Part 2 book “Engineering mathematics” has contents: Complex numbers, presentation of statistical data, measures of central tendency and dispersion, the binomial and poisson distribution, the normal distribution, differentiation of implicit functions, differentiation of implicit functions, integration using algebraic substitutions, and other contents. | Section 6 Complex Numbers This page intentionally left blank Chapter 35 Complex numbers Cartesian complex numbers (i) If the quadratic equation x 2 + 2x + 5 = 0 is solved using the quadratic formula then: (2)2 − (4)(1)(5) x= 2(1) √ √ −2 ± −16 −2 ± (16)(−1) = = 2 2 √ √ √ −2 ± 16 −1 −2 ± 4 −1 = = 2 2 √ = −1 ± 2 −1 √ It is not possible to evaluate −1 in real terms. √ However, if an operator j is defined as j = −1 then the solution may be expressed as x = −1 ± j2. −2 ± √ Since x 2 + 4 = 0 then x 2 = −4 and x = −4 ., (Note that ±j2 may also be written as ± 2j). Problem 2. Problem 1. Solve the quadratic equation: x2 + 4 = 0 Solve the quadratic equation: 2x 2 + 3x + 5 = 0 Using the quadratic formula, −3 ± (3)2 − 4(2)(5) 2(2) √ √ √ −3 ± −31 −3 ± −1 31 = = 4 4 √ −3 ± j 31 = 4 x= (ii) −1 + j2 and −1 − j2 are known as complex numbers. Both solutions are of the form a + jb, ‘a’ being termed the real part and jb the imaginary part. A complex number of the form a + jb is called a Cartesian complex number. (iii) In pure √ mathematics the symbol i is used to indicate −1 (i being the first letter of the word imaginary). However i is the symbol of electric current in engineering, and to avoid possible confusion the√ next letter in the alphabet, j, is used to represent −1 √ √ (−1)(4) = −1 4 = j(±2) √ = ± j2, (since j = −1) x= Hence √ 31 3 or − ± , x=− +j 4 4 correct to 3 decimal places. (Note, a graph of y = 2x 2 + 3x + 5 does not cross the x-axis and hence 2x 2 + 3x + 5 = 0 has no real roots). Problem 3. (a) j3 Evaluate (b) j4 (c) j23 (d) −4 j9 314 Engineering Mathematics (a) j3 = j2 × j = (−1) × j = −j, since j2 = −1 (b) j4 = j2 × j2 = (−1) × (−1) = 1 Imaginary axis B (c) j23 = j × j22 = j × ( j2 )11 = j × (−1)11 j4 j3 = j × (−1) = −j (d) j =j×1=j Hence A j2 j9 = j × j8 = j × ( j2 )4 = j × (−1)4 3 2 1 −4 −4 −4 −j 4j = = × = 2 j9 j j −j −j 4j = = 4 j or j4 −(−1) 0 j 1 2 3 Real axis j 2 j 3 D j 4 Now
