Title: Default for 10.3 The Exponential Form of a Complex Number
Course: HELM Workbooks 2023-2024

Series expansions for exponential and trigonometric functions

We have, so far, considered two ways of representing a complex number: \[\qquad z=a+i b\qquad \text{Cartesian form}\]or \[\qquad z=r(\cos\theta +i\sin\theta)\qquad \text{polar form}\]In this quiz we introduce a third way of denoting a complex number: the exponential form.

If \(x\) is a real number then, as we shall verify in a later quiz, the exponential number \(e\) raised to the power \(x\) can be written as a series of powers of \(x\):\[\qquad e^x=1+x+{x^2\over {2!}}+{x^3\over {3!}}+{x^4\over {4!}}+\cdots\]in which \(n!=n(n-1)(n-2)\dots (3)(2)(1)\) is the factorial of the integer \(n\). Although there are an infinite number of terms on the right-hand side, in any practical calculation we could only use a finite number. For example if we choose \(x=1\) (and taking only six terms) then\[\begin{aligned} e^1 &\approx 1+1+{1\over {2!}}+{1\over {3!}}+{1\over {4!}}+{1\over {5!}}\\&= 2+0.5+0.16666+0.04166+0.00833\\&=2.71666\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{aligned}\] which is fairly close to the accurate value of \(e=2.71828\) (to 5 d.p.)

We ask you to accept that \(e^x\), for any real value of \(x\), is the same as \(\displaystyle{1+x+{x^2\over {2!}}+{x^3\over {3!}}+\cdots }\) and that if we wish to calculate \(e^x\) for a particular value of \(x\) we will only take a finite number of terms in the series. Obviously the more terms we take in any particular calculation the more accurate will be our calculation.

As we shall also see later, similar series expansions exist for the trigonometric functions \(\sin x\) and \(\cos x\): \[\qquad \sin x=x-{x^3\over {3!}}+{x^5\over {5!}}-{x^7\over {7!}}+\cdots\]\[\qquad \cos x=1-{x^2\over {2!}}+{x^4\over {4!}}-{x^6\over {6!}}+\cdots\]in which \(x\) is measured in radians.

The observant reader will see that these two series for \(\sin x\) and \(\cos x\) are similar to the series for \(e^x\). Through the use of the symbol \(i\) (where \(i^2=-1)\) we will examine this close correspondence.

In the series for \(e^x\) replace \(x\) on both left-hand and right-hand sides by \(i\theta\) to give: \[\qquad e^{i\theta}=1+(i\theta)+{(i\theta)^2\over {2!}}+{(i\theta)^3\over {3!}}+{(i\theta)^4\over {4!}}+{(i\theta)^5\over {5!}}+\cdots\]Then, as usual, replace every occurrence of \(i^2\) by \(-1\) to give \[\qquad e^{i\theta}=1+i\theta-{\theta^2\over {2!}}-i{\theta^3\over {3!}}+{\theta^4\over {4!}}+i{\theta^5\over {5!}}+\cdots\]

which, when re-organised into real and imaginary terms gives, finally: \[\begin{aligned} e^{i\theta}&=\left[1-{\theta^2\over {2!}}+{\theta^4\over {4!}}-\cdots \right]+i\left[ \theta-{\theta^3\over {3!}}+{\theta^5\over {5!}}-\cdots \right]\\ &=\cos\theta +i\sin\theta.\end{aligned}\]

Key Point

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

Example

Find complex number expressions, in Cartesian form, for

1. \({e^{i\pi /4}}\qquad\)
2. \(e^{-i}\qquad\)
3. \(e^{i\pi}\)

Solution

We use the above Key Point:

1. \(\displaystyle{e^{i\pi/4} = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}}\)
2. \(e^{-i} = \cos{(-1)} + i\sin{(-1)} \approx 0.540 - i(0.841)\). Don't forget: use radians.
3. \(e^{i\pi} = \cos\pi + i\sin\pi = -1+i(0) = -1\)

Exercise

Express \( \displaystyle{z= [value]e^{i[value]}} \) in Cartesian form. Give your answer in decimal form, rounding to two decimal places where appropriate:

\(z=\quad\)\(\quad+\quad i\)
Feedback: We use the previous key point Key Point: \[ z = [value]e^{i[value]} = [value]\left(\cos[value] + i\sin[value]\right) = [value] + [value]i \]

