1/23/2004 Review of Complex Arithmetic 1/9 A Review of Complex Arithmetic A complex value C has both a real and imaginary component: a = Re {C } and b = Im {C } so that we can express this complex value as: C = a + jb where j 2 = −1 . Just as a real value can be expressed as a point on the real line, a complex value can be expressed as a point on the complex plane: a Im {C } C = a + jb b Re {C } 1/23/2004 Review of Complex Arithmetic 2/9 The values (a,b) are a Cartesian representation of a point on the complex plane. Recall that we can alternatively denote a point on a 2-dimensional plane using polar coordinates: C distance from the origin to the point ∠C φc = rotation angle from the horizontal ( Re {C } ) axis i.e., Im {C } C = a + jb C b φc a Re {C } Using our knowledge of trigonometry, we can determine the relationship between the Cartesian (a,b) and polar ( C ,φc ) representations. From the Pythagorean theorem, we find that: C = a 2 + b2 1/23/2004 Review of Complex Arithmetic 3/9 Likewise, from the definition of sine (opposite over hypotenuse), we find: sin φc = b b = C a 2 + b2 or, using the definition of cosine (adjacent over hypotenuse): cos φc = a a = C a 2 + b2 Combining these results, we can determine the tangent (opposite over adjacent) of φc: tanφc = sinφc b = cosφc a Thus, we can write the polar coordinates in terms of the Cartesian coordinates: C = a 2 + b2 a b φc = tan −1 b a = cos −1 = sin −1 2 2 2 2 a +b a +b ( ) 1/23/2004 Review of Complex Arithmetic 4/9 Likewise, we can use trigonometry to write the Cartesian coordinates in terms of the polar coordinates. For example, we can use the definition of sine to determine b: b = C sinφc and the definition of cosine to determine a: a = C cosφc Summarizing: a = C cosφc b = C sinφc Note that we can explicitly write the complex value C in terms of its magnitude C and phase angle φc : C = a + jb = C cosφc + j C sinφc = C ( cosφc + j sinφc ) Hey! we can use Euler’s equation to simplify this further! 1/23/2004 Review of Complex Arithmetic 5/9 Recall that Euler’s equation states: e j φ = cosφ + j sinφ so complex value C is: C = a + jb = C ( cosφc + j sinφc ) = C e j φc Now we have two ways of expressing a complex value C! C = a + jb and/or C = C e j φc Note that both representations are equally valid mathematically—either one can be successfully used in complex analysis and computation. Typically, we find that the Cartesian representation is easiest to use if we are doing arithmetic calculations (e.g., addition and subtraction). 1/23/2004 Review of Complex Arithmetic 6/9 For example, if: C1 = a1 + j b1 then: C 2 = a2 + j b2 and C1 + C 2 = ( a1 + a2 ) + j (b1 + b2 ) C1 − C 2 = ( a1 − a2 ) + j (b1 − b2 ) Conversely, for geometric calculations (multiplication and division), it is easier to use the polar representation: For example, if: C1 = C1 e j φ then: C2 = C2 e j φ and 1 2 C1 C 2 = C1 e j φ C2 e j φ 1 2 = C1 C 2 e j φ1e j φ2 = C1 C2 e j (φ1 +φ2 ) and: jφ C1 C1 e = C2 C2 e j φ 1 2 C1 e j φ e − j φ = C2 1 = 2 C1 j (φ −φ ) e C2 1 2 1/23/2004 Review of Complex Arithmetic 7/9 Note in the above calculations we have used the general facts: x x =x y z ( y +z ) xy y −z =x( ) z x and Additionally, we note that powers and roots are most easily accomplished using the polar form of C: C n = ( C e j φc ) n =C (e j φc ) n n n = C e jnφc and n C =(C e 1 =C 1 =C Therefore: ( C = C e 2 j φc j φc ) 1 n (e j φc ) n n e ) 2 1 n j φc n 2 =C e j (2φc ) and: C =(C e j φc ) 1 2 = C e j φc 2 1/23/2004 Review of Complex Arithmetic 8/9 Finally, we define the complex conjugate (C*) of a complex value C: C ∗ Complex Conjugate of C = a − jb = C e − j φc A very important application of the complex conjugate is for determining the magnitude of a complex value: C 2 =C C ∗ Typically, the proof of this relationship is given as: C C ∗ = ( a + jb )( a − jb ) = a ( a − jb ) + jb ( a − jb ) = a 2 + jab − jba − j 2b 2 = a 2 + b2 =C 2 However, it is more easily shown as: 1/23/2004 Review of Complex Arithmetic 9/9 C C ∗ = ( C e j φc )( C e − j φc ) 2 =C e j (φc −φc ) 2 = C ej0 =C 2 Another important relationship involving complex conjugate is: C + C ∗ = ( a + jb ) + ( a − jb ) = ( a + a ) + j (b − b ) = 2a Thus, the sum of a complex value and its complex conjugate is a purely real value. Additionally, the difference of complex value and its complex conjugate results in a purely imaginary value: C − C ∗ = ( a + jb ) − ( a − jb ) = ( a − a ) + j (b + b ) = j 2b Note from these results we can derive the relationships: a = Re {C } = C +C ∗ 2 C −C ∗ b = Im {C } = j2