Transcript Complex Numbers Rough Guide

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
bi
z = a + bi
= r(cosq + i sinq)
z = a + 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
b
arctan a +
0
for a > 0
b>0
I

2
for a < 0
for a > 0
b<0
II, III
IV
q=
/2
for a = 0, b > 0
3/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
a22 + b22
+ i
a2b1 – a1b2
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
Euler’s Equation: z = cosq + i sinq = eiq
z = a + ib = r(cosq + isinq) = reiq = elnr + iq
Multiplication: z1 z2 = r1 r2 ei( q1 + q 2 )
z1
r1
i( q  q )
=
z2
r2 e 1 2
Division:
zn = rneinq
Powers:
De Moivre Theorem: (cosq + i sinq)n = cosnq + isinnq
If z = r(cosq + i sinq) = r eiq then 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+2k) + isin(q+2k)) = rei(q+2k)
k = 0,1,2,3,…
The solutions of this equation are:
m
zk = √r
i(q+2k)
e m
q+2k
m
q+2k
= √r(cos m + isin m )
With k = 0,1,2, m-1.
sinq and cosq
cos(q)
=
e(iq)+e-(iq)
2
The hyperbolic functions
cos(iq) =
e(iiq)+e-(iiq)
= cosh(q) =
2
e-q+eq
2
sin(q)
=
e(iq) - e-(iq)
2i
e(iiq)-e-(iiq) i
(e(iiq)-e-(iiq))i
sin(iq) =
=
i
2i
sin(iq) =
2ii
(e-q - eq)i
-2
sin(iq) = isinhq
as sinhq =
(eq - e-q)
2
=
i (eq - e-q)
2