Maths Extension 2 - Complex Numbers

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Complex Number Rules

! z = x + iy

! z = x � iy

! A complex number is represented in thisform

! The conjugate, z , is x � iy

! z = rcis_____OR! z = r(cosθ + isinθ)

! The Modulus-Argument (Mod-Arg) formof representing a complex number

! r = 22 yx +

! θ = tan-1

xy

! The Modulus of a complex numberNote: |z| = r

! The Argument of a complex numberNote: arg z = θ

! |z1z2| = |z1||z2|

! 2

1

zz

||||

2

1

zz=

! arg(z1z2) = argz1 + argz2 π2±

! π2argargarg 212

1 ±−=

zz

zz

! (left) Multiplication and Division rules ofthe Modulus and the Argument

! (below) More complex number rules

! |z| = | z | = 22 yx +

! z + z = 2x

! z - z = 2yi

! 21 zz ± = 21 zz ±

! arg z = �arg z

! z z = |z|2 = | z |2 = x2 + y2

! 21zz = 1z 2z

!

2

1

zz =

2

1

zz

! z-1 = 2

1zz

z=

! zn + 1 = (z + 1)(zn-1 � zn-2 + � - z + 1)_______________n is odd! zn � 1 = (z - 1)(zn-1 + zn-2 + � + z + 1)_______________n is odd! zn � 1 = (z - 1)(z + 1)(zn-2 + zn-4 + � + z + 1)_________-n is even

! arg (zn) = n arg z

Maths Extension 2 - Complex Numbers

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General Ideas of Complex NumbersComplex Numbers are written in the form of a real part and an imaginary part

Complex Number = x + iyReal = xImaginary = y

i = 1−i2 = �1i3 = �ii4 = 1

Modulus|z| = r = 22 yx +

Argument

arg z = θ = tan-1

xy

Modulus-Argument form of a Complex Number

cosθ = rx sinθ =

ry

x = rcosθ y = rsinθ

z = r(cosθ + isinθ)z = rcisθ

iyxz +=∴= rcisθ + isinθ= r(cosθ + isinθ)= rcisθ

Euler�s Formula θθθ isincos ==ie

DeMoivre�s Theorem zn = rncis nθθθθθ nini n sincos)sin(cos +=+

P(x,y)

θ

r

Maths Extension 2 - Complex Numbers

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Proof by Mathematical Inductionθθθθ nini n sincos)sin(cos +=+

Let n = 1LHS

_ 1)sin(cos θθ i+= θθ sincos i+

RHS= θθ )1sin()1cos( i+= θθ sincos i+

∴ true for n = 1

Assume true for n = k_ ki )sin(cos θθ + = θθ kik sincos +

Let n = k + 1_ 1)sin(cos ++ ki θθ= kii )sin(cos)sin(cos 1 ϑθθθ ++= )sin)(cossin(cos ϑθθθ kiki ++= θθθθθθθθ cossinsincossinsincoscos ikikk ++−= )sin()cos( θθθθ kik +++= θθ )1sin()1cos( +++ kik= RHS

= θθ )1sin()1cos( +++ kik

∴ true for n = k + 1

True for n = 1True for n = kTrue for n = k + 1∴ True for all positive integer values of n

Maths Extension 2 - Complex Numbers

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Expressing Complex Numbers in Mod-Arg Form

