sscf crew sscm1023 03 ssch1023 complex-numbers_t2
TRANSCRIPT
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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
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37 of conjugate by ther denominato andnumerator hemultiply tmust We
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MODULUS & ARGUMENT
0 where
alsoImRe
Modulus
Consider
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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
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determine,23
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izzizzd
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iziziz
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Solutions
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iizza
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idic
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3) 26)
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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
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and n,,,d when kave to finhen we h
zrorizSolution
f z roots oFind
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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
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eeie
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sincos
sincos2
sin)
2cos)
thatProve
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