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# The Functions e^x and e^jθ: The Function e^jθ and the Unit Circle

Module by: Louis Scharf. E-mail the author

## Note:

This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

Let's try to extend our definitions of the function e x e x to the argument x = j Θ x=jΘ. Then e j Θ e j Θ is the function

ejθ=limn(1+jθn)nejθ=limn(1+jθn)n
(1)

The complex number 1+jθn1+jθn is illustrated in Figure 1. The radius to the point 1+jθn1+jθn is r=(1+θ2n2)1/2r=(1+θ2n2)1/2 and the angle is φ=tan-1θnφ=tan-1θn This means that the nthnth power of 1+jθn1+jθn has radius rn=(1+θ2n2)n/2rn=(1+θ2n2)n/2 and angle nφ=ntan-1θnnφ=ntan-1θn (Recall our study of powers of z.z.) Therefore the complex number (1+jθn)n(1+jθn)n may be written as

(1+jθn)n=(1+θ2n2)n/2[cos(ntan-1θn)+jsin(ntan-1θn)].(1+jθn)n=(1+θ2n2)n/2[cos(ntan-1θn)+jsin(ntan-1θn)].
(2)

For nn large, (1+θ2n2)n/221(1+θ2n2)n/221, and ntan-1θnnθn=θntan-1θnnθn=θ. Therefore (1+jθn)n(1+jθn)n is approximately

( 1 + j θ n ) n = 1 ( cos θ + j sin θ ) . ( 1 + j θ n ) n = 1 ( cos θ + j sinθ).
(3)

(1+jθn)n=1(1+jθn)n=1( cosθ+jcosθ+j sin θ).

This finding is consistent with our previous definition of ejθejθ !

The series expansion for ejθejθ is obtained by evaluating Taylor's formula at x=jθx=jθ:

ejθ=n=01n!(jθ)n.ejθ=n=01n!(jθ)n.
(4)

When this series expansion for ejθejθ is written out, we have the formula

e j θ = n = 0 1 ( 2 n ) ! ( j θ ) 2 n + n = 0 1 ( 2 n + 1 ) ! ( j θ ) 2 n + 1 = n = 0 ( - 1 ) n ( 2 n ) ! θ 2 n + j n = 0 ( - 1 ) n ( 2 n + 1 ) ! θ 2 n + 1 . e j θ = n = 0 1 ( 2 n ) ! ( j θ ) 2 n + n = 0 1 ( 2 n + 1 ) ! ( j θ ) 2 n + 1 = n = 0 ( - 1 ) n ( 2 n ) ! θ 2 n + j n = 0 ( - 1 ) n ( 2 n + 1 ) ! θ 2 n + 1 .
(5)

It is now clear that cosθcosθ and sinθsinθ have the series expansions

cos θ = n = 0 ( - 1 ) n ( 2 n ) ! θ 2 n cos θ = n = 0 ( - 1 ) n ( 2 n ) ! θ 2 n
(6)
sin θ = n = 0 ( - 1 ) n ( 2 n + 1 ) ! θ 2 n + 1 . sin θ = n = 0 ( - 1 ) n ( 2 n + 1 ) ! θ 2 n + 1 .
(7)

When these infinite sums are truncated at N-1N-1, then we say that we have N-term approximations for cosθcosθ and sinθsinθ:

cos θ n = 0 N - 1 ( - 1 ) n ( 2 n ) ! θ 2 n cos θ n = 0 N - 1 ( - 1 ) n ( 2 n ) ! θ 2 n
(8)
sin θ n = 0 N - 1 ( - 1 ) n ( 2 n + 1 ) ! θ 2 n + 1 . sin θ n = 0 N - 1 ( - 1 ) n ( 2 n + 1 ) ! θ 2 n + 1 .
(9)

The ten-term approximations to cosθcosθ and sinθsinθ are plotted over exact expressions for cosθcosθ and sinθsinθ in Figure 2. The approximations are very good over one period (0θ2π)(0θ2π), but they diverge outside this interval. For more accurate approximations over a larger range of θ'sθ's, we would need to use more terms. Or, better yet, we could use the fact that cosθcosθ and sinθsinθ are periodic in θθ. Then we could subtract as many multiples of 2π2π as we needed from θ θ to bring the result into the range [0,2π][0,2π] and use the ten-term approximations on this new variable. The new variable is called θ θ-modulo 2π2π.

