COMPLEX NUMBERS Consider the quadratic equation; x 0 It has no solutions in the real number system since x or x j j ie. j Similarly x 6 0 gives x 6 or x 6 j Powers of j j j j j. j j j j.j jj j j 5 j j j j 6 j. j j j 7 j. j j Clearly all the powers of j can be described by j, j, j, j. e.g. j 07 j 0. j j 576. j j
General form of complex numbers Consider now the quadratic equation x x 0 0 b ac 0 x 6 6 j j This leads us t oa general form for complex numbers z a bj a and b are real a real part of z Rez b imaginary part of z Imz In the previous example; roots z j, hence a, b For complex numbers : j,, 7j, 6j are all special cases a bj is purely real if b 0, a. a bj is purely imaginary if a 0, bj. Basic Rules of algebra: Consider : z a b j, z a b j Sum : z z a a b b j Difference : z z a a b b j Example : for : z j and : z 7j Product : z z 7j j z z 7j 7 j z. z a b ja b j a a a b j b a j b b j a a b b a b a b j Example : if z j and z j find z z z z z z z j j 5j j 7j
Complex Conjugate: For any complex number a bj there corresponds a complex number a bj obtained by changing the sign of the "imarigary part". a bj is the Complex Conjugate of a bj. Notation : If z a bj then z a bj Clearly and z z _ a z z bj, hence a z z_ Rez b z z_ Imz Also zz a bja bj a j b, hence zz a b, Quotient : To find z z where z a b j, z a b j z z z z z z z z z z Complex number in general form NB : to find the quotient z ie. by z z a b j a b j a b j a b j a a b b b a a b j a b j a a b b a b a b j a b a b in general a bj form we multiply above and below by the complex conjugate of the denominator
Equality Let and z a b j z a b j then z z when a a and b b ie and Rez Rez Imz Imz Complex Number zero : 0 0j a 0 and b 0 ie. both real part and imaginary part are zero. Example : find the real numbers Α and Β such that z z 0, given that z Αj and z Β 5j. z z Β Α 5j 0 if Β 0 or Β and Α 5 0 or Α 5 For complex numbers, z, z and z : Commutative property : z z z z and z z z z Associative property: z z z z z z and z z z z z z Distributive property : z z z z z z z Geometrical Representation: A complex number a b j can be represented either as a point a, b or as a position vector of a, b in the x y plane complex plane or z plane. x axis is called real axis y axis is called imaginary axis. This geographical representation is called Argand Diagram Example :
Modulus of a complex number Let z a bj, the modulus of z is denoted by z and z a b Example: For z j z 9 6 5 5 z length of the vector for z and is always 0 Also zz a bja bj a b z Polar form of z Point x, y represents z x yj, where x r cos Θ, y r sin Θ thus z r cos Θ j r sin Θ r cos Θ j sin Θ This is called the polar form of z Here r x y z Θ is called the "argument of z" denoted by Θ arg z The Θ arg z is not unique since Θ k k an integer produces another value of arg z. However Θ gives the principle value of arg z. Clearly r x y and Θ tan x y Also cos Θ x r sin Θ y r x x y y x y
Determination of pricipal Argument of z : Θ. z x jy in first quadrant x 0, Y 0 Θ argz tan x y 0 Θ Example : determine the modulus and argument of z j r z j Θ arg z tan 0.588radians. z x zy in second quadrant x 0, y 0 Θ argz tan x y, Θ Example : Determine z and arg z for z j r z j Since tan tan thus Θ argz rad.
. z x zy in third quadrant x 0, y 0 : Θ argz tan x, y Θ Example Determine the polar form of complex number r z z 6 j 6 j 6 Also tan x 8 y tan 6 tan 6 rad.
Example: Verify the inequailities z z z z Triangle inequality Also z z 5j 9 5 Polar form of z : 5 z z 9 z 8 cos 5 5 j sin 6 6 8 cos 5 5 j sin 6 6. z x jy in fourth quadrant x, y 0 : Θ arg z tan x y Θ 0 Example : find the priciple argument of z j. Θ arg z tan x y tan tan rad.
