Solution:

← Back to Fundamentals of Electric Circuits Overview In converting from rectangular to polar form using Eq. (B.5), we must exercise care in determining the correct value of . These are the four possibilities:

$$z = x + jy, \qquad \theta = \tan^{-1} \frac{y}{x}$$

(1st Quadrant)
\n

$$z = -x + jy, \qquad \theta = 180^{\circ} - \tan^{-1} \frac{y}{x}$$

(2nd Quadrant)
\n

$$z = -x - jy, \qquad \theta = 180^{\circ} + \tan^{-1} \frac{y}{x}$$

(3rd Quadrant)
\n

$$z = x - jy, \qquad \theta = 360^{\circ} - \tan^{-1} \frac{y}{x}$$

(4th Quadrant)

assuming that x and y are positive.

The third way of representing the complex z is the exponential form:

$$z = re^{j\theta} \tag{B.8}$$

This is almost the same as the polar form, because we use the same magnitude r and the angle .

The three forms of representing a complex number are summarized as follows.

$$z = x + jy, \qquad (x = r \cos \theta, y = r \sin \theta)$$

Rectangular form
\n

$$z = r/\theta, \qquad \left(r = \sqrt{x^2 + y^2}, \theta = \tan^{-1}\frac{y}{x}\right)$$

Polar form
\n

$$z = re^{j\theta}, \qquad \left(r = \sqrt{x^2 + y^2}, \theta = \tan^{-1}\frac{y}{x}\right)$$

Exponential form
\n(B.9)

The first two forms are related by Eqs. (B.5) and (B.6). In Section B.3 we will derive Euler’s formula, which proves that the third form is also equivalent to the first two.

Example B.1 Express the following complex numbers in polar and exponential form: (a) z1 = 6 + j8, (b) z2 = 6 − j8, (c) z3 = −6 + j8, (d) z4 = −6 − j8.

Solution:

Notice that we have deliberately chosen these complex numbers to fall in the four quadrants, as shown in Fig. B.2. (a) For z1 = 6 + j8 (1st quadrant),

$$r_1 = \sqrt{6^2 + 8^2} = 10$$

, $\theta_1 = \tan^{-1}\frac{8}{6} = 53.13^\circ$

Hence, the polar form is 10⧸ 53.13° and the exponential form is 10*e^j*53.13° . (b) For z2 = 6 − j8 (4th quadrant),

$$r_2 = \sqrt{6^2 + (-8)^2} = 10$$

, $\theta_2 = 360^\circ - \tan^{-1}\frac{8}{6} = 306.87^\circ$

In the exponential form, z = rej so that dz∕d = jre ^j = jz.

so that the polar form is 10 ⧸ 306.87° and the exponential form is 10e j306.87*°* . The angle 2 may also be taken as −53.13°, as shown in Fig. B.2, so that the polar form becomes 10 ⧸−53.13° and the exponential form becomes 10e⁻j53.13*°* .

(c) For z3 = −6 + j8 (2nd quadrant),

$$r_3 = \sqrt{(-6)^2 + 8^2} = 10$$

, $\theta_3 = 180^\circ - \tan^{-1}\frac{8}{6} = 126.87^\circ$

Hence, the polar form is 10⧸ 126.87° and the exponential form is 10e j126.87° . (d) For z4 = −6 − j8 (3rd quadrant),

$$r_4 = \sqrt{(-6)^2 + (-8)^2} = 10$$

, $\theta_4 = 180^\circ + \tan^{-1}\frac{8}{6} = 233.13^\circ$

so that the polar form is 10⧸ 233.13° and the exponential form is 10*e ^j*233.13° .

For Example B.1.

Convert the following complex numbers to polar and exponential Practice Problem B.1 forms: (a) z1 = 3 − j4, (b) z2 = 5 + j12, (c) z3 = −3 − j9, (d) z4 = −7 + j. Answer: (a) 5⧸ 306.9°, 5e ^j306.9° , (b) 13⧸ 67.38°, 13ej67.38° , (c) 9.487⧸ 251.6°, 9.487*e^j251.6° , (d) 7.071⧸ 171.9°, 7.071e ^j*171.9° .