Example

If \(z = re^{i\theta}\) and \(w = te^{i\phi}\) then find expressions for

1. \(z^{-1}\)
2. \(z^*\)
3. \(zw\)

Solution

1. If \(z = re^{i\theta}\) then \[z^{-1} = \frac{1}{re^{i\theta}} = \frac{1}{r}e^{-i\theta}\] using the normal rules for indices.
2. Working in polar form: if \(z = re^{i\theta} = r(\cos\theta + i\sin\theta) \) then \[z^* = r(\cos\theta - i\sin\theta) = r(cos(-\theta) + i\sin(-\theta)) = re^{-i\theta}\] since \(\cos(-\theta) = \cos{\theta}\) and \(\sin(-\theta) = -\sin{\theta}\). In fact this reflects the general rule: to find the complex conjugate of any expression simply replace the \(i\) by \(-i\) wherever it occurs in the expression.
3. \( zw = (re^{i\theta})(te^{i\phi}) = rte^{i\theta}e^{i\phi} = rte^{i\theta + i\phi} = rte^{i(\theta + \phi)} \) which is again the result we are familiar with: when complex numbers are multiplied their moduli multiply and their arguments add.

We see that in some circumstances the exponential form is even more convenient than the polar form since we need not worry about cumbersome trigonometric relations.

Exercise

Express the complex number \(z = \displaystyle[value]\)  in exponential form. The argument of \(z\) should be given precisely in terms of \(\pi\), i.e. do not give your answer in decimal form:

\(z=\quad\)
Feedback: We want to express \(z\) in the form \(re^{i\theta}\). We calculate \(r = \sqrt{([value])^2 + ([value])^2} = [value]\) and \(\theta = \tan^{-1}\left([value]/[value]\right) = [value]\). Therefore: \[z = [value]e^{[value]i}.\]

Hyperbolic and trigonometric functions

We have seen previously that \[\qquad e^{i\theta}=\cos\theta +i\sin\theta\] It follows from this that \[\qquad e^{-i\theta}=\cos (-\theta)+i\sin (-\theta)=\cos\theta -i\sin\theta\] Now if we add these two relations together we obtain \[\qquad \cos\theta ={e^{i\theta}+e^{-i\theta}\over {2}}\] whereas if we subtract the second from the first we have\[\qquad \sin\theta ={e^{i\theta}-e^{-i\theta}\over {2i}}\] These new relations are reminiscent of the hyperbolic functions introduced in a previous quiz. There we defined \(\cosh x\) and \(\sinh x\) in terms of the exponential function: \[\qquad \cosh x ={e^{x}+e^{-x}\over {2}}\qquad \sinh x ={e^{x}-e^{-x}\over {2}}\] In fact, if we replace \(x\) by \(i\theta\) in these last two equations we obtain \[\qquad \cosh (i\theta) ={e^{i\theta } +e^{-i\theta}\over 2}\equiv \cos \theta\qquad \text{and}\qquad \sinh (i\theta) ={e^{i\theta } -e^{-i\theta}\over 2}\equiv i\sin \theta\] Although, by our notation, we have implied that both \(x\) and \(\theta\) are real quantities in fact these expressions for \(\cosh\) and \(\sinh\) in terms of \(\cos\) and \(\sin\) are quite general.

Key Point

If \(z\) is any complex number then \[\cosh(iz) \equiv \cos z\quad\text{and}\quad\sinh(iz) \equiv i\sin z\] Equivalently, replacing \(z\) by \(iz\): \[\cosh z \equiv \cos(iz) \quad\text{and}\quad i\sinh z \equiv \sin(iz)\]

Exercise

Given that \(\cos^2z + \sin^2z \equiv 1\) for all \(z\) then, utilising complex numbers, obtain the equivalent identity for hyperbolic functions:

\(\quad\equiv1\)
Feedback: You should obtain \(\cosh^2z − \sinh^2z \equiv 1\) since, if we replace \(z\) by \(iz\) in the given identity then \(\cos^2(iz) + \sin^2(iz) \equiv 1\). But as noted in the previous Key Point \(\cos(iz) \equiv \cosh z\) and \(\sin(iz) \equiv i\sinh z\) so the result follows.

Further analysis similar to that in Exercise 3 leads to Osborne's rule.

