original power point math

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COMPLEX NUMBER



Complex numbers

Operation with complex numbers

Roots of negative numbers

Conjugate of a complex number



To differentiate between real numbers and complex

numbersTo do operations on complex numbersTo find the conjugate of complex number and solveproblem involving the conjugateTo solve the equation involving complex numbers

Learning outcomes



Complex numbers were introduced bymathematicians to solve equation such as x² = -1 .

Such equations have no real solutions since there isno real number whose square is a negative number.

By introducing i = √ -1 or i ² = -1 , we can solve

x² = -1 as follow.

THE COMPLEX NUMBER SYSTEM



A complex number is a number thathas the form a+bi , where a and bare real numbers and i is called asimaginary unit which i = √-1

DEFINITION



If z = a+bi , then a called the real part of z and b is

called the imaginary part of z and are denoted byRe{z} and Im{z} respectively. The symbol z, whichcan stand for any of set of complex numbers, iscalled complex variable.

Re{z} = aIm{z} = b

For example, if z = 2-5i = 2+(-5)i then Re{z} = 2 andIm{z} = -5



Every real number is a complex number with theimaginary part taking the value zero.

For example, 7 = 7+0i and - √2 = -√2+0i. Therefore, theintroduction of complex numbers extended the

number system.The real number R now become subset of the complexnumber set , symbol C.

NOTEThe imaginary part of z = a +bi is b not bi



EXAMPLEState the real and imaginary parts of the followingcomplex numbers.

(a) 5 + 8i(b) -3-√3i (c) 4 - √2i - √3 (d) -1.32



Solution :(a) Let z = 5 + 8iRe(z) = 5 , Im(z) = 8

(b) Let z = -3 - √3i z = -3 + (-√3) iRe(z) = -3 , Im(z) = -√3



(c) Let z = 4 - √2i - √3 z = (4 - √3) + ( - √2)i

Re(z) = 4 - √3 , Im(z) = -√2

(d) Let z = -1.32 = -1.32 + 0iRe(z) = - 1.32 , Im(z) = 0



Algebraic operation on complex numbers

Operations of complex numbers can be divided intofour :

AdditionSubtractionMultiplication

Division



ADDITION OF COMPLEX NUMBERS

To add two complex numbers, add the real parts and theimaginary parts separately.

( x + yi ) + ( u + vi ) = ( x + u ) + ( y + v )i



Example :

If Z 1 = 3 – 4i and Z 2 = 4 – 3i

Z1 + Z2 = 3 – 4i + 4 – 3i= 3 + 4 – 4i – 3i= 7 – 7i



SUBTRACTION OF COMPLEX

NUMBERS

To subtract two complex numbers, subtract the real parts andthe imaginary parts separately.

( x + yi ) -( u + vi ) = x + yi -u -vi = ( x -u ) + ( y -v )i



Example :

Z1 – Z2 = 3 – 4i – (4 – 3i)= 3 – 4 – 4i + 3i= -1 – i

I f Z 1 = 3 – 4i and Z 2 = 4 – 3i



MULTIPLICATION OF COMPLEX NUMBERSIn multiplying two complex numbers, follow the rule formultiplying two binomials and whenever the term i² isencountered, replace it with -1.

Thus :

( x + yi ) . ( u + vi ) = x(u + vi ) + yi (u + vi ) = xu + xvi + yui + yvi²= xu + ( xv + yu )i + yv ( -1 ) = ( xu - yv ) + ( xv + yu )i



Example :If Z 1 = 3 – 4i and Z 2 = 4 – 3i

Z1.Z2 = (3 - 4i)(4 - 3i)= 12 - 9i - 16i + 12i²= 12 - 25i + 12(-1)= -25i



DIVISION OF COMPLEX

NUMBERSIn dividing a complex number 2 + 8i by a real number 2 ( which is acomplex number with zero imaginary part ), we divide the real andimaginary part by 2.That is

( 2 + 8i ) ÷ 2

= 2 + 8i = 2 + 8 i = 1 + 4i

2 2 2



To divide ( 1 — i ) by ( 2 + i ), rewrite 1 — I to an2 + i

an equivalent expression where the denominator is a realnumber.

This is done by multiplying both numerator and

denominator by the conjugate of the denominator.