�3 + 3i

Modz = 22 yx +

= 99 += 18= 23

Arg

θ = tan-1

xy

= tan-1

−33

= tan-1 �1= 135°

=4

3π

Expressing Complex Numbers in x + iy form

232 πcis

=

+

23sin

23cos2 ππ i

=

+−

21

212 i

= �1 + i

Express in x + iy form from a quadratic formula

x =2

411 −±−

= 2

131 −×±−

=2

31 i±−

θ

z = rcisθ

+=+−∴

43sin

43cos2333 ππ ii

= i23

21 +−

= i23

21 −−

Conjugate pairs

Maths Extension 2 - Complex Numbers

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Theory Example

Addition(a + ib) + (c + id) = a + ib + c + id

= (a + c) + i(b + d)(3 + 4i) + (2 + 5i) = 3 + 4i + 2 + 5i

= 5 + 9i

Subtraction(a + ib) � (c + id) = a + ib � c � id

= (a � c) + i(b � d)(3 + 4i) - (2 + 5i) = 3 + 4i � 2 � 5i

= 1 � i

Multiplication(a + ib)(c + id) = ac + iad + ibc � bd

= (ac � bd) + i(ad + bc)(3 + 4i)(2 + 5i) = 6 + 15i + 8i � 20

= �14 + 23i

Division

idciba

++ =

idcidc

idciba

−−×

++

= 22 dcbdibciadac

+++−

= 2222

)()(dcadbci

dcbdac

−−+

++

ii

5243

++ =

ii

ii

5252

5243

−−×

++

= 254

208156+

++− ii

= 297

2926 i−

Maths Extension 2 - Complex Numbers

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Finding Square Roots of Complex Numbers

iyx + = a + ibx + iy = (a + ib)2

= a2 � b2 + 2aib

∴ x = a2 � b2

y = 2aib

Eg. Find the Square Root of 5 + 12i

i125 + = a + ib5 + 12i = a2 � b2 + 2aib

5 = a2 � b2 12 = 2ab6 = ab

a = b6

5 = 2

36b

� b2

5b2 = 36 � b4

0 = b4 + 5b2 � 360 = (b2 � 4)(b2 + 9)

2±=∴ b

! Simultaneous Equations! Find b

26=a

=3

26

−=a

= �3

Because square roots ofcomplex numbers come inconjugate pairs, i125 + =

! 3 + 2i____OR! �3 �2i

To test these, we just square them

(3 + 2i)2 = (3 + 2i)(3 + 2i)= 9 + 12i � 4= 5 + 12i

(�3 �2i)2 = (�3 �2i)(�3 �2i)= 9 + 12i � 4= 5 + 12i

Maths Extension 2 - Complex Numbers

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Complex Numbers on the Argand Diagram

Addition of VectorsAdd tip to tail|z1 + z2| ≤ |z1| + |z2|

Subtraction of VectorsFlip the tail, and do a normal addition|z1 � zz| ≥ |z1| � |z2|

z1 � zzz2 � z1

Multiplication of vectorsarg z3 = arg z1 + arg z2|z3| = |z1||z2|z4 = iz2

Multiplication of i, is a rotation through an

angle of 2π in the anti-clockwise direction.

Division of i, is a rotation through an angle of

2π in the clockwise direction.

Z1

Z2

Z1

Z2

Z3

Z4

Maths Extension 2 - Complex Numbers

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LOCUS|z � (a + ib)| = |z � (x + iy)|

|z � (�4 + 3i)| = |z � (�2 � 5i)|

by first principles:|x + iy + 4 � 3i| = |x + iy + 2 + 5i||(4 + x) + i(�3 + y)| = |(2 + x) + i(5 + y)|

( ) ( )22 34 yx +−++ = ( ) ( )22 52 yx +++256822 +−++ yxyx = 2910422 ++++ yxyx

256y8x + = 4104 ++ yxy16 = 44 −x

41

4−=∴ xy

|z � (a + ib)| = k k is the distance|z � 2| = 3

by first principles:|x + iy � 2| = 3

( ) 222 yx +− = 322)2( yx +− __= 9

θ=−−

)arg()arg(

bzaz

arg(z � 1) � arg(z + 1) = 4π ________OR

4)1arg()1arg( π=

+−

zz

Can either be 2 circlesCheck the angle ± ?Work out the locus by drawing a line andestimating the angle

A = arg(z + 1)B = arg(z � 1)

-4+3i

-2-5i

2

3

A B