Key Point

Hyperbolic function identities are obtained from trigonometric function identities by replacing \(\sin\theta\) by \(\sinh\theta\) and \(\cos\theta\) by \(\cosh\theta\) except that every occurrence of \(\sin^2\theta\) is replaced by \(−\sinh^2\theta\). This is known as Osborne's Rule.

Example

Use Osborne’s rule to obtain the hyperbolic identity equivalent to \(1 + \tan^2\theta \equiv \sec^2\theta\).

Solution

Here \(1 + \tan^2\theta \equiv \sec^2\theta\) is equivalent to \(\displaystyle{1 + \frac{\sin^2\theta}{\cos^2\theta} \equiv \frac{1}{\cos^2\theta}}\) . Hence if \[\sin^2\theta \mapsto −\sinh^2\theta\qquad\text{and}\qquad \cos^2\theta \mapsto \cosh^2\theta\] then we obtain \[1 − \frac{\sinh^2\theta}{\cosh^2\theta} \equiv \frac{1}{\cosh^2\theta}\qquad\text{or, equivalently,}\qquad 1 − \tanh^2\theta \equiv \text{sech}^2\theta.\]

Exercise

Two standard identities in trigonometry are

1. \(\sin2z\equiv2\sin z\cos z\)
2. \(\cos2z\equiv\cos^2z-\sin^2z\)

Use Osborne’s rule to obtain the corresponding identities for hyperbolic functions.

1. \(\sinh2z\quad\equiv\quad\)
2. \(\cosh2z\quad\equiv\quad\)
Feedback: This is a direct application of Osborne's rule. 1. \(\sinh2z\equiv2\sinh z\cosh z\) 2. \(\cosh2z\equiv\cosh^2z+\sinh^2z\)

Exercise

Express \(z=[value](a+ib)\) in Cartesian form:

\(z=\quad\) \(\quad+\quad\) \(\ \ i\)

Hint: the imaginary unit, \(i\), should not appear in your final expressions.

Therefore find the real and imaginary parts of \(z = [value](1 + [value]i)\). Express your answers in decimal form, rounding to 3 decimal places if necessary:

- \(\text{Re}(z)=\quad\)
- \(\text{Im}(z)=\quad\)
Feedback: We use the hyperbolic sum-angle identity, followed by Osborne's rule:\[[value](a+ib) \equiv [value](a)\cosh(ib) +[value](a)\sinh(ib) \equiv [value] + i\,[value].\] We then plug in \(a=1\) and \(b=[value]\) to get:\[[value](1+[value]i) = [value] + [value]i.\]

Exercise

Express the following complex numbers in Cartesian form

1. \([value]e^{i\pi /3}\)
2. \(e^{-[value]\pi i}\)
3. \(e^{i\pi/ [value] }e^{i\pi / [value] }\)

Your answers to (a) and (c) should be given in decimal form, rounding to 3 decimal places where appropriate:

1. \(\quad+\quad i\)
2. \(\quad+\quad i\)
3. \(\quad+\quad i\)
Feedback: The answers here are a direct application of the formula, \(e^{i\theta}\equiv\cos\theta+i\sin\theta\): 1. To 3 d.p.: \( [value]e^{i\pi/3} = [value](\cos(i\pi/3) + i\sin(i\pi/3)) = [value](0.5 + i(0.866)) = [value] + i[value] \) 2. \(e^{-[value]\pi i} = \cos(-[value]\pi) + i\sin(-[value]\pi) = \cos([value]\pi) - i\sin([value]\pi) = \cos(2\pi) + i\sin(2\pi) = 1\) 3. First we need to collect the indices, \[e^{i\pi/[value]}e^{i\pi/[value]} = e^{i\pi/[value] + i\pi/[value]} = e^{i\pi(1/[value] + 1/[value])} = e^{i[value]}.\] We then apply the aforementioned formula to this expression: \[e^{i[value]} = \cos([value]) + i\sin([value]) = [value] + [value]i.\]

Exercise

Find the exponential form of the complex number \(z = \displaystyle [value] + \displaystyle \left([value]\right)i\). Give the argument of \(z\) in terms of \(\pi\), i.e. not in decimal form.

