductor will be

\[V_L = L\frac{dI}{dt},\]

where \(L\) is the inductance of the inductor \((\)in henries, \(\text{H}),\) and \(\frac{dI}{dt}\) is the derivative of the current with respect to time.

The voltage across the capacitor will be

\[V_C = \frac{Q}{C},\]

where \(Q\) is the charge across the capacitor, and \(C\) is its capacitance \((\)in Farads, \(\text{F}).\)

Note that the current is given by

\[I = \frac{dQ}{dt}\implies Q = \int I \, dt.\]

For the differential equation for the above circuit, we simply add up the (time-dependent) voltages across each element, and the sum must be the input voltage. This translates to the following differential equation:

\[\begin{align} V_0e^{i\omega t} &= V_R + V_L + V_C\\ &= IR + L\frac{dI}{dt} + \int \frac{I}{C}dt. \end{align}\]

Now, how do we approach this integro-differential equation? This is where the magic of complex exponentiation comes in!

Note that differentiation of the function \(V_0e^{i\omega t}\) is essentially multiplying by \(i\omega\), and integration is essentially dividing by \(i\omega\). So, the above integro-differential equation can much more simply be written as

\[V_0 = IR + i\omega L + \frac{1}{i\omega C}.\]

Note that \(Z_L = i\omega L\) is called the **impedence** of \(L\), and \(Z_C = \frac{1}{i\omega C}\) the **impedence** of \(C\).

Rewriting,

\[V_0 = IR + i\left(\omega L - \frac{1}{\omega C}\right).\]

At this point, you might say, "Hey wait, this is nonsense, we still have imaginary numbers..." However, we can now treat this voltage as a vector in the complex plane that will have a magnitude and a phase.

Finally, the impedence of each element is now treated as a "complex resistance" as if we had three resistors in series.

So, the solution for the voltage can be obtained from the following complex quantities:

- \(V_R = \frac{Z_RV_0}{Z_R + Z_L + Z_C}\)
- \(V_L = \frac{Z_LV_0}{Z_R + Z_L + Z_C}\)
- \(V_C = \frac{Z_CV_0}{Z_R + Z_L + Z_C}.\)

We then can solve for the real and imaginary part of each waveform. Solving, we can find the real wave forms across each of the elements, where the amplitude of the voltage across each element, \(n\), will be given by

- \(A_n = |V_n| = \sqrt{\text{Re}(V_n)^2 + \text{Im}(V_n)^2}\)
- \(\tan\omega _n = \frac{\text{Im}(V_n)}{\text{Re}(V_n)}.\)

Or in other words,

\[V_n = \left|\frac{Z_nV_0}{Z_{tot}}\right|e^{i(\omega - \omega _n)t},\]

where \(Z_{tot} = Z_R + Z_L + Z_C.\)

Now if we just look at the real part of the result, we can completely determine the resulting waveforms. That is,

- for the resistor, the amplitude is \(\frac{R}{Z_{tot}}\) and it is in phase with the applied voltage;
- for the inductor, the amplitude is \(\frac{Z_L}{Z_{tot}}\) and it is out of phase by \(\omega _L\) with respect to the applied voltage;
- for the capacitor, the amplitude is \(\frac{Z_C}{Z_{tot}}\) and it is out of phase by \(\omega _C\) with respect to the applied voltage.

So, we have seen that by using complex exponentials, we are able to simplify and solve problems, the results of which are very real and tangible. That is, If we were to measure an ideal volt meter across the elements, the resulting waveform would look exactly as we described above.

## See Also

**Cite as:**Complex Exponentiation.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/complex-exponentiation/