sscf crew sscm1023 03 ssch1023 complex-numbers_t2

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COMPLEX NUMBERS

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COMPLEX NUMBERS

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DEFINITION A complex number is made up of both real and imaginary components. It can be represented by an expression of the form (a+bi), where a and b are real numbers and i is imaginary. When defining i we say that i = . Then we can think of i2 as -1. In general, if c is any positive number, we would write:

icc

1

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If we have a complex number z, where z=a+bithen a would be the real component (denoted: Re z) and b would represent the imaginary component of z (denoted Im z). Thus the real component of z=4+3i is 4 and the imaginary component would be 3. From this, it is obvious that two complex numbers (a+bi) and (c+di) are equal if and only if a=c and b=d, that is, the real and imaginary components are equal.

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The complex number (a+bi) can also be represented by the ordered pair (a,b) and plotted on a special plane called the complex plane or the Argand Plane. On the ArgandPlane the horizontal axis is called the real axis and the vertical axis is called the imaginary axis. This is shown in the figure on the next slide:

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PROPERTIES OF THE COMPLEX NUMBERS

The SUM AND DIFFERENCE of complex numbers is defined by adding or subtracting their real components ie:

(a + bi)+(c + di) = (a + c) + (b + d)i(a - bi) - (c - di) = (a - c) - (b - d)i

Eg. (3 + i) + (1-7i) = (3 + 1 ) + (1 – 7)i = (4 – 6i)

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The communitive and distributive properties hold for the PRODUCT of complex numbersie: (a + bi)x(c + di) = a(c + di) + bci + bdi2

= ac + adi + bci + bdi2

We know: i2 = -1Therefore giving us(a + bi)(c + di) = (ac - bd) + (ad + bc)i

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CONJUGATESThe geometric interpretation of a complex conjugate is the reflection along the real axis. ExampleIf z = a+bi is a complex number, the conjugateof z would be………?

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When DIVIDING two complex numbers you are basically rationalizing the denominator of a rational expression. If we have a complex number defined as z =a+bi then the conjugate would be See the following example:Example:

biaii

form in the 37

3 Express

biaz

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i

ii

iiiii

ii

i

291

2912

58224

3737921

3737

373

37 of conjugate by ther denominato andnumerator hemultiply tmust We

22

2

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MODULUS & ARGUMENT

0 where

alsoImRe

Modulus

Consider

22

1

2

12121

22

22

zzz

zzandzzzz

zz

rbaibaz

ibaz

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Additional info:If z1, z2, z3,…..zn

• z1 + z2 = z2 + z1 commutative• z1 + (z2 + z3) = (z1 + z2) + z3 associative• z1z2 = z2z1 commutative• z1(z2z3) = (z1z2)z3 associative• If z1z2 = 1

z2 is the inverse of z1 z2 = (z1)-1

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Exercises:

2

21

12

4

3

121

31

21

321

3252)

)

843)

43)

determine,23

21,23,2 If)1

izzizzd

zc

zzzb

zza

iziziz

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Solutions

84844323232

824232)

15711611681236

2342343)

23223

23

22

21

iiiiii

iiib

iii

iizza

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3

22

2244

3

33

34

3

23

21

23

43

41

23

21

23

43

41

23

21

23

21

23

21 of conjugate

23

21;)

zii

ii

iiz

izz

izzc

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12525

34

4334

433443

3232252232

3252)

222

222

2

22

22

21

12

ii

ii

iiiiii

izzizzd

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idic

ibi

3) 26)

55) 32a)2

forms coordinatepolar in numberscomplex following the2)State

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State the following equations in conjugatecoordinate form. (Given that z = x + iy)

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DeMoivre's TheoremDeMoivre's Theorem is a generalized formula tocompute powers of a complex number in it'spolar form.Looking at

from the earlier formula we can find (z)(z) easily: ( )( ) ( )

( ) ( ) ( )θ3sini+θ3cosr=zz=z

thenθ2sini+θ2cosr=zz=z

323

22

( )θsini+θcosr=z

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Which brings us to DeMoivre's Theorem:Ifand n are positive integers then

Basically, in order to find the nth power of a complex number we take the nth power of the absolute value or length and multiply the argument by n.

( )

( )θnsini+θncosr=z

θsini+θcosr=z

nn

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3π5isin+

3π5cos231:

3πθ Therefore

313tan

231

31z Letting:

31:

55

5

iThen

z

iSolution

iFind

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Working backwards we can also use DeMoivre's Theorem to find the nth root. Let z = r(cos Ө + isin Ө ) and n be a positive integer. Then z has n distinct nth roots given by:

where k = 0, 1, 2, ... , n-1

nki

nkr n 2sin2cos

1

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n2πθα 2θnα

:period 2 a have cosine and sine Sincesinθsinnα and cosθcosnα

rs

isinθcosθrnαisin nα coss:Thenz wwhere

isinθcosθrz andisinαcosαslet w we thisshow To

n1

n

n

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iizw

izw

iizw

izw

nki

nkzw

and n,,,d when kave to finhen we h

zrorizSolution

f z roots oFind

nk

23sin

23cos

1sincos

2sin

2cos

140sin

40cos

2sin2cos

43 2 1 0T

01

0tan

101 011:

14

41

3

41

2

41

1

41

0

1

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The roots in Argand plane?

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i+32e1=

21i+

23e=

6πisin+

6πcose=ee=eThen

e:Evaluate

1-

1-6πi1-6

πi+1-

6πi+1-

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iii

i

ii

ii

eeie

ieieeb

eea

sincos

sincos2

sin)

2cos)

thatProve

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