Convert the following complex numbers into rectangular form: Example B.2 (a) 12⧸ −60°, (b) −50⧸ 285°, (c) 8e ^j10° , (d) 20e⁻jπ∕³ .

Solution:

(a) Using Eq. (B.6),

$$12\angle -60^{\circ} = 12\cos(-60^{\circ}) + j12\sin(-60^{\circ}) = 6 - j10.39$$

Note that = −60° is the same as = 360° − 60° = 300°. (b) We can write

$$-50/285^{\circ} = -50 \cos 285^{\circ} - j50 \sin 285^{\circ} = -12.94 + j48.3$$

(c) Similarly,

$$8e^{j10^{\circ}} = 8 \cos 10^{\circ} + j8 \sin 10^{\circ} = 7.878 + j1.389$$

(d) Finally,

$$20e^{-j\pi/3} = 20\cos(-\pi/3) + j20\sin(-\pi/3) = 10 - j17.32$$

Find the rectangular form of the following complex numbers: Practice Problem B.2 (a) −8⧸ 210°, (b) 40⧸ 305°, (c) 10e ⁻j30° , (d) 50e jπ∕² .

Answer: (a) 6.928 + j4, (b) 22.94 −j32.77, (c) 8.66 − j5, (d) j50.

We have used lightface notation for complex numbers—since they are not time- or frequency-dependent whereas we use boldface notation for phasors.

B.2 Mathematical Operations

Two complex numbers z1 = x1 + jy1 and z2 = x2 + jy2 are equal if and only if their real parts are equal and their imaginary parts are equal,

$$x_1 = x_2, \t y_1 = y_2 \t (B.10)$$

The complex conjugate of the complex number z = x + jy is

$$z^* = x - jy = r \angle -\theta = re^{-j\theta}$$

(B.11)

Thus, the complex conjugate of a complex number is found by replacing every j by −j.

Given two complex numbers z1 = x1 + jy1 = r1^⧸θ¹ and z2 = x2 + jy2 = r^{2 ⧸}θ2, their sum is

$$z_1 + z_2 = (x_1 + x_2) + j(y_1 + y_2)$$

(B.12)

and their difference is

$$z_1 - z_2 = (x_1 - x_2) + j(y_1 - y_2)$$

(B.13)

While it is more convenient to perform addition and subtraction of complex numbers in rectangular form, the product and quotient of the two complex numbers are best done in polar or exponential form. For their product,

$$z_1 z_2 = r_1 r_2 / \theta_1 + \theta_2 \tag{B.14}$$

Alternatively, using the rectangular form,

$$z_1 z_2 = (x_1 + jy_1)(x_2 + jy_2)$$

= $(x_1 x_2 - y_1 y_2) + j(x_1 y_2 + x_2 y_1)$ (B.15)

For their quotient,

$$\frac{z_1}{z_2} = \frac{r_1}{r_2} \underline{\beta_1 - \theta_2}$$

(B.16)

Alternatively, using the rectangular form,

$$\frac{z_1}{z_2} = \frac{x_1 + jy_1}{x_2 + jy_2}$$

(B.17)

We rationalize the denominator by multiplying both the numerator and denominator by z2*.

denominator by

$$z_2^*$$

.
\n

$$\frac{z_1}{z_2} = \frac{(x_1 + jy_1)(x_2 - jy_2)}{(x_2 + jy_2)(x_2 - jy_2)} = \frac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2} + \frac{jx_2y_1 - x_1y_2}{x_2^2 + y_2^2}$$

(B.18)

Example B.3 If A = 2 + j5, B = 4 ⁻ *^j*6, find: (a) A*(A + B), (b) (A + B)∕(A − B).

Solution:

(a) If A = 2 + j5, then A* = 2 − j5 and

$$A + B = (2 + 4) + j(5 - 6) = 6 - j$$

so that

$$A^*(A + B) = (2 - j5)(6 - j) = 12 - j2 - j30 - 5 = 7 - j32$$

(b) Similarly,

$$A - B = (2 - 4) + j(5 - -6) = -2 + j11$$

Hence,

Hence,
\n

$$\frac{A+B}{A-B} = \frac{6-j}{-2+j11} = \frac{(6-j)(-2-j11)}{(-2+j11)(-2-j11)}$$

\n

$$= \frac{-12 - j66 + j2 - 11}{(-2)^2 + 11^2} = \frac{-23 - j64}{125} = -0.184 - j0.512$$

Given that C = −3 + j 7 and D = 8 + j, calculate: Practice Problem B.3 (a) (C − D*)(C + D*), (b) D² ∕C*, (c) 2CD∕(C + D).