1 -i = 1 -i . 2- i = 1 ( 2 -i ) i(2 -i )2 + i 2 + i 2 -i 2² + 1²

= 2 -i -2i + i² = 2 -3i -1 = 1 -3i5 5 5 5

We usually write the result in the form a + bi



If Z 1 = 3 – 4i and Z 2 = 4 – 3i

Z1 = (3 – 4i) • (4 + 3i)Z2 (4 – 3i) (4 + 3i)

=12 + 9i – 16i – 12i²4² – (3i)²

=12 – 7i – 12(-1)4² – 3²(-1)

= 24 – 7i25

=24 _ 7 i25 25

Example :



COMPLEX CONJUGATEIf z = a + b i , then its complex conjugate denoted by

z* = a – b i .

The conjugate of a complex number is obtain by

changing the sign of the imaginary part.



EXAMPLES :

1. z = 2 – i ,

2. z = 3 + 8i ,

3. z = -8 + 6i ,

z* = 2 + i

z* = 3 – 8i

z* = -8 – 6i



NOTES

Any complex number a + b i has a complex conjugate

a – bi and

( a + b i )( a –bi ) = a²+b² is a real number. This fact is used

in simplifying expressions where the denominator of a

quotient is complex.



Product of a Complex Number

And Its Conjugate

( a + b i ) ( a – bi ) = a² – ab i + ab i – b² i²

= a² – b² i² ( real number )= a² + b²

Eg : ( 2 + 3i ) ( 2 – 3i ) = 4 – 9i = 4 + 9= 13



Conjugate of a complex numberconjugate of a complex number z = x + yi is z* or z = x – yi

EXAMPLE :z = 3 + 4i , z* = 3 – 4iz = 4 – 5i , z* = 4 + 5iz = 2i - 3 , z* = -2i – 3 or -3 -2i



If z = x +yi and z* = x – yizz* = ( x + yi )(x – yi)

= x²-(yi)²= x²-(y²i²)= x²-y²(-1)= x²+y² (a real number)



Solve Problem Involving The Conjugate

If z=1-2i , evaluatea) z + z*

=(1+2i) + (1-2i)

b) zz*=(1-2i)(1+2i)

= 1²+ 2²= 5



c) (z*) ² = (1+2i ) ²= -3+4i

d) (z²)* = [(1-2i ) ²]*= (-3-4i )*

= -3+4i

e) ( i z-4)(z*-6 i )= [i (1-2i )-4] [1+2i -6i ]= (i -2i ²-4)(1-4i )= (i -2)(1-4i )= 2+9i



Given Z1 = 2- √3i and Z 2 =-5 + i .

Find Z1 + 2Z2

Solution :

Z1 + 2Z2 = 2 - √3i + 2( -5 + i )= 2 - √3i - 10 + 2i= -8 + ( 2- √3 ) i

ADDITION :

PROBLEM SOLVING



SUBTRACTION :Given Z1 = 2 - √ 3i and Z 2 = -5 +i

Find Z1- Z2

Solution :

Z1 - Z2 = 2 - √3i -(- 5 + i )

= 2 - √ 3 i + 5 - i=7 - ( √3 + 1 ) i



MULTIPLICATION :Given v = - 6i , z = 5 -5i , w = -2 + 3i . find the following product.

(a) zw(b) zvw(c) v(z + w )

solution :

(a) zw = ( 5 - 5i ) ( - 2 + 3i ) = 5 ( - 2 + 3i ) - 5i ( - 2 + 3i )= - 10 + 15i + 10i - 15i²= - 10 + 15i + 10i - 15(- 1)= 5 + 25i



(b) zvw = ( 5 - 5i ) ( -6i ) (-2+ 3i ) = (-30i + 30i ) ( -2 + 3i )= ( -30i - 30 ) ( -2 + 3i ) = 60i - 90i + 60 - 90i

= 60i - 90 (-1) + 60 - 90i = 150 - 30i

(b) v( z + w ) = -6i( 5 - 5i -2 + 3i ) =-6i (3 - 2i )=-18i + 12i² =-18i + 12 (-1) =-12 - 18i



DIVISION :Perform the following operation and simplify

(a) 5 - 3i (b) 9i (c) 1 + 5i3 + i 2 - i -i



Solution :

(a) 5 - 3i = 5 - 3i . 3 - i = 5( 3 - i ) - 3i(3 - i )3 + i 3 + i 3 - i 3² + 1²

= 15 - 5i - 9i + 3i = 15 - 14i - 3

9 + 1 10

= 12 - 14 i = 6 - 7 i10 10 5 5

original power point math

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