Go back to EE-201 Page 1 Complex Numbers Review for EE-201 (Hadi saadat) In our numbering system, positive numbers correspond to distance measured along a horizontal line to the right. Negative numbers are represented to the left of origin. Numbers corresponding to distances along the line shown in Figure 1 are called real numbers and the horizontal axis is known as the real axis. −6−5−4−3−2−1 0121212 Figure 1. The system of real numbers. Å Consider a point c in a two-dimensional plane located a distance along a line at an angle , taken counterclockwise from the positive horizontal axis. ↑ b c(a, b) |M| θ → a Ý Figure 2. Graphical representation of a point in the Ü- plane (known as complex plane). a The projection of c along the horizontal axis or the so called real axis is shown by . This component is known b c as the real component of c. The projection on the vertical axis is shown by . Thus, we may show by its two ´a; bµ coordinates as c . To differentiate between the two components, it is unfortunately common practice to call the vertical component of c the imaginary component. In this context, the vertical axis is known as the imaginary axis. Operator j Ý Consider a circle of radius unity with center at the origin of the Ü- plane as shown in Figure 3. ↑ 1∠90 =j 1∠180 θ 1∠0 → j2 =−1 j4=1 1∠270=j3=−j Figure 3. Graphical representation of the operator j. 6 ¼ The radius along the positive real axis may be written as ½ . If this line of unity length is rotated counterclockwise Æ Æ 6 ½ 9¼ j by 9¼ , it will be located along the vertical axis, i.e., . This operation is represented by the operator ,andwe write Æ 6 j =½ 9¼ 2 Æ 9¼ This means that if any real number is multiplied by j it is rotated by in a counterclockwise direction. Rotating through another 9¼,wehave Æ Æ Æ 6 6 6 ´j µ´j µ=´½ 9¼ µ´½ 9¼ µ=½ ½8¼ = ½ i.e., ¾ j = ½ Taking the square root, we may write Ô j = ½ ¾ = ½ The equation j is a statement that the two operations are equivalent. However, and that is the beauty of the concept, in all algebraic computation imaginary numbers can be handled as if j had a numerical value. Æ Æ ¾7¼ Continuing the rotation by another 9¼ , i.e., a total of , we write ¿ ¾ Æ Æ 6 6 j =´j µ´j µ= j =½ ¾7¼ =½ 9¼ and 4 ¾ ¾ Æ Æ 6 6 j =´j µ´j µ=´ ½µ´ ½µ = ½ ¿6¼ =½ ¼ We can write the reciprocal operation ½ as j ½ ´j µ j j Æ 6 = = = j =½ 9¼ = ¾ j ´j µ´j µ j ½ Rectangular and Polar Forms c´a; bµ With the introduction of the operator j , the point in Figure 2 may be represented as c = a · jb This representation indicates that the real part, a, is measured along the real axis (the abscissa) and the so called imag- inary part, b, is reckoned along the imaginary axis (the ordinate). This representation is known as the rectangular form of a complex number c. = a · jb jÅ j The complex number c may also be represented as the length or magnitude of a line segment, ,atan angle, , as indicated in Figure 2. Thus, 6 c = a · jb = jÅ j 6 = jÅ j jÅ j c The form c is called the polar form, is the magnitude and is called the angle or the argument of . The conversion from rectangular to polar form can be deduced immediately from Figure 2 in conjunction with the Pythagorean theorem, i.e., Ô ¾ ¾ jÅ j = a · b and the angle is given by b ½ = ØaÒ a For conversion from polar to rectangular from, a = jÅ j cÓ× b = jÅ j ×iÒ Thus, c can be written as c = jÅ j´cÓ× · j ×iÒ µ 3 The Euler’s identity – Exponential Form Å j Consider a complex number with the Magnitude j =1, 6 c =½ =cÓ× · j ×iÒ Taking derivative of c with respect to ,resultin dc = ×iÒ · j cÓ× = j ´cÓ× · j ×iÒ µ d or dc = jc d Separating the variable, ½ dc = jd c Integrating the above equation, we get ÐÒ c = j · à c = ¼ where à is the constant of integration. The magnitude of is unity regardless of the angle, therefore, at , j ´¼µ · à à =¼ ÐÒ´½µ = ,or ,and ÐÒ c = j or j c = e c = cÓ× · j ×iÒ With c given by the equation ,theexponential form also known as the Euler’s identity is