Complex numbers – A very rough guide

1. Complex numbers – A very rough guide The imaginary unit : i2= -1 The complex number: z = a + bi Re(z)= a = real part Im(z) = b = imaginary part Argand Diagram Im(z) i z = a + bi = r(cosq + i sinq) z = a + bi bi r q a = Re(z) = r cosq b = Im(z) = r sinq r = |z| = √(a2+b2) Re(z) a z* Quadrant z* is the complex conjugate of z z* = a - bi • 0 for a > 0 I • b > 0 • for a < 0 II, III 2p for a > 0 IV b < 0 b a arctan + q = p/2 for a = 0, b > 0 3p/2 for a = 0, b < 0 Some basic Algebra of complex numbers with z1 = a1 + i b1, z2 = a2 + i b2 Addition z1± z2 = (a1 ± a2) + (b1 ± b2)i Multiplication z1 z2 = (a1 a2 - b1 b2) + (a1 b2 + a2 b1)i Division z1 z2 a1 a2 + b1 b2 a2b1 – a1b2 = + i a22 + b22 a22 + b22 Two complex numbers are identical when both their real and their imaginary parts are identical: a1 + b1i = a2 + b2i when a1 = a2 b1 = b2

2. z = cosq + i sinq = eiq Euler’s Equation: z = a + ib = r(cosq + isinq) = reiq = elnr + iq z1 z2 r1 r2 Multiplication: z1 z2 = r1 r2 ei( ) Division: = ei( ) Powers: zn = rneinq De Moivre Theorem: (cosq + i sinq)n = cosnq + isinnq If z = r(cosq + i sinq) = r eiqthen its complex conjugate z* = r(cosq - i sinq) = r e-iq Solution of the Equation zm = r(cosq + i sinq) - Roots Recall: zm = r(cos(q) + isin(q)) = reiq= r(cos(q+2pk)+ isin(q+2pk)) = rei(q+2pk) k = 0,1,2,3,… The solutions of this equation are: q+2pk m q+2pk m i(q+2pk) m m m zk = √r e = √r(cos + isin ) With k = 0,1,2, m-1. sinq and cosq e(iq)+e-(iq) e(iq) - e-(iq) cos(q) sin(q) = = 2 2i e(iiq)-e-(iiq) (e(iiq)-e-(iiq))i i i sin(iq) = = The hyperbolic functions 2i 2ii e(iiq)+e-(iiq) i (eq - e-q) cos(iq) (e-q - eq)i = sin(iq) = = 2 2 -2 e-q+eq = cosh(q) sin(iq) = isinhq = 2 (eq - e-q) as sinhq = 2