## Exercise 1

Write out the first several terms in the series expansions for cosθcosθ and sinθsinθ.

Demo 2.1 (MATLAB). Create a MATLAB file containing the following demo MATLAB program that computes and plots two cycles of cosθcosθ and sin θ versus θ. You should observe Figure 3. Note that two cycles take in 2(2π)2(2π) radians, which is approximately 12 radians.

clg;
j = sqrt(-1);
theta = 0:2*pi/50:4*pi;
s = sin(theta);
c = cos(theta);
plot(theta,s);
ylabel('sine and cosine');
hold on
plot(theta,c);
hold off

## Exercise 2

(MATLAB) Write a MATLAB program to compute and plot the ten-term approximations to cos θ cosθ and sin θ sinθ for θ θ running from 0 to 2 ( 2 π ) 2 ( 2 π ) in steps of 2 π / 50 2 π / 50 . Compute and overplot exact expressions for cos θ cosθ and sin θ sinθ. You should observe a result like Figure 2.

The Unit Circle. The unit circle is defined to be the set of all complex numbers z z whose magnitudes are 1. This means that all the numbers on the unit circle may be written as z=ejθz=ejθ. We say that the unit circle consists of all numbers generated by the function z=ejθz=ejθ as θ θ varies from 0 to 2π2π. See Figure 4.

A Fundamental Symmetry. Let's consider the two complex numbers z 1 z 1 and 1z1*1z1*, illustrated in Figure 4. We call 1z1*the1z1*the “reflection of z z through the unit circle” (and vice versa). Note that z1=r1ejθ1z1=r1ejθ1 and 1z1*=1r1ejθ11z1*=1r1ejθ1. The complex numbers z1-ejθz1-ejθ and 1z1*-ejθ1z1*-ejθ are illustrated in Figure 4. The magnitude squared of each is

| z 1 - e j θ | 2 = ( z 1 - e j θ ) ( z 1 * - e - j θ ) | z 1 - e j θ | 2 = ( z 1 - e j θ ) ( z 1 * - e - j θ )
(10)
|1z1*-ejθ|2=(1z1*-ejθ)(1z1-e-jθ).|1z1*-ejθ|2=(1z1*-ejθ)(1z1-e-jθ).
(11)

The ratio of these magnitudes squared is

β 2 = ( z 1 - e j θ ) ( z 1 * - e - j θ ) ( 1 z 1 * - e j θ ) ( 1 z 1 - e - j θ ) β 2 = ( z 1 - e j θ ) ( z 1 * - e - j θ ) ( 1 z 1 * - e j θ ) ( 1 z 1 - e - j θ )
(12)

This ratio may be manipulated to show that it is independent of θ θ, meaning that the points z 1 z 1 and 1z1*1z1* maintain a constant relative distance from every point on the unit circle:

β 2 = e j θ ( e - j θ z 1 - 1 ) ( z 1 * e j θ - 1 ) e - j θ 1 z i ( 1 - e j θ z 1 * ) ( 1 - z 1 e - j θ ) 1 z 1 = | z 1 | 2 , independent of θ ! β 2 = e j θ ( e - j θ z 1 - 1 ) ( z 1 * e j θ - 1 ) e - j θ 1 z i ( 1 - e j θ z 1 * ) ( 1 - z 1 e - j θ ) 1 z 1 = | z 1 | 2 , independent of θ !
(13)

This result will be of paramount importance to you when you study digital filtering, antenna design, and communication theory.

## Exercise 3

Write the complex number z-ejθz-ejθ as rejφrejφ. What are rr and φφ?

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