j e =cÓ× · j ×iÒ for an angle , we obtain j e =cÓ× j ×iÒ ×iÒ Adding and subtracting the above two equations lead to the representation for cÓ× and , j j e · e = cÓ× and ¾ j j e e ×iÒ = ¾j With the addition of the Euler’s identity, the three way of representing a complex number, rectangular, polar and exponential forms ar all equivalent and we may write j 6 c = a · jb = jÅ j = jÅ je Mathematical Operations £ = a · jb c The conjugate of a complex number c denoted by and is defined as £ c = a jb 6 = jÅ j or in polar form for c ,wehave £ 6 c = jÅ j 4 To add or subtract two complex numbers, we add (or subtract) their real parts and their imaginary parts. For two c = a · jb c = a · jb ½ ½ ¾ ¾ ¾ complex numbers designated by ½ and ,theirsumis c · c =´a · a µ·j ´b · b µ ½ ¾ ½ ¾ ½ ¾ ¾ c c j = ½ ¾ The multiplication of ½ and in rectangular form is obtained as follows (note ) c c =´a · jb µ´a · jb µ=a a b b · j ´a b · b a µ ½ ¾ ½ ½ ¾ ¾ ½ ¾ ½ ¾ ½ ¾ ½ ¾ 6 6 c = jÅ j c = jÅ j c c ½ ½ ¾ ¾ ¾ ½ ¾ or in polar form for ½ ,and , The multiplication of and is 6 6 c c = ´jÅ j µ´jÅ j µ ½ ¾ ½ ½ ¾ ¾ j j ½ ¾ = ´jÅ je µ´jÅ je µ ½ ¾ j ´ · µ ½ ¾ = jÅ jjÅ je ½ ¾ 6 = jÅ jjÅ j · ½ ¾ ½ ¾ £ = a · jb c = a jb if a complex number c is multiplied by its conjugate , the result is £ ¾ ¾ ¾ cc =´a · jbµ´a jbµ=a · b = jÅ j or in polar form £ ¾ 6 6 cc = jÅ j jÅ j = jÅ j c c ¾ For the division of ½ by in rectangular form, we multiply numerator and denominator by the conjugate of the denominator, this results in c ´a · jb µ ½ ½ ½ = c ´a · jb µ ¾ ¾ ¾ ´a · jb µ´a jb µ ½ ½ ¾ ¾ = ´a · jb µ´a jb µ ¾ ¾ ¾ ¾ a a · b b b a a b ½ ¾ ½ ¾ ½ ¾ ½ ¾ = · j ¾ ¾ ¾ ¾ a · b a · b ¾ ¾ ¾ ¾ 6 6 c = jÅ j c = jÅ j c c ½ ½ ¾ ¾ ¾ ½ ¾ or in polar form for ½ ,and , the division of by is 6 c jÅ j ½ ½ ½ = 6 c jÅ j ¾ ¾ ¾ j ½ jÅ je ½ = j ¾ jÅ je ¾ jÅ j ½ j ´ µ ½ ¾ = e jÅ j ¾ jÅ j ½ 6 = ½ ¾ jÅ j ¾ Example 1 For the complex numbers Æ Æ 6 6 6 6 c =¾¼ ¿6:87 ; c =4¼ 5¿:½¿ ; c =4¼·j 8¼; c =½¾ ; c =5 ; c =¿¼·j 4¼ ¾ ¿ 4 5 6 ½ and 6 6 c =´c · c · c µ=´c £ c c µ ¾ ¿ 4 5 6 Find ½ c c ¾ Substituting for the values and converting ½ and to rectangular form, we have Æ Æ 6 6 ¾¼ ¿6:87 ·4¼ 5¿:½¿ ·4¼·j 8¼ c = 6 6 ´½¾ µ´5 µ ´¿¼ · j 4¼µ 6 6 ´½6 · j ½¾µ · ´¾4 j ¿¾µ · ´4¼ · j 8¼µ 8¼ · j 6¼ = = ´6¼ · j ¼µ ´¿¼ · j 4¼µ ¿¼ j 4¼ Æ 6 ½¼¼ ¿6:87 Æ 6 = =¾ 9¼ = j ¾ Æ 6 5¼ 5¿:½¿ 5 Use MATLAB To evaluate the complex number c described in Example 1. In MATLAB, we use the following commands: c1 = 20*exp(j*36.87*pi/180); % In MATLAB angles must be in radian c2 = 40*exp(-j*53.13*pi/180); c3 = 40 + j*80; c4 = 12*exp(j*pi/6); c5 = 5*exp(-j*pi/6); c6 = 30 + j*40; c = (c1+c2+c3)/(c4*c5-c6) M=abs(c) % Magnitude theta = angle(c)*180/pi % Angle in degree Save in a file with extension m, and run to get the result c= 0.0000 + 2.0000i M= 2.0000 theta = 90.0000 Example 2 A complex function is described by ¾5¼¼´j!µ g = ´¾5 · j!µ´½¼¼ · j!µ ! Write an script m-file to evaluate the magnitude and phase angle of g for from 0 to 200 in step of 1, and obtain the magnitude and phase angle plots versus ! . We use the following statements: w=0:1:200; g= (2500*j*w)./((25+j*w).*(100+j*w)); %Array Multiplication & division use .* & ./ M=abs(g); % Magnitude theta = angle(g)*180/pi; % Angle in degree subplot(2,1,1), plot(w, M), grid ylabel(’M’), xlabel(’\omega’) subplot(2,1,2), plot(w, theta), grid ylabel(’\theta, degree’), xlabel(’\omega’) The result is 6 20 15 M 10 5 0 0 50 100 150 200 ω 100 50 0 , degree θ −50 −100 0 50 100 150 200 ω Figure 4. Magnitude and phase angle plots for complex function in Example 2. Homework Problems ! = ½½ ! = 5¼ ! = ½½¾ 1. Using your calculator evaluate g for the function in Example 2 for (a) ,(b) ,and(c) . Express your answers both in rectangular and polar forms. 2. A complex function is described by ½¼¼¼¼ g = ¾ ´½¼ · j!µ´½¼¼ ! ·½¼´j!µµ ! Write an script m-file to evaluate the magnitude and phase angle of g for from 0 to 50 in step of 1, and obtain the magnitude and phase angle plots versus ! .