if and only if
a
=
c
and
b
=
d
5
Complex
Numbers
√
1

by:
Ryan Henderson
Brian Ng
Alex Chamberlain
Kathryn Shim
The Basics
3
Complex Numbers
By:
Ryan Henderson
Brian Ng
Alex Chamberlain
Kathryn Shim
Pure Imaginary Number
Equality of Complex Numbers
Complex Numbers in Trigonometry
Multiplying and Dividing with Complex Numbers
Multiplication of Complex Numbers
To multiply two complex numbers, multiply the two R's together and add the cosine and sine θ's together respectively
z1 x z2
=
r1 r2[cos( θ1 + θ2 ) + i sin( θ1 + θ2 )]
Division of Complex Numbers
To divide two complex numbers, divide R1 by R2 and subtract the cosine θ's and sine θ's from their respective θ's.
z1 / z2
=
r1 / r2[cos( θ1  θ2 ) + i sin( θ1  θ2 )]
De Moivre's Theorem
Instead of multiplication or division, you're raising a complex number to a power
Z
=
r [cos( n x θ ) + i sin( n x θ )]
n
n
Real Number
i
+
1

√
Complex Number
a
b
Standard Form
a
+
b
i=
c
+
d
i
Operations with Complex Numbers
Adding and Subtracting Complex Numbers
(a+bi)+(c+di)=(a+c)+(b+d)i
Ex:
(3i)+(2+3i)
=(3+2)+(1+3)i
=
5+2i
Multiplying Complex
Numbers
(a+bi)(c+di)
Multiply just like two polynomials (FOIL)
Example:
(2i)(4+3i)
=8+6i4i3i^2
=8+6i4i3(1)
=8+3+6i4i
=
11+2i
Remember...
i^1=i
i^2=1
i^3=i
i^4=1
C mplex Conjugates
Every complex number has a conjugate....
Example
Multiplying two complex conjugates will produce a real number. This is useful for taking i out of the denominator. For instance...
This can be found by inverting the sign of i in a complex number.
Convert to standard form:
2+3i
_____
42i
2+3i
_____
42i
( )
4+2i
_____
4+2i
nth Roots of Complex Numbers
=
8 + 4i + 12i + 6i
________________
16  4i + 4i  4i
=
2 + 16i
_______
20
2
2
=
1 + 8i
_____
10
Absolute value of a complex number
The absolute value of a complex number
can be found by adding the squares of the coefficients of the real and imaginary numbers and then square rooting the quantity. This is very similar to the Pythagorean theorem.
A Graphical Representation
√a + b
2
2
____________
r = a + bi =
Trigonometric Form of a Complex Number
a.k.a. Polar Form
The trig form of a complex number makes finding powers and roots more efficient.
The trig form of z = a + bi is defined as ...
z = r(cosθ + isinθ)
r is called the modulus of z
θ is called the argument of z
Finding the trig form of a complex number
z = a + bi
The modulus of z is the absolute value of z.
The argument of z is the angle formed between the initial side and the terminal side.
√a + b
2
2
____________
r = a + bi =
The math...
a = rcosθ
b = rsinθ
therefore
θ = tan (b/a)
1
*Note the quadrant of the complex number
Complex number u = a + bi is an nth root of complex number z if
z = u^n = (a + bi)^n
Quiz
Find all the zeros in x(x3i)=0
x=0,3i,3i
Determine the value of i^353
i^353
= ((i^4)^88)i
=i
Solve (87i) / (12i)
Find the zeros of x^4x^3+49x^249x=0
x^3(x1)+49x(x1)=0
x(^3+49x)(1)=0
(x)(x1)(x^2+49)=0
x=0,1,+7i
Find the absolute value of 63i
sqrt(6^2+(3)^2)
=sqrt(45)
=3sqrt(5)
Find the standard form of 3(cos120°+isin120°)
=3[(1/2)+(sqrt(3)/2)]
=(3/2)+(3sqrt(3)/2)i
Solve: (3(cos(π/3)+isin(π/3)))*(4(cos(π)+isin(π/6)))
=(3*4)[cos(π/3+π/6)+isin(π/3+π/6)]
=12[cos(π/2+isin(π/2)]
=12(0+i)
=0+12i
Solve:
[2(cos120°+isin120°)]/[4(cos60°+isin60°)]
=(2/4)[cos(12060)isin(12060)]
=(1/2)[cos60+isin60]
=(1/2)[(1/2)+(sqrt(3)/2)]
=(1/4)+(isqrt(3)/4)
Solve:[2(cos50+isin50)]^3
=(2^3)[cos(50*3)+isin(50*3)]
=8[cos150+isin150]
=4sqrt(3)+4i
Solve: [27(cos270+isin270)]^(1/3)
To find a formula for an nth root of a complex number, let u be an nth root of z where:
=3√27[(cos(270+360(0))/3)+isin(270+360(0))/3)]
= 3i
Properties that are valid for complex numbers:
Associative Properties of Addition and Multiplication
Commutative Properties of Addition and Multiplication
Distributive Property of Multiplication over Addition
u = s(cosβ + isinβ)
z = r(cosθ + isinθ)
That's All Folks!
Thanks for Watching!
= ((87i)(1+2i)) / ((12i)(1+2i))
RELEVANT XKCD
= (8+16i7i14i^2) / ((12i+2i4i^2))
= (22+9i) / 5
Using DeMoivre's Theorem...
s^n(cosnβ + isinnβ) = r(cosθ + isinθ)
s^n = r
Substitute r with s^n into the previous equation and divide by r...
cosnβ + isinnβ = cosθ + isinθ
cosnβ = cosθ
sinnβ =sinθ
Because sine and cosine both have a period of 2π,
cosnβ = cosθ and sinnβ =sinθ must differ by a multiple of
2π.
Therefore, integer k only exists when
nβ = (θ +
2πk)
β = (θ +
2πk)/n
Finally, by substituting β into the trigonometric form of u, you can get this formula:
n is the distinct # of roots
k = 0, 1, 2, ..., n1
=3√27[(cos(270+360(1))/3)+isin(270+360(1))/3)]
= [(3√3)/2](3/2)i
=3√27[(cos(270+360(2))/3)+isin(270+360(2))/3)]
= [(3√3)/2](3/2)i
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