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COMPLEX NUMBERS

Reference: Croft & Davision, Chapter 14, p.621 http://www.math.utep.edu/sosmath 1. Introduction Extended the set of real numbers to find solutions of greater range of equations. Let j be a root of the equation Then and. COMPLEX NUMBERS. Example 1.1.

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COMPLEX NUMBERS

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  1. Reference: Croft & Davision, Chapter 14, p.621 http://www.math.utep.edu/sosmath 1. Introduction Extended the set of real numbers to find solutions of greater range of equations. Let j be a root of the equation Then and COMPLEX NUMBERS V-01

  2. Example 1.1 Write down (a) , (b) , (c) Solution: (a) (b) (c) The number , and are called imaginary number V-01

  3. Example 1.2 Simplify (a) , (b) Solution: or (a) (b) V-01

  4. Example 1.3 Solve Solution The solution are known as complex number. is a real part and (or ) is an imaginary part V-01

  5. 16th Century Italian Mathematician – Cardano z = a + bj (Rectangular Form) a: real number; real part of the complex number b: real number; imaginary part of the complex number bj: imaginary number Im a + bj b Re a Complex Number V-01

  6. Addition and Subtraction of Complex Numbers If and then 2. Algebra of Complex Numbers V-01

  7. If and , find and Solution: Example 2.1 V-01

  8. Multiplicationof complex numbers If and then V-01

  9. Find if and . Solution: Example 2.2 V-01

  10. Find if and . Solution: Example 2.3 V-01

  11. θ -b = a - bj Conjugate If then the complex conjugate of z is = and . Im z = a + bj b Example 2.4 If , find and . Solution: θ Re a V-01

  12. Division of two complex numbers i.e. V-01

  13. Simplify Solution: Example 2.5 V-01

  14. Simplify Solution: Example 2.6 V-01

  15. Example 2.7 Simplify Solution: V-01

  16. Equality of Complex Numbers If then and Example 2.7 Find the values of x and y if . Solution: V-01

  17. The representation of complex numbers by points in a plane is called an Argand diagram. Example 3.1 Represent , and on an Argand diagram. 3. Argand Diagram, Modulus and Argument V-01

  18. Imaginary axis 3 2 1 Real axis 3 1 2 -2 -3 -1 -1 -2 -3 Im a + bj b θ Re Remark: a Solution: The angle θ is called the argument V-01

  19. Find the modulus and argument of (a) , (b) , (c) and (d) . Example 3.2 Solution: Wrong ! V-01

  20. Argument of a complex numberThe argument of a complex number is the angle between the positive x-axis and the linerepresenting the complex number on an Argand diagram. It is denoted arg (z). Important Note !!! V-01

  21. Im a + bj b θ Re a 4. Polar Form, Product and Quotient in Polar Form which is the polar form expression V-01

  22. Express the complex number in polar form. Solution: 4 θ α -3 Example 4.1 Wrong! V-01

  23. A 3300 300 -300 C Example 4.2 Express in true polar form. Solution: V-01

  24. Let and . V-01

  25. Example 4.3 If and find and . Solution: V-01

  26. Express the conjugate of in true polar form. Solution: Example 4.4 V-01

  27. 5. Exponential FormTo derive the exponential form we shall need to refer to the power series expansions of cos x, sin x … V-01

  28. Euler’s Formula !!! V-01

  29. Define and  = = which is the exponential form expression and  is in radian. V-01

  30. Express the complex number and its conjugate in exponential form. Solution: Im a + bj b  θ Re which is the exponential form expression and  is in radian. a Example 5.1 V-01

  31. and which is the exponential form expression and  is in radian. Example 5.2 Find (a) and (b) . Solution: V-01

  32. De Moivre’s Theorem Example 6.1 Use De Moivre’s theorem to write in an alternative form.Solution V-01

  33. If z = r(cos  + j sin ), find z4 and use De Movire’s theorem to write your result in an alternative form. Solution: Example 6.2 z4 = r4 (cos  + j sin )4 = r4 (cos 4 + j sin 4) V-01

  34. More concise form: If z = r then zn = rnn For example, z4 = r44 Example 6.3 If z = 2/8 write down z4. Express your answer in both polar and Cartesian form. Solution (a) z4 = 24  (4)(/8) = 16 /2 • i) z4 = 16 (cos /2 + j sin /2) ii) a = 16 cos /2 = 0 b = 16 sin /2 = 16 V-01

  35. Example 6.4 If z = 3 (cos /12 + j sin /12) find z3 in Cartesian form Solution z3 = 33(cos(3)(/12) + j sin(3)(/12)) = 27(cos /4 + j sin /4) • a = 27cos /4 = 19.09 • b = 27sin /4 = 19.09 V-01

  36. Example 6.5 (a) Express z = 3 + 4j in polar form. (b) Hence, find (3 + 4j)10, leaving your answer in polar form. Solution V-01

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