Exponential form of z : Recall Let x jθ refer to "Tables of mathmatical formulas" e x x x j x x... e jθ jθ jθ cos Θ jsin Θ Θ jθ jθ Θ... j Θ... Θ Θ5 5... Thus z r cos Θ jsin Θ has a new form z r e jθ the exponential form Important Points :. For the correct value in exponential form Θ must be in radians not in degrees.. Negative Θ is measured in the clockwise direction. Replacing Θ by Θ in e jθ cos Θ jsin Θ we get e jθ cos Θ sinθ, or e jθ cos Θ jsin Θ. e jθ and e jθ are complex cojugates hence z re jθ and z re jθ Consider e jθ cos Θ sin Θ and e jθ cos Θ sinθ Adding and substracting e jθ e jθ j sin Θ gives two important results : cos Θ ejθ e jθ sin Θ ejθ e jθ
Log of complex numbers : Let z re jθk k 0,,,... ln z ln re jθk ln r jθ k ln e ln r jθ k The principal value of ln z is ln z ln r jθ k 0 Example : Principal value of ln e j k is ln j 0.69 0.785j
Multiplication in polar form Let z r cos Θ j sin Θ z r cos Θ j sin Θ then z z r r cos Θ j sin Θ cos Θ j sin Θ r r cos Θ cos Θ sin Θ sin Θ j sin Θ cos Θ r r cosθ Θ jsin Θ Θ r r e jθ Θ Similarly z r cosθ Θ jsin Θ Θ z r cos Θ sin Θ r r e jθ Θ Clearly z z r r z z z z r r z z and arg z z Θ Θ arg z arg z arg z z Θ Θ arg z arg z Example : z z cos for z cos and z cos j sin j sin j sin cos 7 7 j sin and z z cos j sin
De Moivre s Theorem : If z r e jθ z r e jθ z n r n e jθ n then z z z n r r r n e j Θ Θ..Θ n r r r n cos Θ Θ.Θ n j sinθ Θ.Θ n Set then, z z... z n z r e jθ r cos Θ j sin Θ z n r n cos Θ j sin Θ n r n e jnθ r n cos nθ j sin nθ Set r to give De Moivre s Theorem cos Θ j sin Θ n cos nθ j sin nθ Application: If z then z cos Θ j sin Θ e jθ z cos Θ j sin Θ cos Θ j sin Θ cos Θ j sin Θ cos Θ j sin Θ cos Θ sin Θ cos Θ j sin Θ e jθ Hence and z n e jnθ cos nθ j sin nθ z n e jnθ cos nθ j sin nθ Adding and subtracting gives cos nθ z n z n and j sin nθ z n z n
Example : Evaluate i cos 6 ii j 5, j sin 6 iii j. 8 Solution : By De Moivre s Theorem i cos 6 j sin 6 cos 6 j sin 6 cos j sin j 0 j ii Let z j, then r z and Θ arg z tan Thus z 5 cos j sin 5 5 cos 5 j sin 5 5 cos 6 j sin 6 cos j sin j j iii Set z j thus from ii r z and Θ arg z tanh Thus z 8 z8 cos j sin 8 8 cos 8 j sin 8 cos j sin 6
Powers of cos Θ and sin Θ : Example : cos 6 Θ Binomial expansion gives z z cos 6 Θ 6 z6 6z 5z 0 5z 6z z 6 cos6θ 6 6 6 z6 6 z 6 z 5 z z 0 z cos 6Θ cos Θ 0 cos Θ 0 6 cos Θ 5 cos Θ 5 6 Example : Express cos 6Θ and sin 6Θ in terms of cos Θ and sin Θ Example : Sketch z j on an Argand Diagram Polar form of z is and evaluate z z arg z 6 z cos j sin 6 6 cos 6 e j 6 j sin 6 Hence z e j 6 e j cos j sin
Roots of complex numbers: Recall de Moivre s theorem; For z rcos Θ j sin Θ z n r n cos nθ j sin nθ ffor any value of n integer or fraction, positive or negative This is a very important result. When n is a fraction we are finding roots of complex numbers Consider w n z integer The n different solutions w 0, w, w,,..., w n of this equation are the n th roots of z denoted by n z or z n Let z rcos Θ j sin Θ and w Ρcos Φ j sin Φ Equation w n z gives Ρ n cos nφ j sin nφ rcos Θ j sin Θ Equality of two complex numbers in polar form means that; i their modulus are equal i.e. Ρ n r or Ρ r n ii their arguments, arg w n and arg z may differ by a multiple of, say k, k 0,,... Thus nφ Θ k or Θ k Φ, k 0,,,..., n n all other choicesof k duplicate the values of Φ Thus n th roots of z, w 0, w, w,,..., w n are given by w k r n cos Θ k n j sin Θ k n r n exp Θk j n k 0,,..., n
Example : Find all the cube roots of 8 Set z 8, then r 8 8, arg 8 Hence 8 8cos j sin and 8 8 exp 8 j k There are cube roots of w z 8 Thus the cube roots of z 8 are w 0 cos cos k j sin k j j k 0,, since n j sin w cos j sin 0 w cos 5 j j j sin 5 w 0, w, w are equally spaced radians 0 apart on circle of radius 8, centered at 0, 0
For w n z roots are spaced by n n rad and lie on a circle of radius r Example : Find the values of j There are two ways depending upon how you express j i j j ii j j i.e find cube roots of i.e find the cube roots of j and square them. Here it appears sensible to use method i i polar form of : r and Θ arg w k cos k j sin k k 0,, w 0 cos j sin j w cos j sin w cos 5 5 j sin j Exercise : Solve the equation w j