Answer: (a) −103 − j26, (b) −5.19 + j 6.776, (c) 6.045 + j11.53.

Evaluate: Example B.4

Evaluate:
\n(a)

$$\frac{(2+j5)(8e^{j10^{\circ}})}{2+j4+2(-40^{\circ})}$$

(b) $\frac{j(3-j4)^{*}}{(-1+j6)(2+j)^{2}}$

Solution:

(a) Because there are terms in polar and exponential forms, it may be best to express all terms in polar form:

$$2 + j5 = \sqrt{2^2 + 5^2} / \tan^{-1} 5/2 = 5.385 / 68.2^\circ$$ $$(2 + j5)(8e^{j10^\circ}) = (5.385 / 68.2^\circ)(8 / 10^\circ) = 43.08 / 78.2^\circ$$ $$2 + j4 + 2 / \frac{-40^\circ}{2} = 2 + j4 + 2 \cos(-40^\circ) + j2 \sin(-40^\circ)$$ $$= 3.532 + j2.714 = 4.454 / 37.54^\circ$$

Thus,

$$\frac{(2+j5)(8e^{j10^{\circ}})}{2+j4+2 \angle -40^{\circ}} = \frac{43.08 \angle 78.2^{\circ}}{4.454 \angle 37.54^{\circ}} = 9.672 \angle 40.66^{\circ}$$

(b) We can evaluate this in rectangular form, because all terms are in that form. But

$$j(3 - j4)* = j(3 + j4) = -4 + j3$$

\n

$$(2 + j)^2 = 4 + j4 - 1 = 3 + j4$$

\n

$$(-1 + j6)(2 + j)^2 = (-1 + j6)(3 + j4) = -3 - 4j + j18 - 24$$

\n

$$= -27 + j14$$

Hence,

$$= -27 + j14$$

Hence,

$$\frac{j(3 - j4)^{*}}{(-1 + j6)(2 + j)^{2}} = \frac{-4 + j3}{-27 + j14} = \frac{(-4 + j3)(-27 - j14)}{27^{2} + 14^{2}}$$ $$= \frac{108 + j56 - j81 + 42}{925} = 0.1622 - j0.027$$

Practice Problem B.4 Evaluate these complex fractions:

Evaluate these complex fractions:
\n(a)

$$\frac{6/30^{\circ} + j5 - 3}{-1 + j + 2e^{j45^{\circ}}}$$

\n(b)

$$\left[ \frac{(15 - j7)(3 + j2)^{*}}{(4 + j6)^{*}(3/70^{\circ})} \right]^{*}$$

Answer: (a) 3.387 ⧸ −5.615°, (b) 2.759 ⧸ −287.6°.

B.3 Euler’s Formula

Euler’s formula is an important result in complex variables. We derive it from the series expansion of ex , cos , and sin . We know that

$$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \cdots$$

(B.19)

Replacing x by j gives

$$e^{j\theta} = 1 + j\theta - \frac{\theta^2}{2!} - j\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots$$

(B.20)

Also,

$$\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \frac{\theta^6}{6!} + \cdots$$

\n

$$\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \frac{\theta^7}{7!} + \cdots$$

\n(B.21)

so that

$$\cos \theta + j \sin \theta = 1 + j\theta - \frac{\theta^2}{2!} - j\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + j\frac{\theta^5}{5!} - \cdots$$

(B.22)

Comparing Eqs. (B.20) and (B.22), we conclude that

$$e^{j\theta} = \cos\theta + j\sin\theta$$

(B.23)

This is known as Euler’s formula. The exponential form of representing a complex number as in Eq. (B.8) is based on Euler’s formula. From Eq. (B.23), notice thatθ

$$\cos \theta = \text{Re}(e^{j\theta}), \qquad \sin \theta = \text{Im}(e^{j\theta})$$