\(z =\quad\)

Therefore, find the Cartesian form of \(z^{-1}\). Give your answers in decimal form, rounding to 3 decimal places if necessary:

\(z^{-1} =\quad\)\(\quad+\quad i\)
Feedback: We want to calculate \(z\) in the form \(re^{i\theta}\). We calculate \(\displaystyle{r = \sqrt{\left([value]\right)^2 + \left([value]\right)^2} = [value]}\) and \(\displaystyle{ \theta = \tan^{-1}\left([value]/[value]\right) = [value] }\). Therefore \[z = [value]e^{[value]i}.\] To calculate the inverse, \(z^{-1}\), in Cartesian form we use laws of indices and then apply a Key Point from this workbook: \[z^{-1} = [value]^{-1} = [value]^{-1}e^{-i([value])} = \dfrac{1}{[value]}\left(\cos\left(-[value]\right) + i\sin\left(-[value]\right)\right) = [value] + [value]i.\]

The exponential form

Since \(z=r(\cos\theta +i\sin\theta )\) and since \(e^{i\theta}=\cos\theta +i\sin\theta\) we therefore obtain another way in which to denote a complex number: \(z=re^{i\theta}\), called the exponential form.

Key Point

The exponential form of a complex number is \[z = re^{i\theta}\quad\text{in which}\quad r=|z|\quad\text{and}\quad\theta=\text{arg}(z)\] so \[z=re^{i\theta}=r(\cos\theta + i\sin\theta)\]

Feedback applied to an amplifier

Feedback is applied to an amplifier such that \[\qquad A'=\dfrac{A}{1-\beta A}\] where \(A', A\) and \(\beta\) are complex quantities. \(A\) is the amplifier gain, \(A'\) is the gain with feedback and \(\beta\) is the proportion of the output which has been fed back.

[Image: Diagram of amplifier with feedback]
An amplifier with feedback

1. If at 30 Hz, \(A=-500\) and \(\beta=0.005e^{8\pi i /9}\), calculate \(A'\) in exponential form.
2. At a particular frequency it is desired to have \(A'=300e^{5\pi i /9}\) where it is known that \(A=400e^{11\pi i /18}\). Find the value of \(\beta\) necessary to achieve this gain modification.

Mathematical statement of the problem

1. Substitute \(A=-500\) and \(\beta=0.005e^{8\pi i /9}\) into \(A'=\dfrac{A}{1-\beta A}\) in order to find \(A'\).
2. We need to solve for \(\beta\) when \(A'=300e^{5\pi i /9}\) and \(A=400e^{11\pi i /18}\).

Mathematical analysis

1. We have \[A'=\dfrac{A}{1-\beta A}=\dfrac{-500}{1-0.005e^{8\pi i /9}\times (-500)}=\dfrac{-500}{1+2.5e^{8\pi i /9}}\] Expressing the bottom line of this expression in Cartesian form this becomes: \[\qquad A'=\dfrac{-500}{1+2.5\cos\left(\dfrac{8\pi}{9}\right)+2.5i\sin\left(\dfrac{8\pi}{9}\right)}\approx \dfrac{-500}{-1.349+0.855i}\] Expressing both the top and bottom lines in exponential form we get: \[\qquad A'\approx\dfrac{500e^{i\pi}}{1.597e^{i 2.576}}\approx 313e^{0.566i}\]
2. We have \[A'=\dfrac{A}{1-\beta A}\rightarrow A'(1-\beta A)=A\rightarrow -\beta AA'=A-A'\] and \[\text{i.e.}\qquad \beta=\dfrac{A'-A}{AA'}\rightarrow\beta=\dfrac{1}{A}-\dfrac{1}{A'}\] So \[\beta=\dfrac{1}{A}-\dfrac{1}{A'}=\dfrac{1}{400e^{11\pi i /18}}-\dfrac{1}{300e^{5\pi i /9}}\approx 0.0025e^{-11\pi i/18}-0.00333e^{-5\pi i /9}\] Expressing both complex numbers in Cartesian form gives \[\beta = 0.0025\cos\left(-\dfrac{11\pi}{18}\right)+0.0025i\sin \left(-\dfrac{11\pi}{18}\right)\] \[\qquad-0.00333\cos \left(-\dfrac{5\pi}{9}\right)-0.00333i\sin \left(-\dfrac{5\pi}{9}\right)\] \[=-2.768\times 10^{-4}+9.3017\times 10^{-4}i=9.7048\times 10^{-4}e^{1.86i}\] So to 3 significant figures \(\beta=9.70\times 10^{-4}e^{1.86i}\)