(B.24)

and that

$$|e^{j\theta}| = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1$$

Replacing by − in Eq. (B.23) gives

$$e^{-j\theta} = \cos\theta - j\sin\theta \tag{B.25}$$

Adding Eqs. (B.23) and (B.25) yields

$$\cos \theta = \frac{1}{2} (e^{j\theta} + e^{-j\theta})$$

(B.26)

Subtracting Eq. (B.25) from Eq. (B.23) yields

$$\sin \theta = \frac{1}{2j} (e^{j\theta} - e^{-j\theta})$$

(B.27)

Useful Identities

The following identities are useful in dealing with complex numbers. If z = x + jy = r⧸, then

$$zz^* = x^2 + y^2 = r^2$$

(B.28)

$$\sqrt{z} = \sqrt{x + jy} = \sqrt{r}e^{j\theta/2} = \sqrt{r} \angle{\theta/2}$$

(B.29)

$$zn = (x + jy)n = rn / n\theta = rn ejn\theta = rn (cos n\theta + j sin n\theta)$$

(B.30)

$$z^{1/n} = (x + jy)^{1/n} = r^{1/n} \sqrt{\theta/n + 2\pi k/n}$$

(B.31)

$$k = 0, 1, 2, ..., n - 1$$ $$\ln(re^{j\theta}) = \ln r + \ln e^{j\theta} = \ln r + j\theta + j2k\pi$$

(B.32)

$$(k = \text{integer})$$ $$\frac{1}{j} = -j$$

\n

$$e^{\pm j\pi} = -1$$

\n(B.33)\n

$$e^{\pm j2\pi} = 1$$

\n

$$e^{j\pi/2} = j$$

\n

$$e^{-j\pi/2} = -j$$

\n

$$Re(e^{(\alpha + j\omega)t}) = Re(e^{at}e^{j\omega t}) = e^{at} \cos \omega t$$

\n

$$Im(e^{(\alpha + j\omega)t}) = Im(e^{at}e^{j\omega t}) = e^{at} \sin \omega t$$

\n(B.34)

If A = 6 + j8, find: (a) ^√ Example B.5 __ A , (b) A⁴ .

Solution:

(a) First, convert A to polar form:

$$r = \sqrt{6^2 + 8^2} = 10$$

, $\theta = \tan^{-1} \frac{8}{6} = 53.13^\circ$ , $A = 10/53.13^\circ$

Then

$$\sqrt{A} = \sqrt{10}/53.13^{\circ}/2 = 3.162/26.56^{\circ}$$

(b) Because A = 10 ⧸ 53.13°,

$$A^4 = r^4 / 4\theta = 10^4 / 4 \times 53.13^\circ = 10,000 / 212.52^\circ$$

If A = 3 − j4, find: (a) A¹∕³

Answer: (a) 1.71 ⧸ 102.3°, 1.71 ⧸ 222.3°, 1.71 ⧸ 342.3°,

(b) 1.609 + j5.356 + j2nπ (n = 0, 1, 2, … ).

(3 roots), and (b) ln A. Practice Problem B.5

Appendix C

Mathematical Formulas

This appendix—by no means exhaustive—serves as a handy reference. It does contain all the formulas needed to solve circuit problems in this book.

C.1 Quadratic Formula

The roots of the quadratic equation ax² + bx + c = 0 are

$$x_1, x_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

C.2 Trigonometric Identities

1 csc x = ____ sin x cot x = _____ 1 tan x
(law of sines)
(law of cosines)
(law of tangents)
± cos x sin y
± y) = cos x cos y ∓ sin x sin y
− y) − cos(x + y)
2 sin x cos y = sin(x + y) + sin(x − y)
2 cos x cos y = cos(x + y) + cos(x − y)
1 ∓ tan x tan y

$$\cos 2x = \cos^2 x - \sin^2 x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$$

\n

$$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$$

\n

$$\sin^2 x = \frac{1}{2} (1 - \cos 2x)$$

\n

$$\cos^2 x = \frac{1}{2} (1 + \cos 2x)$$